The mathematical logic that we’ve studied in the previous two chapters
is foundational to all types of math. However, as demonstrated in the
last few sections of Chapter 2, mathematicians rarely lay out all of
the full, gory details when writing proofs. Instead, they rely on
axioms, definitions, and previous theorems to work out the desired
result. Occasionally, propositional logic may be used when doubts arise
about the validity of a given proof, but that too is rare.
Something we’ve hinted at in the previous two chapters is that when we
want to validate an argument, we start with some number of premises.
Another way of stating this is that we start off with some initial
collection, or set, of premises, and from those premises, we
hopefully arrive at the desired conclusion. The group of premises we
start off with may change from one argument to another, but in any
case, we have a starting collection of premises no matter what argument
we’re trying to build up.
The idea of a collection, or set, of objects underlies almost all of
mathematics, whether that be a collection of premises (in mathematical
logic), a collection of points (in geometry), a collection of possible
outcomes for an experiment (in probability), or a collection of outputs
for some given collection of inputs (mathematical relations). In this
chapter, we start to define and work with these collections of
objects, and what we can do with collections of objects in general.
Even though mathematicians rarely describe all of the formal logic
used in their arguments, any set-theoretic aspects are almost always
explicitly laid out. Hence, getting a good understanding of Set Theory
will be vital in learning to not only do mathematics, but to read
mathematics as well.
Subsections of Set Theory
Sets
Trying to precisely define a set can be rather tricky. While there are
formal definitions and axioms, here we will rely on our intuition.
Even without a formal specification, we can still derive many useful
results that hold up under scrutiny.
In this section, we’ll learn what a set is (intuitively), and how to
describe what kinds of things are in a set.
Intuitively Defining a Set
Sets, or collections, of objects abound in daily life. We could speak
of the set of fruits available for purchase at a local grocery store,
the set of birds native to North America, the set of components used
to build a specific computer, the set of roads from New York to Los
Angeles, and so on. We can make note of a couple of things here:
The order in which we list the items in the collection doesn’t
appear to be relevant. If we simply want to know what kinds of birds
are native to North America, we can list the Wild Turkey before the
Sandhill Crane, or the Sandhill Crane before the Wild Turkey.
If we accidentally list an item more than once, that also doesn’t
affect the set we’re dealing with. For example, if we list I-80 W
twice by accident, it doesn’t change the fact that I-80 W is still a
road between New York and Los Angeles. Sometimes we may care about
this when planning a trip, but not if we just want to know what
roads exist between New York and Los Angeles.
However, if we miss an object — if it’s listed zero times — that
does affect our set, because then our set is not complete. For
example, if a local grocery store sells mangos but forgets to
advertise them, they may lose out on customers who want to purchase
mangos.
As such, the order in which we list elements from a set is irrelevant.
The number of times we list an element from a set is also irrelevant.
All we care about is whether or not an object is in a set or not.
Sets of objects also appear frequently in mathematics. We could speak
of the set of numbers used to count objects, the set of points that
make up a line in space, the collection of parabolas with real roots,
the set of solutions to a system of equations, and so forth. Again, we
don’t really care about the order in which these objects can be
listed out, and we also don’t care if some objects from the set are
listed more than once. However, in order to be a complete description,
every possible object must be listed at least once.
SET, ELEMENT, MEMBER
A set is an unordered collection of well-defined objects.
Each object within the set is called an element, or member, of
that set. We write
$$x \in A$$
to denote that object $x$ is a member of the set $A$. On the other
hand, if some object $y$ is not in $A$, then we write
$$y \notin A.$$
By well-defined, what we mean is that, if presented with a
description of some set $A$, we’re able to determine whether some
given object — which we’ll refer to as $x$ — is an element of $A$,
without any ambiguity.
Example 3.1.1: A set that isn’t well-defined
Suppose we were interested in forming sets of Major League Baseball
players.
What kinds of players would we include if we wanted to form the set of
“outstanding left-fielders”? The first thing we’d need to do is figure
out which players are outstanding left-fielders. Well, who qualifies
as an outstanding left-fielder? Are they the players who can throw
with the most accuracy? If so, what’s the accuracy cutoff?
If accuracy isn’t the metric, what about running speed? How fast a
player can run to catch a ball may be important. What about the ratio
of caught balls to missed balls? How exactly do we decide who to
include in the set of outstanding left-fielders?
The problem with a “set” such as “outstanding left-fielders” is that
the term outstanding is vague. Since it’s not exactly clear how to
determine who an outstanding left-fielder is, we have no well-defined
criterion. As such, the above description does not define a set.
Let’s do something different. This time, let’s niche down into
something like this:
$$\text{The set of all Major League Baseball players who hit at least 300 home runs in the 1990s.}$$
Here, all we have to do is check the records of players from the
1990s, and see who hit at least 300 home runs. Here are those players:
Notice the difference: one description was vague and open to
interpretation, so not everyone would agree on what players to
include. The other description was specific, and could be verified
against the available records — it wasn’t open to interpretation at
all, since we offered a very specific metric, and only players meeting
that requirement are included.
Typically, we use capital letters such as $A$, $B$, $C$, $\dots$, $X$,
$Y$, and $Z$ to represent sets. That isn’t always the case, but it’s
fairly common. In contrast, we typically use lowercase letters such as
$a$, $b$, $c$, $\dots$, $x$, $y$, and $z$ to denote elements.
Something else worth noting is that we enclose the elements of a set
within curly braces $\{\ \}$, as demonstrated in the next example.
Example 3.1.2: Testing membership in a well-defined set
Let’s revisit the set defined in the previous example — the only
description that actually defined a set, that being the set of “Major
League Baseball players who hit at least 300 home runs in the 1990s.”
Instead of repeatedly saying “the set of Major League Baseball players
who hit at least 300 home runs in the 1990s,” we’ll use the capital
letter $M$ to refer to that set:
$$M = \text{the set of Major League Baseball players who hit at least 300 home runs in the 1990s.}$$
We can use lowercase letters to refer to specific players:
We can go through and determine set membership for each of these
players.
$a \notin M$, because Joe DiMaggio played baseball from 1936 to 1951,
and so didn’t play baseball in the 1990s, meaning he didn’t hit at
least 300 home runs in the 1990s.
$b \in M$, because Ken Griffey Jr. did hit at least 300 home runs in
the 1990s. Even though he played baseball for a brief time in 1989, as
well as for a long time in the 2000s, that’s irrelevant — we can still
record how many home runs he hit in the 1990s alone, which was 382.
$c \notin M$, because Babe Ruth didn’t play baseball in the 1990s, and
so didn’t hit at least 300 home runs in the 1990s.
$d \in M$, because Mark McGwire meets the requirement to be in $M$.
$e \notin M$, because even though Mike Piazza did play Major League
Baseball in the 1990s, he didn’t hit enough home runs to be in $M$ (he
hit 240, 60 short of the required 300).
$f \in M$, because Barry Bonds hit at least 300 home runs in the MLB
in the 1990s. As such, he meets the requirement for membership in $M$.
$g \notin M$, because Rocky Marciano didn’t hit at least 300 home runs
in the MLB in the 1990s. As a matter of fact, Rocky didn’t play
baseball professionally at all — instead, he was the Heavyweight World
Champion Boxer, active from 1947 to 1951. So not only is this the
wrong sport, but the time frame is incorrect too.
Building a Set
There are two primary ways to describe what kinds of elements are
within a set.
One way is to simply list them all. Of course, this method is only
practical if the desired set contains a small number of elements,
though it’s perhaps the most specific, and leaves absolutely nothing
to the imagination.
Example 3.1.3: Listing every element of a set
Let $A$ denote the set
$$A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}.$$
Let $B$ denote the set
$$B = \{-12, -9, -6, -3\}.$$
Since order and repetition don’t matter when building a set, we also
have that
Since we explicitly state which elements are in each set, it’s
extremely easy to determine if an element is in either set simply by
inspection.
$3.14 \notin A$, because it isn’t one of the listed numbers in $A$.
$11 \in A$, because it’s explicitly listed within the set’s
definition.
$-6 \in B$, because it’s explicitly listed in $B$’s definition.
$13 \notin B$, because it isn’t one of the listed numbers in $B$.
This method of building, or defining, a set is commonly called either
the Roster Method, or the Exhaustive Method.
However, nothing stops us from defining sets that have infinitely many
elements. When dealing with such a set, the Roster Method is somewhat
inadequate on its own. We could list enough elements to make the
pattern obvious, and finish with an ellipsis.
Example 3.1.4: Using an ellipsis for an infinite set
Let $A$ denote the set
$$A = \{2, 4, 6, 8, 10, 12, \dots\}.$$
When defining set $A$, all the numbers we listed are positive even
integers, so we can reasonably assume that $A$ refers to the set of
all positive even integers. Since there are infinitely many positive
even integers, we have to use an ellipsis.
Let $B$ denote the set
$$B = \{1, 2, 4, 8, 16, 32, 64, \dots\}.$$
For $B$, it looks like all the numbers we listed are the non-negative
integer powers of $2$:
Hence, we can reasonably assume that $B$ refers to the set of all
powers of $2$ where the exponent is a non-negative integer.
When using the Roster Method to list the first few elements of a set,
it should be pretty clear what the pattern is. Anything that requires
an elaborate setup to describe should be explicitly mentioned and
described, so as to prevent confusion. Remember, it should be obvious
what elements are in the set.
A few listed elements can still be ambiguous
Consider the set
$$\square = \{1, 2, 4, 8, 16, \dots\}$$
(here, we use the symbol $\square$ instead of a capital letter, for
reasons that will become clear).
This almost looks like set $B$ from the previous example, defined
using powers of $2$ — so we might think $\square$ is the set of all
powers of $2$ where the exponent is a non-negative integer.
Let’s reveal the next number in this particular set:
$$\square = \{1, 2, 4, 8, 16, 31, \dots\}.$$
Hold on — $31$ isn’t an integral power of $2$! So $\square$ isn’t the
set of all non-negative integer powers of $2$ after all.
So what is it, exactly? As it turns out, this particular sequence
describes the maximum number of regions a circle can be divided into
by placing points around its circumference, and connecting every pair
of those points with a straight line — a classic, deceptively tricky
combinatorics problem. Continuing the sequence a bit further:
There is a formula for this sequence, but it involves something
called “binomial coefficients,” which we’ll discuss later in this
book — for now, it would look like complete gibberish.
The point is that, even though the first few terms of a listed set
may seem to follow one pattern, there may be multiple patterns that
fit, so listing only a few numbers at the start may not adequately
describe the desired set. That’s exactly why we used the $\square$
symbol here — to signal that this is a misleading, ambiguous way to
describe a set. Never be ambiguous when describing a set!
It’s because of this kind of ambiguity that another method of
defining sets is more commonly used, called the Set-Builder Method.
This method requires specifying a rule, or a list of conditions, that
determine whether an element is a part of the set. Typically, this
method looks something like this:
$$A = \{x \mid \text{condition}\}.$$
There are a few different parts to this notation, worth going over
one at a time:
$A$ — the letter, or general symbol, we use to refer to the set.
$\{\ \}$ — the curly braces that enclose the definition of the set.
Every set is enclosed by curly braces.
$x$ — the letter, or again some general symbol, we use to refer to
some element within the set.
$\mid$ — this bar separates the listing of the element from the
condition. When reading out the definition of a set, the vertical
bar can be translated as “such that.”
$\text{condition}$ — specifies what must be true about $x$ for it to
be an element of $A$. In other words, the condition is what we use
to test whether some object is contained within $A$.
As such, when we see something like $A = \{x \mid \text{condition}\}$,
we can read this as “$A$ refers to the set of all elements $x$ such
that the condition is satisfied.” There’s some flexibility in how this
method can be used, as demonstrated below.
Example 3.1.5: Using the Set-Builder Method
Here’s a typical example of the Set-Builder Method:
$$A = \{n \mid n \text{ is an even integer}\}.$$
Based on this definition, we see that $6 \in A$, $-12 \in A$,
$100238 \in A$, and $-12300450678 \in A$, along with infinitely many
other numbers. On the other hand, $1 \notin A$, $-13 \notin A$,
$100343 \notin A$, $3.1415 \notin A$, and $-2.718 \notin A$, among
infinitely many other numbers.
Sometimes, multiple conditions can be placed on a set using the
conjunction operator discussed in Chapter 1, like so:
$$B = \{x \mid (x \text{ is an integer}) \land (x > 10)\}.$$
So, any number that’s simultaneously an integer and larger than $10$
is included in $B$. Thus $90 \in B$, $92 \in B$, and $100100 \in B$,
and so on. On the other hand, $-2 \notin B$, $10 \notin B$, and
$0.4335 \notin B$, because $-2$ isn’t larger than $10$, because $10$
isn’t larger than itself, and because $0.4335$ is neither an integer
nor larger than $10$, so it fails both conditions.
It’s possible to specify one condition before the vertical bar in a
set’s description, like so:
$$C = \{x \text{ is an integer} \mid x \leq -4\}.$$
Here, we’re implying that, before we even consider the usual
condition, we’re presupposing that only integers can be included in
$C$. In other words, this definition is equivalent to saying
$$C = \{x \mid (x \text{ is an integer}) \land (x \leq -4)\}.$$
Of course, we could use the disjunction operator as well:
$$D = \{x \mid (x \text{ is a multiple of } 0.4) \lor (x \text{ is a negative integer})\}.$$
As such, we find that $2.4 \in D$, $-12 \in D$, and $-100.4 \in D$,
because $2.4$ is a multiple of $0.4$, because $-12$ is a negative
integer (and a negative multiple of $0.4$), and because $-100.4$ is a
multiple of $0.4$. However, $1 \notin D$, because $1$ is neither a
multiple of $0.4$ nor a negative integer.
The Set-Builder Method is extremely useful for describing sets with
infinitely many elements — but that doesn’t mean it isn’t useful for
describing sets with finitely many elements too. This is especially
true when there’s a large number of elements in the set.
Example 3.1.6: Using the Set-Builder Method for a large finite set
Consider the set
$$X = \{n \text{ is an integer} \mid n^2 < 100\}.$$
First, we know that we only have integers in $X$, since that’s a
condition for set inclusion. The second condition is that the square
of the integer must be less than $100$. We can check a few numbers to
be sure:
As a matter of fact, we know that the first integer $n$ where
$n^2 \geq 100$ is $10$, since $10^2 = 100$. As such, all the positive
integers less than $10$ are included in the set, and since negative
integers have positive squares too, all the negative integers greater
than $-10$ are included as well. We also verified $0 \in X$ above.
Any integer larger than $9$, or less than $-9$, won’t be included,
because the smallest integer larger than $9$ is $10$, and $10^2$ is
not less than $100$; similarly, the largest integer smaller than
$-9$ is $-10$, and $(-10)^2$ is also not less than $100$.
As such, we know that all of the elements in $X$ are
Even though we could just list them, the Set-Builder Method at least
gives us a rule to decide whether some number belongs to the set — if
we had simply listed the numbers, we wouldn’t know if there was some
underlying condition, or if the numbers were chosen arbitrarily.
Of course, with a set like
$$Y = \{n \text{ is an integer} \mid n^2 \leq 1000000000000\},$$
it would take a long time to fully list out the elements. We still
prefer the Set-Builder Method to describe $Y$, since it’s typically
more succinct.
Example 3.1.7: Combining a formula with a condition
Consider the set
$$\Psi = \{n^3 \mid (n \text{ is an integer}) \land (n^2 < 100)\}$$
(here, we use the capital Greek letter psi, $\Psi$, just to add some
variety to the names we’ve been giving our sets).
It looks like $\Psi$ consists of perfect cubes, but one of the
conditions is that the square of the underlying integer must be
less than $100$. Let’s examine the possibilities:
Any integer with absolute value $10$ or greater has a square of at
least $100$, so we don’t need to examine any more integers. Since
$\Psi$ doesn’t consist of these perfect squares, but instead the
corresponding cubes, we have
In the previous section, we saw sets that can have vastly different
numbers of elements. Some sets have a finite number of elements,
meaning that if we started to list out all of that set’s elements,
we’d eventually be able to stop, having written down every element. In
contrast, some sets have infinitely many elements, meaning that if we
start listing elements and stop at any point in time, there would
still be elements missing.
For finite sets, knowing the number of elements they contain can be
useful.
CARDINALITY
For any finite set $A$, we use the notation
$$|A|$$
to refer to the number of elements in $A$. We refer to the number of
elements in $A$ as the cardinality of $A$.
Even though we don’t yet speak of the cardinality of an infinite set,
it’s technically correct to say that the cardinality of an infinite
set like $\mathbb{Z}$ is countably infinite. This idea will be
explored later, when we talk about the different special sets of
numbers.
Example 3.2.1: Cardinality of finite and infinite sets
Consider the following sets:
\[
\begin{array}{rl}
A = \{x^3 \mid (x \text{ is an integer}) \land (|x^3| < 100)\} & = \{1, -1, 8, -8, 27, -27, 64, -64\} \\
B = \{x^2 \mid (x \text{ is an integer}) \land (x^2 < 100)\} & = \{1, 4, 9, 16, 25, 36, 49, 64, 81\} \\
C = \{2n \mid n \text{ is an integer}\} & = \{0, 2, -2, 4, -4, 6, -6, 8, -8, 10, -10, \dots\}
\end{array}
\]
Both $A$ and $B$ are finite sets, so we can speak of their
cardinalities: $|A| = 8$ and $|B| = 9$. We see that $C$ is an infinite
set, so we don’t speak of its cardinality (yet).
Care must be taken when we’re dealing with sets that contain a wide
variety of items. If a set has an element that is itself a set with
multiple items, the elements of that inner set do not count towards
the outer set’s cardinality.
This set has a wide variety of different types of objects: numbers,
letters from the English alphabet (the $a$, $b$, $c$, $x$, $y$, and
$z$ aren’t referring to variables in this particular case — just the
letters themselves), and even a few sets!
Let’s list out each element of $X$ on its own line:
\[
\begin{array}{l}
a \\
b \\
c \\
1 \\
2 \\
3 \\
x \\
y \\
z \\
\{1, 2, 3\} \\
\{\{1\}, 2, 3\} \\
\{a\}
\end{array}
\]
Notice that the element $a$ is not the same thing as the element
$\{a\}$: one is simply a letter, and the other is a set (containing
that letter), so they’re entirely different elements. The same is
true for the elements $1$, $2$, $3$, $\{1, 2, 3\}$, and
$\{\{1\}, 2, 3\}$. The element $\{1, 2, 3\}$ is a set, and even though
it contains several elements itself, it only counts as one element
of $X$.
As such, there are no repeated elements in $X$, and since all twelve
are distinct, they all count towards the cardinality of $X$, meaning
$$|X| = 12.$$
Of course, the set $\{1, 2, 3\}$ has its own cardinality:
$|\{1, 2, 3\}| = 3$. But it’s still just $1$ set, and it only counts
for $1$ element when considering the cardinality of $X$. Similarly,
$|\{\{1\}, 2, 3\}| = 3$.
Contrast this with $|\{\{1\}, \{2, 3\}\}| = 2$, because the $2$ and
the $3$ are included together in a single set within
$\{\{1\}, \{2, 3\}\}$.
Subsets
A set can contain a wide variety of objects. They can contain objects
that most people interact with on a daily basis — automotive parts
that can be used to service a 1967 Camaro, tools used to carve statues
out of wood, or art supplies needed to paint a picture. Likewise, they
can contain a wide variety of mathematical objects, like numbers,
shapes, or axioms.
Sometimes, we only care about some of the objects in a set. For
example, we may only be interested in automotive parts needed to
service a car’s headlights, or we may only be interested in art
supplies needed for a fresco painting. Regardless, we’re able to build
a lot of structure by constructing new sets from old ones — simply by
restricting what elements are included in the new set. Here, we
explore this kind of relationship between sets.
Starting with a Base Set
If we’re interested in selecting only some of the elements from a set,
we’ll need to know what the original, or underlying, set contains.
This is exactly the same idea behind the Universe of Discourse
discussed when we studied open propositions in Chapter 1. There, a
Universe of Discourse (or simply Universe) was the set of all
values we allow to be substituted in for the variables of a
propositional function.
It’s the exact same situation here. We even commonly use the same
script letter $\mathcal{U}$ to denote the base set, though we can
still use any symbol we want.
Example 3.3.1: Why a base set removes ambiguity
Suppose we were presented with the following set:
$$X = \{x \mid 1 \leq x \leq 10\}.$$
Is $1.5 \in X$?
It’s hard to tell without knowing what kinds of numbers we even want
included in this set. Does it include all the real numbers between $1$
and $10$ inclusive? Or does it only include whole numbers from $1$ to
$10$ inclusive?
Let’s remedy the situation by stating that set $X$ is only going to
draw elements from the set $\mathcal{U}$ of all integers. Then we know
that
$$X = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.$$
Now that we have an established universe, we can simply write
$X = \{x \mid 1 \leq x \leq 10\}$ without any ambiguity. We could also
write $X = \{x \in \mathcal{U} \mid 1 \leq x \leq 10\}$, or even
$X = \{x \mid (1 \leq x \leq 10) \text{ and } (x \in \mathcal{U})\}$.
Because we’ve established a base set, all of these different ways of
describing a set yield the exact same set $X$ of integers between $1$
and $10$ inclusive.
As such, since we now know what kinds of numbers we’re including in
$X$, we can definitively say that $1.5 \notin X$.
Example 3.3.2: Changing the base set changes the resulting set
Reconsider the previous example, where we redefine $\mathcal{U}$ to be
$$\mathcal{U} = \{x \mid x \text{ is an even integer}\}.$$
In this case, since the allowable elements are the even integers, we
have that
The definition of a subset is closely related to what we’ve discussed
so far in this section. We start with a universe of discourse, and
then construct a set using only elements from that universe. We could
go even further, constructing a third set by only taking elements from
that second set.
SUBSET
If $A$ and $B$ are sets constructed using only elements of some given
universe $\mathcal{U}$, we say that $A$ is a subset of $B$ if (and
only if) every element of $A$ is also an element of $B$. When $A$ is a
subset of $B$, we write
$$A \subseteq B.$$
Logically, we would write
$$\forall x\ [(x \in A) \Longrightarrow (x \in B)].$$
Alternatively, if $A$ is not a subset of $B$, we write
$$A \not\subseteq B.$$
Notice that the above definition uses a universally quantified logical
implication. This offers us a chance to review logical implications,
and what they mean in the context of sets and subsets. Since there’s a
logical implication involved, the proposition
$$(x \in A) \to (x \in B)$$
is always true; thus, when we have $A \subseteq B$, there are only
three scenarios to worry about:
\[
\begin{array}{c|c|c|l}
x \in A & x \in B & (x \in A) \to (x \in B) & \textbf{Interpretation} \\
\hline
0 & 0 & 1 & \text{The element referred to by } x \text{ is in neither } A \text{ nor } B. \\
0 & 1 & 1 & \text{The element referred to by } x \text{ is not in } A \text{, but is in } B. \\
1 & 1 & 1 & \text{The element referred to by } x \text{ is in both } A \text{ and } B.
\end{array}
\]
Notice that there’s no row where $x \in A = 1$ and $x \in B = 0$ —
since we’re asserting $A \subseteq B$, that scenario simply can’t
happen.
Example 3.3.3: Confirming a subset relationship
Consider the universe $\mathbb{N}$ of all positive integers
$\{1, 2, 3, 4, 5, 6, 7, 8, 9, \dots\}$. Now consider the following
sets:
In order for $Y \subseteq X$, we need to verify that
$\forall n\ [(n \in Y) \Longrightarrow (n \in X)]$.
If $n$ refers to anything that isn’t in $Y$, then $n \notin Y$,
meaning $n \in Y = 0$, and hence $(n \in Y) \to (n \in X) = 1$ — so we
don’t have to worry about checking elements that aren’t in $Y$. Let’s
examine each element of $Y$, and see if it’s in $X$ as well:
The column representing $(n \in Y) \to (n \in X)$ is all $1$s for each of
the elements of $Y$. Hence it’s a tautology, meaning it’s a logical
implication, and we can write $(n \in Y) \Longrightarrow (n \in X)$. Thus,
we’ve shown that $Y$ is a subset of $X$, and we can write
$Y \subseteq X$.
It’s also worth noting that we also have $X \subseteq \mathbb{N}$ and
$Y \subseteq \mathbb{N}$.
Example 3.3.4: A set that is not a subset
Consider the universe $\mathbb{N}$ of all positive integers
$\{1, 2, 3, 4, 5, 6, 7, 8, 9, \dots\}$. Now consider the following
sets:
In order for $Y \subseteq X$, we’d need to verify that
$\forall n\ [(n \in Y) \Longrightarrow (n \in X)]$. However, suppose we
substituted $11$ in for $n$. We see that $11 \in Y$, but
$11 \notin X$, meaning $(11 \in Y) \to (11 \in X) = 0$ — so we don’t have
a logical implication, and $Y \not\subseteq X$.
However, we do have $X \subseteq Y$. Remember that a subset is a set
in its own right — it just happens that all of that subset’s elements
are also contained within some other set as well.
Every Set Is a Subset of Itself
It’s also natural to wonder whether a set counts as a subset of
itself. Recall that $A \subseteq B$ requires every element of $A$ to
also be an element of $B$ — in other words, that $(x \in A) \to (x \in B)$
holds for every $x$. When $B$ is $A$ itself, that condition becomes
$(x \in A) \to (x \in A)$, which is always true, since any proposition
implies itself. So every element of $A$ automatically satisfies the
requirement for membership in $A$, which gives us the following
theorem.
Theorem 3.3.1: Every set is a subset of itself
For any universe set $\mathcal{U}$, and set $A \subseteq \mathcal{U}$,
we have that
$$A \subseteq A.$$
Proof 3.3.1
Consider some arbitrary element $x$ from $\mathcal{U}$. If we have
$x \in A$, then obviously $x \in A$ (because we literally just
considered that case). Thus, since every element contained in $A$ is
(obviously) contained within $A$, the definition of subset tells us
that $A \subseteq A$, as desired.
Example 3.3.5: Applying the theorem
Consider the set $B = \{x, y, z\}$, where $x$, $y$, and $z$ aren’t
being used as variables here, but are simply the 24th, 25th, and 26th
letters of the alphabet. By the previous theorem, we have that
$\{x, y, z\} \subseteq \{x, y, z\}$.
Proper Subsets
Of course, when we assert $A \subseteq B$, all we’re saying is that
whenever $x \in A$, we also have $x \in B$. However, this definition
doesn’t say whether every single element of $B$ must be contained
within $A$ too. We have a special term to describe when $A$ is a
subset of $B$, but $B$ has elements not contained within $A$.
PROPER SUBSET
Consider two sets $A$ and $B$. $A$ is said to be a proper subset
of $B$ if (and only if) every element contained within $A$ is also in
$B$, but $B$ has at least one element that is not contained within
$A$. We denote this by writing
$$A \subset B.$$
Logically, we would write this as
$$(\forall a \in A\ [a \in B]) \land (\exists b \in B\ [b \notin A]).$$
Example 3.3.6: Confirming a proper subset relationship
Consider the following sets:
\[
\begin{array}{rl}
M\text{: } &\{x \mid x \text{ is an odd integer}\} \\
N\text{: } &\{x \mid x \text{ is an integer}\}
\end{array}
\]
Because every odd integer is a specific kind of integer, every element
contained within $M$ is also an element of $N$, so we can write
$M \subseteq N$. However, notice that $0 \in N$, but $0 \notin M$,
meaning $N$ has at least one number not contained within $M$. So $M$
is a proper subset of $N$, and we can also write $M \subset N$.
Notice that by the previous example, for two sets $A$ and $B$, it’s
possible for both $A \subseteq B$ and $A \subset B$ to be true.
However, that’s not always the case, as demonstrated by the next
example.
Example 3.3.7: A subset that is not a proper subset
Consider the set $\Psi = \{a, b, c, 1, 2, 3, x, y, z, \{1, 2, 3\}\}$.
By the theorem above, we clearly have $\Psi \subseteq \Psi$. However,
notice that we don’t have $\Psi \subset \Psi$: obviously any element
contained within $\Psi$ is going to be contained within $\Psi$ (that’s
tautologically true), but there’s no element within $\Psi$ that isn’t
also contained within $\Psi$ (again, tautologically true). Thus, it’s
not the case that $\Psi$ is a proper subset of itself — no set can be
a proper subset of itself.
Now consider the set $\Omega = \{b, 1, z, \{1, 2, 3\}\}$. We can
clearly see that every element contained within $\Omega$ is also an
element of $\Psi$, so we can write $\Omega \subseteq \Psi$. Notice
though that $a \in \Psi$, but $a \notin \Omega$, so we also have
$\Omega \subset \Psi$.
Based on the previous two examples, it seems that if we know
$A \subset B$, we also know $A \subseteq B$. Let’s formally show this
is the case.
Theorem 3.3.2: A proper subset is also a subset
If $A \subset B$, then $A \subseteq B$.
Proof 3.3.2
Since we know $A \subset B$, we have by definition that for any
element $a$ in $A$, $a$ will also be in $B$. Thus, by definition, we
have $A \subseteq B$ as well.
Notice that if we know $A \subseteq B$, we don’t automatically know
$A \subset B$. For example, we have $A \subseteq A$, but
$A \not\subset A$.
The Empty Set
Just because we can put just about anything in a set doesn’t mean we
need to have something in a set. There’s a special and unique kind
of set that has no elements in it at all!
EMPTY SET, NULL SET
The empty set is the unique set containing no elements at all. The
empty set is sometimes referred to as the null set, and is often
symbolized by $\emptyset$.
The empty set is a very special set, and occurs throughout all of set
theory, but it does have some interesting quirks that are a bit
strange at first, though they become more natural as we work more and
more with this kind of set.
First, note that because it’s a set containing nothing, we can write
$$\emptyset = \{\ \}.$$
There’s absolutely nothing written in between the curly braces, so we
have, for example, that $\emptyset \neq \{0\}$ — above, we wrote a
single number between the braces, and since $\{0\}$ is a set
containing one item, it’s not equal to the empty set.
The second thing to note is that a set can have the empty set as one
of its elements:
$$S = \{\emptyset\}.$$
Because $S$ is a set containing one element (that element itself being
a set), we have $|S| = 1$. Thus, since $S$ contains at least one
element, it’s not the same thing as the empty set, even though it’s
the set containing the empty set: $S \neq \emptyset$. Comparing $S$
to $\emptyset$, since $\emptyset$ contains no elements, we have
$|\emptyset| = 0$. If we substitute $\{\emptyset\}$ in for $S$, we see
that $\{\emptyset\} \neq \emptyset$.
The Empty Set Is Always a Subset
Recall the definition of subset, which states that $A \subseteq B$
when every element in $A$ is also an element of $B$. In other words,
$(n \in A) \to (n \in B)$ is always true. We use that definition, as well
as the definition of the empty set, in our next theorem:
Theorem 3.4.1: The empty set is a subset of any set
For any universe $\mathcal{U}$, let $A$ be any set such that
$A \subseteq \mathcal{U}$.
$$\emptyset \subseteq A$$
Proof 3.4.1
Let $x$ be any arbitrary element from $\mathcal{U}$. Because
$\emptyset$ contains no elements, it’s impossible for $x \in
\emptyset$, so it’s always true that $x \in \emptyset = 0$.
Notice that since $x \in \emptyset$ is always equal to $0$, we have
that $(x \in \emptyset) \to (x \in A) = 0 \to (x \in A) = 1$. It doesn’t
matter whether $x$ is an element of $A$ or not, because the hypothesis
of the implication $x \in \emptyset$ is always false, meaning the
overall implication always evaluates to true. Thus,
$(x \in \emptyset) \Longrightarrow (x \in A)$, and so
$\emptyset \subseteq A$, as desired.
Notice that in this theorem, we didn’t place any special requirements
on set $A$ — it just had to be an arbitrary set constructed from some
universe $\mathcal{U}$. Thus, by the Rule of Universal Generalization,
since $A$ was arbitrary, and we have $\emptyset \subseteq A$, the
empty set must be a subset of every possible conceivable set —
including the universe $\mathcal{U}$ itself!
Example 3.4.1: Applying the theorem
By the previous theorem, for the set $A = \{1, 2, 3\}$, we have that
$$\{\ \} \subseteq A.$$
Proof Technique: Element Arguments
In the previous two sections, we provided proofs for two theorems
(Theorem 3.4.1 and Theorem 3.3.1). Both proofs required that we select an
arbitrary element from some set $A$, and show that because that
arbitrarily chosen element satisfies some property, all elements of
set $A$ satisfy that property. This is simply a rehash of the concept
of Universal Specification and Universal Generalization as discussed
in Chapter 2.
In this section, we adapt the methods of Universal Specification and
Universal Generalization to a new, powerful proof technique we can use
for sets.
The Underlying Argument
Here, we show off the basic structure of the argument that underlies
an element argument.
Let’s suppose $p(x)$ and $q(x)$ are open statements defined on some
set $A$ (here the set $A$ is acting like our universe set
$\mathcal{U}$).
\[
\begin{array}{lll}
\textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\
(1) & \forall x \in A\ [p(x)] & \text{Premise} \\
(2) & c \in A & c \text{ is an arbitrarily chosen element from } A \\
(3) & p(c) & \text{Universal Specification on (1) and (2)} \\
(4) & p(c) \to q(c) & \text{Axiom, Definition, Premise, or Theorem} \\
(5) & q(c) & \text{Modus Ponens on (3) and (4)} \\
(6) & \therefore \forall x \in A\ [q(x)] & \text{Universal Generalization on (2) and (5)}
\end{array}
\]
Even though every element argument follows this basic structure, pay
close attention to step (4). Notice that the reason given was either
an axiom, definition, premise, or theorem. This is the part of the
argument that changes with each and every property we’re trying to
show is true for every element of some given set $A$. Depending on
what we’re trying to show, we’ll need to bring in multiple axioms,
definitions, premises, or theorems in order to reach the desired
conclusion.
Typically, step (4) will be expanded into multiple steps, depending on
how many axioms, definitions, and theorems are needed to get from
proposition $p(c)$ to $q(c)$.
Some Straightforward Theorems
Here is one basic theorem whose proof relies on a fairly typical
element argument.
Theorem 3.5.1: Transitivity of the subset relation
If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$.
Proof 3.5.1
Let $x$ be an arbitrarily chosen element from $A$.
Because $x \in A$, and because $A \subseteq B$, we know that
$x \in B$ as well.
Furthermore, because $x \in B$ and $B \subseteq C$, we must have that
$x \in C$.
Thus, we’ve determined that for an arbitrarily chosen element $x$ from
$A$, $x$ must also be an element of $C$, and so we must have that
$A \subseteq C$ as desired.
Let’s rewrite this proof using the structure presented above.
\[
\begin{array}{lll}
\textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\
(1) & \forall x \in A\ [x \in B] & \text{Premise } A \subseteq B \\
(2) & x \in A & x \text{ is an arbitrarily picked element from } A \\
(3) & x \in B & \text{Universal Specification on (1) and (2)} \\
(4) & \forall x \in B\ [x \in C] & \text{Premise } B \subseteq C \\
(5) & x \in C & \text{Modus Ponens on (3) and (4)} \\
(6) & \therefore \forall x \in A\ [x \in C] & \text{Universal Generalization on (2) and (5)}
\end{array}
\]
This particular proof was a one-to-one translation of the basic
structure of an element argument. Let’s look at a theorem that takes a
little more work to prove.
Theorem 3.5.2: Transitivity of the proper-subset relation
If $A \subset B$ and $B \subset C$, then $A \subset C$.
Proof 3.5.2
General Strategy: Looking at the definition of proper subset,
there are essentially two things we need to do. First, we show that
every element of $A$ is also an element of $C$, meaning
$A \subseteq C$. Next, we show that there is an element in $C$ that is
not in $A$. Thus, the definition of proper subset will be satisfied.
Step 1: Show that $A \subseteq C$.
Let $x \in A$ be an arbitrarily picked element of $A$. Because
$A \subset B$, Theorem 3.3.2 also tells us that $A \subseteq B$. As
such, we also have that $x \in B$.
Furthermore, because $B \subset C$, Theorem 3.3.2 tells us that
$B \subseteq C$ as well. As such, since $x \in B$ and $B \subseteq C$,
we also have that $x \in C$.
Thus, since any arbitrarily picked element of $A$ is also an element
of $C$, we know that $A \subseteq C$.
Step 2: Show that there is an element in $C$ that is not in $A$.
Because we know that $B \subset C$, there exists an element $y$ that
is in $C$ that is not in $B$; in other words,
$$\exists y \in C\ [y \notin B].$$
Furthermore, since we know that $A \subseteq B$ and $y \notin B$, we
also know that $y \notin A$ as well.
Conclusion. At this point, we have shown that $A \subseteq C$, and
that $\exists y \in C\ [y \notin A]$. Thus by definition, we have that
$$A \subset C$$
as desired.
This proof required that we give a name to an element we knew existed
— namely, an element that was contained within $C$, but was not
contained within $B$, and as such not contained in $A$ either. There’s
a special name we give to this situation, which we describe below.
Existential Specification and Existential Generalization
When we discussed quantifiers, we talked about two different kinds:
those of the universal variety, and those of the existential variety.
The Universal Quantifier has its own specification and generalization
schemes, and the Existential Quantifier naturally has its own
counterparts.
The concept of Existential Specification comes into play when we
know some element exists, and in order to manipulate that element, we
simply give it a name. In other words, if we know that
$$\exists x \in \mathcal{U}\ [p(x)] = 1,$$
then we can simply use a symbol to refer to that particular element
where $p(x)$ is a true statement. The exact symbol chosen doesn’t
matter: it could be the letter $c$, a Greek letter like $\beta$, or
some geometrical symbol like $\square$ — just as long as we’re
consistent with what symbol is used. For example, knowing that
$\exists x \in \mathcal{U}\ [p(x)]$ is true, we could use the letter
$c$ to denote the particular element within $\mathcal{U}$ such that
$$p(c) = 1.$$
The concept of Existential Generalization is essentially the
reverse of Existential Specification. If we know that the particular
element $c \in \mathcal{U}$ makes the open statement $p(x)$ true,
meaning $p(c) = 1$, then we know that some element exists that makes
$p(x)$ a true statement (because we identified such an element),
meaning we know that
$$\exists x \in \mathcal{U}\ [p(x)] = 1.$$
We implicitly used Existential Specification and Existential
Generalization in the proof for Theorem 3.5.2. Since we’re discussing
a general proof strategy, let’s write out a formal argument in
tabular format, as we’ve previously done for Theorem 3.5.2:
\[
\begin{array}{lll}
\textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\
(1) & A \subset B & \text{Premise} \\
(2) & A \subseteq B & \text{Theorem 3.3.2 on (1)} \\
(3) & B \subset C & \text{Premise} \\
(4) & B \subseteq C & \text{Theorem 3.3.2 on (3)} \\
(5) & A \subseteq C & \text{Theorem 3.5.1 on (2) and (4)} \\
(6) & (\forall b \in B\ [b \in C]) \land (\exists c \in C\ [c \notin B]) & \text{Definition of } B \subset C \text{ on (3)} \\
(7) & \exists c \in C\ [c \notin B] & \text{Conjunctive Simplification on (6)} \\
(8) & (y \in C) \land (y \notin B) & \text{Existential Specification on (7)} \\
(9) & y \notin B & \text{Conjunctive Simplification on (8)} \\
(10) & y \notin A & \text{Modus Tollens on (2) and (9)} \\
(11) & \exists c \in C\ [c \notin A] & \text{Existential Generalization on (10)} \\
(12) & (A \subseteq C) \land (\exists c \in C\ [c \notin A]) & \text{Conjunction on (5) and (11)} \\
(13) & (\forall a \in A\ [a \in C]) \land (\exists c \in C\ [c \notin A]) & \text{Definition of } A \subseteq C \text{ on (12)} \\
(14) & \therefore A \subset C & \text{Definition of } A \subset C \text{ on (13)}
\end{array}
\]
As can be seen, even though we can formally lay out every step of our
proof in a tabular format, the paragraph format presented above is
perhaps more readable and less cumbersome to write.
Two Additional Theorems
We’ve already examined the theorems
\[
\begin{array}{l}
[(A \subset B) \land (B \subset C)] \Longrightarrow (A \subset C)
\end{array}
\]
and
\[
\begin{array}{l}
[(A \subseteq B) \land (B \subseteq C)] \Longrightarrow (A \subseteq C).
\end{array}
\]
We could naturally examine situations involving both regular subsets
and proper subsets.
Theorem 3.5.3: Combining subset and proper-subset transitivity, part one
If $A \subseteq B$ and $B \subset C$, then $A \subset C$.
Proof 3.5.3
Because $B \subset C$, invoking Theorem 3.3.2 yields the fact that
$B \subseteq C$. Thus, because we know that $A \subseteq B$ and
$B \subseteq C$, Theorem 3.5.1 tells us that $A \subseteq C$.
Now, because $B \subset C$, we know there exists some $c \in C$ such
that $c \notin B$. But because we also know that $A \subseteq B$, we
also know that $c \notin A$ as well.
Thus, because we know that $A \subseteq C$ and there is an element
$c \in C$ such that $c \notin A$, we know that $A \subset C$ by
definition as desired.
Theorem 3.5.4: Combining subset and proper-subset transitivity, part two
If $A \subset B$ and $B \subseteq C$, then $A \subset C$.
Proof 3.5.4
Just as in the proof for Theorem 3.5.3, we know that $A \subseteq C$.
Because $A \subset B$, we know there is some element $b \in B$ such
that $b \notin A$. But because $B \subseteq C$, we know that $b \in C$
as well.
Thus, because we know that $A \subseteq C$ and there is an element
$b \in C$ such that $b \notin A$, we know that $A \subset C$ by
definition as desired.
Element arguments represent a powerful proof technique because they
give us a way to account for what elements are in what sets, which
will be an important skill in the upcoming sections when we start to
talk about how we can combine and operate on sets.
Remember that when we talk about a set, what we care about is whether
some given object is a member of that set or not, so by picking an
arbitrary element, we can use the subset relationships discussed so
far to determine if that element is a member of any other set.
A Set of Operations on Sets
We’ve discussed how to construct new sets by simply taking some of the
elements from one set and putting them into a new set. While subsets
are vitally important not only in Set Theory but throughout all of
mathematics, subsets are formed by only considering one set at a time.
We can construct a wide variety of sets by considering two or more
sets at a time. In this section, we learn what kinds of sets we can
construct by considering more than one set at a time.
Complement
Before we talk about more than one set at a time, let’s talk about one
more operation that only relies on one set. Technically, it does
require two sets, because this particular operation requires knowing
what universe is being considered — every element in the universe that
isn’t in $A$ ends up in $A$’s complement.
COMPLEMENT
Consider a set $A$ consisting of elements taken from some universal
set, denoted $\mathcal{U}$.
The complement of $A$, often denoted $A^C$ or $\overline{A}$, is
the set of all elements in $\mathcal{U}$ that are not in $A$. In other
words, we have that
The idea of set complement is really simple: we take all elements
within $\mathcal{U}$ that are not in $A$, and put all of those
elements in a new set we usually call either $A^C$ or $\overline{A}$,
though occasionally other names are used.
Example 3.6.1: Finding a complement
Consider a universal set $\mathcal{U}$ consisting of all the whole
numbers from $0$ to $10$ inclusive, meaning
Now consider the set $S$ of all perfect squares contained within
$\mathcal{U}$, meaning we have
$$S = \{0, 1, 4, 9\}.$$
We then get that
$$S^C = \{2, 3, 5, 6, 7, 8, 10\}.$$
Of course, the essential property we care about when it comes to sets
is whether a specific element is a member of that set or not. How do
we tell that an element $x$ is in $A^C$ or not? Based on the
definition of complement, we see that the following logical
equivalency is used to test if $x$ is in $A$’s complement:
Is $5$ in $S$ or in $S^C$? Of course $5 \in \mathcal{U}$, so it will
either be in $S$ or $S^C$. $S$ was defined to be the set of all
elements from $\mathcal{U}$ that are perfect squares. Since
$5 \in \mathcal{U}$ and $5$ is not a perfect square, we know that
$$5 \in S^C.$$
Now consider the number $9$. Because $9 \in \mathcal{U}$ and $9$ is a
perfect square ($9 = 3^2$), we have that
$$9 \notin S^C.$$
What about a number like $1.5$? Remember that $\mathcal{U}$ only
consists of whole numbers between $0$ and $10$. Since $1.5$ is not a
whole number, we have that $1.5 \notin \mathcal{U}$, and as such we
have that
For clarity, the set $S$ is the set of all perfect squares contained
in $\mathcal{N}$, and $C$ is the set of all perfect cubes contained
within $\mathcal{N}$.
The union of sets $S$ and $C$ is simply the set containing all the
elements from $S$ or $C$, or even both. As such, we get that
Even though in some sense the above set is correct, remember that when
dealing with a set, order and repetition are irrelevant. As such, we
can simplify how the set is written:
In simplifying how we wrote the set referred to by $S \cup C$, we
started by ordering all the numbers (because order is irrelevant),
which makes it easy to see any duplicate entries. After seeing where
the duplicates were, we were able to eliminate them (because
repetition is irrelevant).
One situation not yet considered is when one set is entirely contained
within another. Because a union consists of elements contained in
either set, this means that the union will actually just be whichever
set contains the other one — since every element of the smaller set is
already an element of the larger one, unioning them adds nothing new.
Example 3.6.4: A union where one set contains the other
where $\mathcal{N}$ is the universal set under consideration. Here,
$S$ represents perfect squares within $\mathcal{N}$, and $Q$ is the
set of all perfect fourth powers contained in $\mathcal{N}$. Because
every element in $Q$ is also in $S$, there’s no difference between
Another kind of set we can compose from two other sets is to take
elements that are in both sets. Here, we’re not taking elements that
are present in either set; instead, we demand that the elements are
present in both sets.
INTERSECTION
Consider two sets $A$ and $B$ composed of elements from some universal
set $\mathcal{U}$.
The intersection of sets $A$ and $B$, denoted $A \cap B$, is the
set
$$A \cap B = \{x \in \mathcal{U} \mid (x \in A) \land (x \in B)\}.$$
Just like with the union of two sets, we’d like to determine if an
element $x$ is contained within the intersection of two sets. There’s
a logical equivalency we can use to determine if element $x$ is
contained within $A \cap B$:
$$(x \in A \cap B) \Longleftrightarrow ((x \in A) \land (x \in B)).$$
Before continuing on, it’s worth pointing out the similarity between
the union and intersection symbols and the logical or and and
operators, respectively:
\[
\begin{array}{rl}
\cup \text{ pairs with } &\lor \\
\cap \text{ pairs with } &\land
\end{array}
\]
Example 3.6.5: Computing an intersection
Going back to the example above, we have the following sets:
To figure out what elements are in $S \cap C$, we simply look for
elements that are in both $S$ and $C$:
$$S \cap C = \{0, 1, 64\}.$$
When two sets in question don’t overlap at all, there’s nothing
in the intersection. Of course, this means that the set $A \cap B$ has
nothing in it, or in other words,
$$A \cap B = \{\ \} = \emptyset.$$
This means that the intersection of two sets with nothing in common is
simply the empty set.
where $X$ is the set of odd perfect squares in $\mathcal{N}$, and $Y$
is the set of even perfect cubes in $\mathcal{N}$. Since $X$ and $Y$
share no elements in common, we have that
$$X \cap Y = \emptyset.$$
When one of $A$ or $B$ is a subset of the other, the intersection will
end up just being whichever set was the subset.
Example 3.6.7: An intersection where one set contains the other
It’s possible that we may want to exclude certain elements from the
universe within any sets we construct. Of course, any given set can
contain any element from the universe, whether or not we want it
included. One thing we can do to remedy such a situation is to collect
all of the elements we want excluded in a special set, which we’ll
refer to as $B$ for now (an “exclusion set”). Then, for some given set
we’ll call $A$, we can make a new set by removing all elements from
$A$ that are also in $B$. This gives us a new set containing all
elements from $A$ that we don’t want excluded.
DIFFERENCE
Consider two sets $A$ and $B$ containing elements from some universal
set $\mathcal{U}$.
The difference from $B$ to $A$, which we’ll denote $A - B$, is the
set of all elements in $A$ that are not in $B$. In other words,
$$A - B = \{x \in \mathcal{U} \mid (x \in A) \land (x \notin B)\}.$$
$A - B$ consists of whatever’s left of $A$ once every element $A$
shares with $B$ — that is, every element of $A \cap B$ — has been
removed.
Once again, there’s a logical equivalency we can use to determine if a
given object is contained in the set $A - B$:
$$(x \in A - B) \Longleftrightarrow ((x \in A) \land (x \notin B)).$$
where $\mathcal{N}$ is the universal set, from which sets $S$ and $C$
share elements.
In order to determine which elements are in $S - C$, we start with
$S$, and take away any elements that are also in $C$. What we could do
is look at the intersection of $S$ and $C$:
$$S \cap C = \{0, 1, 64\}.$$
Now, all we have to do is copy $S$, but remove all those elements that
are in $S \cap C$:
$$S - C = \{4, 9, 16, 25, 36, 49, 81, 100\}.$$
We can do the same thing for $C - S$. We simply start with $C$, and
take away elements that are also in $S$ (remember that all the common
elements are in $S \cap C$):
$$C - S = \{8, 27\}.$$
Care should be taken when one of the sets is a subset of the other
(in other words, when one of the sets is entirely contained within the
other). If $B$ is a subset of $A$, then $A - B$ consists of exactly
those elements of $A$ that aren’t also in $B$. But if $A$ is entirely
contained within $B$, we’d have to remove every element of $A$,
leaving us with the empty set $\emptyset$.
Example 3.6.9: A difference resulting in the empty set
where $\mathcal{N}$ is the universal set being used.
By definition, we have that
$$S - E = \{1, 9, 25, 49, 81\}$$
since we’re trying to remove all of the even perfect squares from $S$.
However, notice that since every element in $E$ is also an element in
$S$, when we determine $E - S$, we must remove every element in $E$.
Basically, we start with every element in $E$, and since we’re
removing every element in $E$, we’re left with nothing in $E - S$,
which is just the empty set:
$$E - S = \emptyset.$$
Based on the previous example, it seems that if $A \subseteq B$, then
$A - B = \emptyset$. Is that true for every single set? In a later
section, we’ll start to use element-argument-style proofs to determine
if that’s indeed the case, but for now, our intuition on the matter
seems clear.
Symmetric Difference
With set difference, we only consider elements that are in $A$ but not
in $B$. With a symmetric difference, we instead want elements that are
either in $A$ and not in $B$, or in $B$ and not in $A$.
SYMMETRIC DIFFERENCE
Consider two sets $A$ and $B$ containing elements from some universal
set $\mathcal{U}$.
The symmetric difference between $A$ and $B$, denoted
$A \triangle B$, is the set
$$A \triangle B = \{x \in \mathcal{U} \mid ((x \in A) \land (x \notin B)) \lor ((x \notin A) \land (x \in B))\}.$$
The symmetric difference consists of everything covered by the union
of $A$ and $B$, except whatever’s also covered by their intersection —
in other words,
$$A \triangle B = (A \cup B) - (A \cap B).$$
Of course, we can examine given objects to determine membership in
such a set:
$$(x \in A \triangle B) \Longleftrightarrow [((x \in A) \land (x \notin B)) \lor ((x \notin A) \land (x \in B))].$$
Example 3.6.10: Testing membership in a symmetric difference
When one set is a subset of the other, the symmetric difference
essentially carves the smaller set out of the larger one — it’s called
a symmetric difference because it’s essentially a difference between
the larger set and the subset.
Example 3.6.11: A symmetric difference is commutative
where $\mathcal{N}$ is the universal set in question.
We see that
$$S \triangle E = \{1, 9, 25, 49, 81\}.$$
Since we only want elements that are in $S$ and not $E$, or elements
that are in $E$ and not $S$, we also have that
$$E \triangle S = \{1, 9, 25, 49, 81\}$$
as well.
We saw in the previous example that $S \triangle E$ and
$E \triangle S$ ended up being the exact same set. Is that always the
case? In a later section, we’ll use an element argument to
demonstrate that this observation does indeed hold in general.
Multiple Unions and Intersections
We can combine as many sets into a new set as we want. However, before
we start combining multiple sets into one, we should recognize that
dealing with multiple sets can be cumbersome. First of all, imagine
having to deal with multiple sets with a wide variety of names that
were hard to track. We can mitigate this issue by coming up with a
naming scheme that makes each set we’re dealing with easy to track.
INDEX SET, INDEX
Consider some universal set $\mathcal{U}$.
Let $I$ be a non-empty set of objects that are not necessarily taken
from $\mathcal{U}$, and for each $i \in I$, let
$$A_i \subseteq \mathcal{U}.$$
The set $I$ is called an index set, and every $i \in I$ is called
an index.
Before looking at an example, let’s stress the fact that an index set
is basically used to enumerate and track sets, and so doesn’t
necessarily need to be taken from some universal set (though it could
be taken from its own particular universal set).
Example 3.6.12: Naming several sets with an index set
Consider the universe $\mathbb{Z}$ of all whole numbers, including $0$
and all negative whole numbers:
We could deal with a number of sets called $A$, $B$, $C$, $D$, and $E$
where
\[
\begin{array}{rl}
A &= \{2, 4, 6, 8, 10\} \\
B &= \{-1, 0, 1, 2, 3, 4, 5\} \\
C &= \{1, 2, 7, 8\} \\
D &= \{1, 3, 4\} \\
E &= \{-5, -12, -13\}
\end{array}
\]
We could even have more exotic names, like $\Lambda$, $M$, $\Phi$,
$\Psi$, and $\clubsuit$, but this would be unnecessarily tedious
(though there may be times where such names would be useful).
Instead, what we could do is recognize that since we’re dealing with
$5$ sets, we could come up with an index set containing $5$ elements,
like so:
$$I = \{1, 2, 3, 4, 5\}.$$
Then, instead of using $5$ different set names, we can simply use one
symbol, marked with the indices from the index set:
Notice that in this case, we have that
$A_1, A_2, A_3, A_4, A_5 \subseteq \mathbb{Z}$, but we also have that
$I \subseteq \mathbb{Z}$, even though we don’t need that to be the
case.
Example 3.6.13: An index set that is not a subset of the universe
Reconsider the set $\mathbb{Z} = \{0, 1, -1, 2, -2, 3, -3, 4, -4, \dots\}$.
Now consider the index set
where $A_1, A_a, A_x, A_\boxminus, A_\star \subseteq \mathbb{Z}$. This
time, even though we have $I \not\subseteq \mathbb{Z}$, there’s at
least one element in $I$ that’s also in $\mathbb{Z}$.
It’s worth noting that this kind of index set is impractical, and so
won’t be used going forward unless there’s some niche application
where it would be useful.
With an index set to help us, working with multiple sets is easy.
MULTIPLE UNION
Consider a universal set $\mathcal{U}$, along with an index set $I$
where for each $i \in I$, we have that $A_i \subseteq \mathcal{U}$.
The multiple union of all $A_i$ is denoted
$$\bigcup_{i \in I} A_i.$$
If the index set is defined using the whole numbers
$I = \{1, 2, 3, \dots, n\}$, then we can write this multiple union as
$$\bigcup_{i=1}^{n} A_i.$$
As always, if we want to check to see if a given object is in this
set, all we have to do is check to see if that object is in at least
one of the sets making up the multiple union:
Naturally, there’s an intersection counterpart to the multiple union.
MULTIPLE INTERSECTION
Consider a universal set $\mathcal{U}$, along with an index set $I$
where for each $i \in I$, we have that $A_i \subseteq \mathcal{U}$.
The multiple intersection of all $A_i$ is denoted
$$\bigcap_{i \in I} A_i.$$
If the index set is defined using the whole numbers
$I = \{1, 2, 3, \dots, n\}$, then we can write this multiple
intersection as
$$\bigcap_{i=1}^{n} A_i.$$
Checking if a given object is in the multiple intersection of a
collection of sets works nearly the same as checking if it’s in the
multiple union, but we need to make sure the object is in all sets,
which we can check with multiple and operations:
In the previous section, we defined the complement, union,
intersection, difference, symmetric difference, multiple union, and
multiple intersection of sets purely symbolically. While the symbolic
definitions are precise, it’s often easier to build an intuition for
what these operations actually do by picturing them.
The standard way to picture a set is as a circle, with each circle
drawn inside a rectangle standing for the universe $\mathcal{U}$.
Whenever an operation includes some region formed by these circles, we
shade that region green.
Complement
Figure 3.7.1:The complement of a set $A$, denoted $A^C$, is the set of all elements within $\mathcal{U}$ that are not within $A$.
Since the complement of $A$ consists of everything in the universe
that isn’t in $A$, the circle for $A$ itself is left unshaded, while
everything else within the surrounding rectangle is shaded to
represent $A^C$.
Union
Figure 3.7.2:The union of two sets is the set containing any element contained in either set. $A$ and $B$ share some overlap here, but that overlap is still included — any part of either set is shaded green to indicate it’s in the union.
As long as an element is in $A$ or $B$, or even both, it’s included in
the union — so both circles are shaded in their entirety, overlap
included.
Intersection
Figure 3.7.3:The intersection of two sets is the set containing all elements that are in both sets. Only the overlap between the two circles is shaded green, since that’s the only region representing elements contained in both $A$ and $B$.
Here, only the sliver shared by both circles is shaded, since that’s
the only region representing elements common to both $A$ and $B$.
Difference
Figure 3.7.4:The difference $A - B$ is the set of all elements in $A$ that are not in $B$. The overlap with $B$ is left unshaded, since those elements are excluded — only the exclusive part of $A$ is shaded green.
Only the part of $A$’s circle left over once the overlapping sliver
shared with $B$ has been carved away is shaded.
Symmetric Difference
Figure 3.7.5:As long as an element is in one of $A$ or $B$, but not both, it’s included in the symmetric difference of $A$ and $B$ — everything covered by either circle is shaded green, except the overlap.
The symmetric difference shades everything covered by either circle
except the overlapping sliver shared by both — everything a union
would shade, minus whatever an intersection would shade.
Multiple Union
Figure 3.7.6:When dealing with multiple sets $A_1$, $A_2$, and $A_3$, we can take every single element from every set and stick them into a new set, which we refer to as the union of all the sets.
This same idea of shading extends naturally to more than two sets at
once — every region covered by at least one of $A_1$, $A_2$, or $A_3$
is shaded to represent their multiple union.
Multiple Intersection
Figure 3.7.7:When dealing with multiple sets $A_1$, $A_2$, and $A_3$, we can take all elements contained in all three and stick them into a new set, which we refer to as the intersection of all the sets.
Here, only the single region shared by all three circles at once —
where $A_1$, $A_2$, and $A_3$ all overlap — is shaded, representing
their multiple intersection.
The Laws of Set Theory
Understanding the laws of set theory is essential because they solve
the problem of determining when two or more sets are equal, through
fundamental equations including associativity, commutativity, and
distribution. These laws provide a structured framework for defining
and manipulating sets, ensuring consistency and precision in
operations such as subset, difference, and symmetric difference.
In other words, by developing laws of set theory, we can essentially
manipulate equations involving sets as if they were algebraic
equations. In addition, the laws of set theory will allow us to
convert complicated expressions involving sets into simpler ones, much
like how we took complicated logical expressions and simplified them
into smaller, logically equivalent expressions.
By studying these laws, one can formulate and prove mathematical
concepts and theorems accurately. Furthermore, the foundations of many
branches of mathematics are developed using the language of sets and
set theory. In addition to the further development of mathematics,
these laws also prove crucial to subjects like computer science and
data science.
Equal Sets
We’ve briefly described some aspects of set equality when we first
introduced what sets were. We talked about how repetition and order
are irrelevant with respect to what elements are contained within a
set. As such, even if some elements are repeated in various
descriptions, they still represent the same set.
However, as we’ve also seen, there are a wide variety of methods to
describe a set and what elements are contained within it. Just because
we come up with different descriptions, that doesn’t necessarily mean
we’re describing different sets — sometimes we may be describing
exactly the same set.
Example 3.8.5: Two different descriptions of the same set
Consider the set
$$M = \{x \mid x \text{ is an even number between } -3 \text{ and } 10 \text{ inclusive}\}.$$
The description of $N$ is a little more complicated, but we can still
determine what elements are in $N$. The only tricky part is
recognizing that the elements of $N$ are based off elements that are
in the set $\{-1.25, -0.75, -0.25, 0.25, 0.75, 1.25, 1.75\}$, so we
just have to iterate over those elements to figure out what’s in $N$.
We’ll tabulate all values below:
Since $N$ consists of the values $4x + 3$, we now know that
$$N = \{-2, 0, 2, 4, 6, 8, 10\}.$$
Notice that $M$ and $N$ have the same number of elements ($7$), and
all elements are exactly the same. Thus, since $M$ and $N$ represent
the exact same set, we could write
$$M = N.$$
It’s worth pointing out that since every element in $M$ is also in
$N$, and every element in $N$ is also in $M$, we have that
\[
\begin{array}{rl}
M &\subseteq N \\
N &\subseteq M
\end{array}
\]
However, defining set equality by having the same elements isn’t
exactly rigorous. Two sets can share some number of elements while
also having different elements. As such, we don’t define set equality
by simply having the same elements — instead, we look to the end of
the previous example for inspiration.
EQUAL
Consider the two sets $A$ and $B$ taken from some universal set
$\mathcal{U}$.
Sets $A$ and $B$ are called equal, and we write $A = B$, when
$$A \subseteq B$$
and
$$B \subseteq A.$$
This definition is precise because we can determine whether a set is a
subset of another set under scrutiny. As such, because the definition
of set equality requires we show two things to be true, proofs showing
two sets to be equal are usually broken up into two parts.
It’s this definition of set equality that makes the next two theorems
somewhat straightforward to prove.
Theorem 3.8.9: A subset relationship gives an intersection
For any $A, B \subseteq \mathcal{U}$, if $A \subseteq B$, then
$A \cap B = A$.
Proof 3.8.9
Step 1: Show that $A \cap B \subseteq A$.
Let $x$ be an arbitrarily picked element within $A \cap B$ (meaning
$x \in A \cap B$).
By definition, $x \in A$ and $x \in B$, and as such we know that
$x \in A$.
Thus, we have shown that if $x \in A \cap B$, then $x \in A$. This
means that $A \cap B \subseteq A$, completing Step 1.
Step 2: Show that $A \subseteq A \cap B$.
Let $x$ be an arbitrarily picked element of $A$. Because we have by
premise that $A \subseteq B$, we also have that $x \in B$.
Thus, since $x \in A$ and $x \in B$, we have that $x \in A \cap B$ by
definition.
We’ve just shown that when $x \in A$, we also have that
$x \in A \cap B$. We then have by definition that
$A \subseteq A \cap B$, completing Step 2.
Conclusion. Because we’ve shown that $A \subseteq A \cap B$ and
that $A \cap B \subseteq A$, we have by definition that
$$A \cap B = A$$
as desired.
Thus, by this theorem, whenever we know that $A \subseteq B$, we can
replace $A \cap B$ with just $A$ anywhere we see $A \subseteq B$ in an
equation. It’s worth stressing that we can only make such a
replacement when we know $A \subseteq B$ — otherwise, doing so may not
be correct.
There are a few more such equalities we can make use of, which we now
show in the next theorem.
Theorem 3.8.10: Four equivalent ways to say a subset relationship
For any $A, B \subseteq \mathcal{U}$, we have that
$$(A \subseteq B) \Longleftrightarrow (A \cap B = A) \Longleftrightarrow (A \cup B = B) \Longleftrightarrow (B^C \subseteq A^C).$$
Proof 3.8.10
We’ll break this proof up into four steps.
Step 1: $(A \subseteq B) \Longrightarrow (A \cap B = A)$.
This is simply the previous theorem.
Step 2: $(A \cap B = A) \Longrightarrow (A \cup B = B)$.
The premise of this logical implication is that $A \cap B = A$, so we
can bring that fact in whenever needed.
First we show that $A \cup B \subseteq B$.
Let $x$ be an arbitrarily picked element in $A \cup B$. Thus, by
definition, $x \in A$ or $x \in B$. If $x$ happens to be an element of
$B$, then it’s tautologically true that $x \in B$.
If $x \in A$, then the premise tells us that $x \in A \cap B$, meaning
$x \in B$ as well.
Either way, we have that whenever $x \in A \cup B$, we also have that
$x \in B$, meaning $A \cup B \subseteq B$.
Second, we show that $B \subseteq A \cup B$.
Let $x$ be an arbitrarily picked element within $B$. Thus, because
$x \in B$, we also have that $x \in A \cup B$.
We’ve just shown that whenever $x \in B$, we also have that
$x \in A \cup B$. This means that $B \subseteq A \cup B$.
Finally, because we’ve shown that $A \cup B \subseteq B$ and
$B \subseteq A \cup B$, we have that
$$A \cup B = B$$
as desired.
Step 3: $(A \cup B = B) \Longrightarrow (B^C \subseteq A^C)$.
Remember that $A^C$ is the complement of $A$ within $\mathcal{U}$.
Also remember that $A \cup B = B$ is a premise of this logical
implication, so we can use that fact whenever we want (we can’t use
$A \cap B = A$ as a premise here, since that isn’t included in the
implication we’re considering).
We’re trying to show that $B^C \subseteq A^C$, so let $x$ be an
arbitrarily picked element of $B^C$.
Because $x \in B^C$, we have by definition that $x \notin B$. Now,
because $A \cup B = B$ and $x \notin B$, we also have that
$x \notin A \cup B$.
Next, because $x \notin A \cup B$, we also know that $x \notin A$.
Therefore, since $x \notin A$, we must have that $x \in A^C$.
We’ve just shown that whenever $x \in B^C$, we also have that
$x \in A^C$ (assuming $A \cup B = B$ as a premise).
Thus by definition, whenever we know $A \cup B = B$, we have that
$B^C \subseteq A^C$ as desired.
Step 4: $(B^C \subseteq A^C) \Longrightarrow (A \subseteq B)$.
We have $B^C \subseteq A^C$ as a premise, so we’ll invoke that fact
when needed.
Let $x$ be an arbitrarily picked element of $A$. Since $x \in A$, we
have by definition that $x \notin A^C$.
Now, because $B^C \subseteq A^C$ and $x \notin A^C$, we have (by Modus
Tollens) that $x \notin B^C$.
Now, because $x \notin B^C$, we have by definition that $x \in B$.
Thus, we’ve shown that, assuming $B^C \subseteq A^C$ as a premise,
whenever $x \in A$, we must also have that $x \in B$.
Therefore, assuming $B^C \subseteq A^C$ as a premise, we have that
$A \subseteq B$ as desired.
Conclusion. Because we’ve shown that
\[
\begin{array}{l}
(A \subseteq B) \Longrightarrow (A \cap B = A) \\[0.75em]
(A \cap B = A) \Longrightarrow (A \cup B = B) \\[0.75em]
(A \cup B = B) \Longrightarrow (B^C \subseteq A^C) \\[0.75em]
(B^C \subseteq A^C) \Longrightarrow (A \subseteq B)
\end{array}
\]
we have that they’re all logically equivalent, meaning we have that
$$(A \subseteq B) \Longleftrightarrow (A \cap B = A) \Longleftrightarrow (A \cup B = B) \Longleftrightarrow (B^C \subseteq A^C)$$
as desired.
Because of this theorem, whenever we know one of
$$A \subseteq B \qquad A \cap B = A \qquad A \cup B = B \qquad B^C \subseteq A^C$$
is true, we automatically know that the other three are true as well.
Disjoint Sets
The total opposite of two sets $A$ and $B$ being equal is when $A$ and
$B$ have absolutely no elements in common. Of course, if two sets
share no element in common, then their intersection is a set with no
elements in it — the empty set.
DISJOINT
Consider the two sets $A$ and $B$ taken from some universal set
$\mathcal{U}$.
$A$ and $B$ are called disjoint whenever
$$A \cap B = \emptyset.$$
Many of the concepts discussed so far may seem disparate. However,
they’re more connected than may initially appear.
Theorem 3.8.11: Disjoint sets and the symmetric difference
Let $A, B \subseteq \mathcal{U}$.
$$(A \cap B = \emptyset) \Longleftrightarrow (A \cup B = A \triangle B)$$
Proof 3.8.11
Because we have a logical biconditional, we break this proof into two
steps. First, we assume $A \cap B = \emptyset$ as a premise and show
that $A \cup B = A \triangle B$. Next, we assume
$A \cup B = A \triangle B$ as a premise and show that
$A \cap B = \emptyset$.
Step 1: $(A \cap B = \emptyset) \Longrightarrow (A \cup B = A \triangle B)$.
First, we try to show that $A \cup B \subseteq A \triangle B$.
Let $x$ be an arbitrarily picked element within $A \cup B$. Thus, we
have by definition that $x \in A$ or $x \in B$, but because
$A \cap B = \emptyset$, $x$ can’t be in both $A$ and $B$.
Thus we either have that $x \in A$ and $x \notin B$, meaning
$x \in A \triangle B$, or we have that $x \notin A$ and $x \in B$,
meaning we still have that $x \in A \triangle B$.
Either way, we have that $x \in A \triangle B$. Hence, we’ve just
shown that (assuming $A \cap B = \emptyset$) whenever
$x \in A \cup B$, we also have that $x \in A \triangle B$ as well, and
so we have that
$$A \cup B \subseteq A \triangle B.$$
Second, we try to show that $A \triangle B \subseteq A \cup B$.
Now, let $x$ be an arbitrarily picked element in $A \triangle B$.
Thus, by definition, either $x \in A$ and $x \notin B$, meaning
$x \in A \cup B$, or $x \notin A$ and $x \in B$, meaning
$x \in A \cup B$.
Either way, we know that whenever $x \in A \triangle B$, we also know
that $x \in A \cup B$ as well.
Thus, by definition we have that
$$A \triangle B \subseteq A \cup B.$$
Finally, because we know that $A \cup B \subseteq A \triangle B$ and
$A \triangle B \subseteq A \cup B$, we have by definition that
$$A \cup B = A \triangle B.$$
This shows that $(A \cap B = \emptyset) \Longrightarrow (A \cup B = A \triangle B)$,
completing Step 1.
Step 2: $(A \cap B = \emptyset) \Longleftarrow (A \cup B = A \triangle B)$.
Here, we’re assuming $A \cup B = A \triangle B$ as a premise, and are
trying to show that $A \cap B = \emptyset$.
Because we know that $A \cup B = A \triangle B$, we have by the
definition of set equality that $A \cup B \subseteq A \triangle B$ and
$A \triangle B \subseteq A \cup B$.
Thus, for any element $x \in A \cup B$, we also know that
$x \in A \triangle B$. However, since $x \in A \triangle B$, we know
that either
$$x \in A \text{ and } x \notin B \qquad \text{or} \qquad x \notin A \text{ and } x \in B.$$
Either way, no element $x \in A \cup B$ can be in both $A$ and $B$.
Thus, because no element $x$ exists such that $x \in A \cap B$, we
must have that $A \cap B$ is empty, meaning $A \cap B = \emptyset$.
This shows that $(A \cup B = A \triangle B) \Longrightarrow (A \cap B = \emptyset)$,
completing Step 2.
Conclusion. Because we’ve shown that
$(A \cap B = \emptyset) \Longrightarrow (A \cup B = A \triangle B)$ and
$(A \cup B = A \triangle B) \Longrightarrow (A \cap B = \emptyset)$, we
must have that
$$(A \cap B = \emptyset) \Longleftrightarrow (A \cup B = A \triangle B)$$
as desired.
This theorem connects the concepts of equal sets, disjoint sets,
unions, and symmetric differences.
Membership Tables
Remember that $x \in A$ is a statement about $x$ being a member of set
$A$. Because it’s a statement, it has a truth value of either $0$ or
$1$.
One powerful tool we had at our disposal for analyzing logical
expressions was a truth table. We can adapt truth tables to analyze
set relationships as well. In this case, membership tables offer a
tabular way of representing element arguments.
\[
\begin{array}{l|l|l|l}
x \in A & x \in B & x \in A \cup B & x \in A \cap B \\
\hline
0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 1 & 1
\end{array}
\]
Instead of using some generic symbol for a generic object whose
membership in some set we want to analyze, we could simply list
whatever sets we’re analyzing. As such, we could rewrite the above
table like so:
\[
\begin{array}{l|l|l|l}
A & B & A \cup B & A \cap B \\
\hline
0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 1 & 1
\end{array}
\]
All of the set operations have their corresponding tables, which we
can include inside one large table:
\[
\begin{array}{l|l|l|l|l|l|l}
A & B & A^C & A \cup B & A \cap B & A - B & A \triangle B \\
\hline
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 & 0 & 0
\end{array}
\]
Based on the values above, we see that there are obvious analogues
with logical operators:
\[
\begin{array}{rl}
A^C &\text{ pairs with } \neg a \\
A \cup B &\text{ pairs with } a \lor b \\
A \cap B &\text{ pairs with } a \land b \\
A - B &\text{ pairs with } \neg(a \to b) \\
A \triangle B &\text{ pairs with } a \veebar b
\end{array}
\]
Notice that the set difference resembles the logical complement of an
implication.
In addition to the set operations, we can determine subset
relationships. Let’s take a look at one more table to see an example:
\[
\begin{array}{l|l|l|l|l|l}
x \in A & x \in B & x \in A \cup B & x \in A \cap B & (x \in A \cap B) \to (x \in A) & (x \in A) \to (x \in A \cup B) \\
\hline
0 & 0 & 0 & 0 & 1 & 1 \\
0 & 1 & 1 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1
\end{array}
\]
Notice that the columns
$(x \in A \cap B) \to (x \in A)$ and $(x \in A) \to (x \in A \cup B)$ have all
$1$s. As such, we can write
$$(x \in A \cap B) \Longrightarrow (x \in A) \qquad \text{and} \qquad (x \in A) \Longrightarrow (x \in A \cup B)$$
which is exactly the definition of subset. As such, we can determine
that a set $A$ is a subset of some other set $B$ by comparing their
respective columns. If $A$ has $1$s in every row where $B$ also has a
$1$, then we know that $A \subseteq B$.
Let’s prove a simple observation we may have had in our study of set
relationships and operations.
Theorem 3.8.12: Intersection and union sandwich a set
Let $A, B \subseteq \mathcal{U}$.
$$(A \cap B) \subseteq A \subseteq (A \cup B)$$
Proof 3.8.12
Let’s examine a membership table:
\[
\begin{array}{l|l|l|l}
A & B & A \cup B & A \cap B \\
\hline
0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 1 & 1
\end{array}
\]
Now, we can compare specific columns we want, but we’ll do so in two
separate membership tables:
\[
\begin{array}{l|l}
A \cap B & A \\
\hline
0 & 0 \\
0 & 0 \\
0 & 1 \\
1 & 1
\end{array}
\qquad
\begin{array}{l|l}
A & A \cup B \\
\hline
0 & 0 \\
0 & 1 \\
1 & 1 \\
1 & 1
\end{array}
\]
In the table on the left, every row of $A \cap B$ that has a $1$ also
has a $1$ in the $A$ column. As such, we know that
$(A \cap B) \subseteq A$.
Additionally, from the table on the right, every row in the $A$
column that has a $1$ also has a $1$ in the $A \cup B$ column. As
such, we know that $A \subseteq (A \cup B)$.
Because we know that $(A \cap B) \subseteq A$ and
$A \subseteq (A \cup B)$, we can combine them into the single
expression
$$(A \cap B) \subseteq A \subseteq (A \cup B)$$
as desired.
Membership tables can be a convenient way to represent an element
argument in tabular form. However, membership tables won’t totally
replace element arguments, as sometimes the element argument is
simpler to make for more esoteric situations where tables can be
cumbersome to use.
Some Laws of Set Theory
We’ve already examined a number of simple laws (referred to as
theorems) above, and have demonstrated their truth by way of element
arguments and membership tables. What’s demonstrated above only
scratches the surface of the multitude of equal-set relationships we
can exploit. Below, we present a large table offering even more such
laws.
As you look over this table, it would be wise to compare it with the
table presented in Chapter 1, Section 5. You should see many
analogues.
For any sets $A$, $B$, and $C$ taken from some universal set
$\mathcal{U}$, we have the following equivalencies:
Law of Double Complement
$(A^C)^C = A$
DeMorgan’s Laws
$\begin{array}{c} (A \cup B)^C = A^C \cap B^C \\ (A \cap B)^C = A^C \cup B^C \end{array}$
Commutative Laws
$\begin{array}{c} A \cup B = B \cup A \\ A \cap B = B \cap A \end{array}$
Associative Laws
$\begin{array}{c} A \cup (B \cup C) = (A \cup B) \cup C \\ A \cap (B \cap C) = (A \cap B) \cap C \end{array}$
Distributive Laws
$\begin{array}{c} A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \\ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \end{array}$
Idempotent Laws
$\begin{array}{c} A \cup A = A \\ A \cap A = A \end{array}$
Identity Laws
$\begin{array}{c} A \cup \emptyset = A \\ A \cap \mathcal{U} = A \end{array}$
Inverse Laws
$\begin{array}{c} A \cup A^C = \mathcal{U} \\ A \cap A^C = \emptyset \end{array}$
Domination Laws
$\begin{array}{c} A \cup \mathcal{U} = \mathcal{U} \\ A \cap \emptyset = \emptyset \end{array}$
Absorption Laws
$\begin{array}{c} A \cup (A \cap B) = A \\ A \cap (A \cup B) = A \end{array}$
Certainly, many more such laws and equivalencies exist, but the table
above represents some of the more ubiquitous laws that are often used.
These laws present us an opportunity to directly compare three
different proof techniques. First, we use a membership table.
Theorem 3.8.13: Law of Double Complement
For any $A \subseteq \mathcal{U}$,
$$(A^C)^C = A$$
Proof 3.8.13
General Strategy: We’ll use a membership table to compare the
columns for $A$ and $(A^C)^C$.
Since the columns for $A$ and $(A^C)^C$ are exactly the same, they’re
equal, and so we have
$$A = (A^C)^C$$
as desired.
Now, compare such a simple strategy with a standard element argument
used in the next proof.
Theorem 3.8.14: DeMorgan’s Laws
For any $A, B \subseteq \mathcal{U}$,
\[
\begin{array}{rl}
(A \cup B)^C &= A^C \cap B^C \\
(A \cap B)^C &= A^C \cup B^C
\end{array}
\]
Proof 3.8.14
General Strategy: We’ll split this proof up into two steps, where
each step is dedicated to a different part of DeMorgan’s Laws. Both
steps will make use of standard element arguments, as well as the
definition of set equality.
Step 1: $(A \cup B)^C = A^C \cap B^C$.
First, we show that $(A \cup B)^C \subseteq A^C \cap B^C$.
Suppose that element $x \in \mathcal{U}$ was also in $(A \cup B)^C$,
meaning $x \in (A \cup B)^C$.
Since $x \in (A \cup B)^C$, we know that $x \notin ((A \cup B)^C)^C$,
and by the Law of Double Complement, we also know that
$x \notin A \cup B$.
Now we simultaneously know that $x \notin A$ (otherwise $x$ would be
in $A \cup B$, which we just established isn’t the case) and
$x \notin B$ (again, because then $x$ would be in $A \cup B$).
Thus, since $x \notin A$ and $x \notin B$, we know that $x \in A^C$
and $x \in B^C$, meaning $x \in A^C \cap B^C$, which is what we’re
trying to show.
Thus, whenever we have that $x \in (A \cup B)^C$, we know that
$x \in A^C \cap B^C$, meaning
$$(A \cup B)^C \subseteq A^C \cap B^C.$$
Second, we show that $A^C \cap B^C \subseteq (A \cup B)^C$.
Now, suppose we knew that $x \in A^C \cap B^C$. Thus we simultaneously
know that $x \in A^C$ and $x \in B^C$.
This means we have that $x \notin A$ and $x \notin B$. Thus, since $x$
is in neither $A$ nor $B$, $x$ can’t possibly be in $A \cup B$,
meaning $x \notin A \cup B$. This means that $x \in (A \cup B)^C$.
Thus, whenever we know that $x \in A^C \cap B^C$, we also know that
$x \in (A \cup B)^C$, which means we have that
$$A^C \cap B^C \subseteq (A \cup B)^C.$$
Finally, because we’ve shown that $(A \cup B)^C \subseteq A^C \cap B^C$
and $A^C \cap B^C \subseteq (A \cup B)^C$, we have by definition that
$$(A \cup B)^C = A^C \cap B^C$$
as desired.
Step 2: $(A \cap B)^C = A^C \cup B^C$.
Using nearly the exact same process as was used in Step 1, we come to
the conclusion that
$$(A \cap B)^C = A^C \cup B^C.$$
Conclusion. By now, we’ve shown that both
$(A \cup B)^C = A^C \cap B^C$ and $(A \cap B)^C = A^C \cup B^C$, and as
such have completely established DeMorgan’s Laws.
The last of these laws we’ll prove in this section will be the
Distributive Laws. Instead of making two separate element arguments to
establish that
$A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and
$(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$, we’ll use
logical equivalencies to show that $A \cup (B \cap C)$ and
$(A \cup B) \cap (A \cup C)$ are subsets of each other simultaneously.
Theorem 3.8.15: Distributive Laws
For any $A, B, C \subseteq \mathcal{U}$,
\[
\begin{array}{rl}
A \cup (B \cap C) &= (A \cup B) \cap (A \cup C) \\
A \cap (B \cup C) &= (A \cap B) \cup (A \cap C)
\end{array}
\]
Proof 3.8.15
General Strategy: We’ll use logical equivalencies to show that any
element of $A \cup (B \cap C)$ is also an element of
$(A \cup B) \cap (A \cup C)$. Since we use logical equivalencies, this
also applies the other way simultaneously.
\[
\begin{array}{lll}
& \boldsymbol{x \in A \cup (B \cap C)} & \textbf{Reason} \\
\Longleftrightarrow & (x \in A) \lor (x \in B \cap C) & \text{Definition of Set Union} \\
\Longleftrightarrow & (x \in A) \lor [(x \in B) \land (x \in C)] & \text{Definition of Set Intersection} \\
\Longleftrightarrow & [(x \in A) \lor (x \in B)] \land [(x \in A) \lor (x \in C)] & \text{Distribution of } \lor \text{ over } \land \\
\Longleftrightarrow & (x \in A \cup B) \land (x \in A \cup C) & \text{Definition of Set Union} \\
\Longleftrightarrow & x \in (A \cup B) \cap (A \cup C) & \text{Definition of Set Intersection}
\end{array}
\]
We’ve just established that
$$(x \in A \cup (B \cap C)) \Longleftrightarrow (x \in (A \cup B) \cap (A \cup C)).$$
Thus, whenever $x \in A \cup (B \cap C)$, we also simultaneously know
that $x \in (A \cup B) \cap (A \cup C)$. As such, we know that
$$A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C) \qquad \text{and} \qquad (A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$$
are simultaneously true, and so we have that
$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
as desired. The same logic also shows that
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ as well.
Of course, the rest of the Laws of Set Theory can be proven in many
different ways. Because it’s a good exercise to provide a proof for
all the above laws, proofs for the remaining laws are asked for in
this chapter’s practice questions.
Using the Laws of Set Theory
Just as we can use the Laws of Logic to simplify complicated logical
expressions, we can also use the Laws of Set Theory to simplify
complicated expressions involving sets.
Example 3.8.6: Simplifying an expression with the Laws of Set Theory
Suppose we were dealing with the expression
$$X \cap (Y - X)$$
where $X, Y \subseteq \mathcal{U}$.
We can simplify the above expression in the following way:
For any two sets $A$ and $B$ taken from some universal set
$\mathcal{U}$, we have that
$$A \triangle B = (A \cap B^C) \cup (B^C \cap A).$$
This just goes to show how important the union, intersection, and
complement operations are when it comes to sets, because no matter how
we want to combine any combination of sets, we can always do so using
just those three operations.
Example 3.8.7: Rewriting an expression with only Union, Intersection, and Complement
Can we rewrite the expression
$$(X - Y)^C$$
using only union, intersection, and complement? Let’s try and see.
It stands to reason that repeated applications of DeMorgan’s Laws
would work for intersection as well.
Furthermore, no matter how many sets we have, it seems as if we can
just repeatedly apply DeMorgan’s Laws and the Associative Law. Based
on this observation, we make the following observations, where $I$ is
an index set over which the multiple union or intersection is taken:
Of course, mere intuition doesn’t prove anything to be true. Instead,
we must rely on rigorous tools, such as element arguments and logical
equivalencies, to establish truth without any doubt.
Theorem 3.8.16: Generalized DeMorgan’s Laws
For a universal set $\mathcal{U}$ and index set $I$, the following
equalities hold:
General Strategy: We’ll make use of various definitions of Set
Theory, including Complement. We also make use of the logical rules
that apply when negating quantified statements, as discussed in
Chapter 1, Section 10.
\[
\begin{array}{lll}
& \boldsymbol{x \in \left(\bigcup_{i \in I} A_i\right)^C} & \textbf{Reason} \\
\Longleftrightarrow & x \notin \bigcup_{i \in I} A_i & \text{Definition of Set Complement} \\
\Longleftrightarrow & \neg \exists i \in I\ [x \in A_i] & \text{Definition of Multiple Union} \\
\Longleftrightarrow & \forall i \in I\ [x \notin A_i] & \text{Negating a Quantified Statement} \\
\Longleftrightarrow & \forall i \in I\ [x \in A_i^C] & \text{Definition of Set Complement} \\
\Longleftrightarrow & x \in \bigcap_{i \in I} A_i^C & \text{Definition of Multiple Intersection}
\end{array}
\]
We’ve just shown why the first equivalence is true. By similar logic,
the second is also shown to be true. This establishes both
equivalencies as desired.
The generalized versions of DeMorgan’s Laws will prove extremely
useful in our further study of math.
Proof Technique: Exhaustion
All of the proof techniques discussed thus far work on sets that have
infinitely many elements. For example, we’ve talked about theorems
that apply to all even numbers, not merely some of them. For
example, for all even numbers, adding one yields an odd number. As
another example, no matter which two even numbers are added together,
the sum is always another even number.
We’ve even talked about theorems that apply to all sets in general,
not merely some of them. These include the set operations and the set
equalities.
It’s not hard to see why methods that apply to sets of infinitely many
objects are powerful. On the other hand, there may be times when there
are only a finite number of objects we’re interested in examining.
Here, we discuss a method for proving theorems dealing with only a
finite number of elements.
A Contrived Example
Admittedly, dealing with only a finite number of elements seems hardly
useful, as we mostly want results that apply generally, and not
specifically. Here, we provide a contrived example simply to
demonstrate the technique.
Example 3.9.1: Every even number from 2 to 20 as a sum of squares
Suppose we wanted to prove that the integers in the set
$$E = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$$
could all be expressed as a sum of no more than three perfect squares.
We could do this by simply tabulating the results.
Notice that we didn’t say the numbers in $E$ could be written as a
unique sum. There may be more than one way to write some of the
numbers as a sum of perfect squares, like in the following example:
$$18 = 1^2 + 1^2 + 4^2.$$
We could perhaps call the result derived in the previous example a
theorem, since we did provide a proof for the fact, but we prefer to
reserve the term theorem for “major” results. That result was merely
an exercise.
The Method of Exhaustion
Based on the previous example, we can see why the method presented in
this section is called the Method of Exhaustion: it’s because we
exhaustively check every single element in the desired set for a
desired property. Note that while the term exhaustion is used to
describe the thoroughness of checking every element, it doesn’t
necessarily refer to the feeling one may get whilst performing the
checking — though it should be said that for rather large sets, this
method can be tiresome.
Let’s lay out the argument in logical fashion, same as we’ve done
before. The reason this method works is because we verify that every
element in the desired set has some desired property.
Suppose set $A$ is a finite subset consisting of the $n$ elements
$$A = \{a_1, a_2, a_3, \dots, a_n\}$$
taken from some universal set $\mathcal{U}$. Furthermore, suppose $p$
is an open statement defined on $\mathcal{U}$. The Method of
Exhaustion is simply an argument of the form
Computer-Assisted Proofs and the Four Color Theorem
Just because a set is finite doesn’t mean it’s feasible to hand-check
every single element within it. Sometimes tasks can be automated by
writing computer programs to check thousands or even millions of
elements in a timely fashion.
While the example presented above was rather contrived, one major
result proved using the Method of Exhaustion was the Four Color
Theorem. This theorem states that any planar (basically meaning
flat) map only needs at most four different colors so that no
adjacent regions on the map share the same color. For centuries,
cartographers have been making maps of various places around the
world, and one way to draw maps is to color regions so they’re easily
distinguishable. It’s typically expected that adjacent regions on a
map are colored differently, helping them stand out visually.
It was shown at some point in the late 1800s that no more than five
colors were needed for any map so that no adjacent regions shared a
color, but it was suspected that no more than four colors were needed
as well. Though many mathematicians attempted to prove that no more
than four colors were needed, many efforts proved fruitless.
Progress marched forward in the 1960s with the advent of computing
technology. Two mathematicians, Kenneth Appel and Wolfgang Haken,
further developed results derived from Heinrich Heesch to complete a
proof. Essentially, one major step in the proof was showing that all
maps could be “reduced” to a small number of essential configurations
(basically, all maps are distortions of some simpler underlying map
that represent the same types of regions). Appel and Haken found that
there were only $1834$ such configurations.
Once Appel and Haken found these configurations, it was simply a
matter of checking each and every one of them — using the Method of
Exhaustion.
This proof was the first major proof to be verified by computer. As
such, it was a controversial result: not all mathematicians readily
accepted the proof, because it was hard and time-consuming for a human
to check.
Nevertheless, the proof has mostly withstood scrutiny, with a minor
error being corrected by Appel and Haken sometime in the late 1980s.
The point is that while the Method of Exhaustion may seem overly
simple and contrived, being restricted to only finite sets instead of
infinite sets, it’s still a valid proof technique that can come in
handy.
Set Partitions
When working with a set of related objects, we may want to split that
set up into smaller, more manageable sets.
For example, we may want to split up the integers based on parity:
even integers, and odd integers. When dealing with the real numbers,
we may want to split them up into three separate sets: the positive
real numbers, the negative real numbers, and the number $0$. One more
example may be the positive rational numbers, where we split them up
based on how big they are — we may have all of the positive rational
numbers less than $1$ in one set, and all of the positive rational
numbers greater than or equal to $1$ in the other set.
No matter how we split up a set, we’ll want to make sure that all
elements of the original set are accounted for in one of the smaller
subsets we form.
Partitioning a Set
In the introduction to this section, we mentioned three examples of
how we may split a large set into smaller subsets, making sure every
element within the original set is in one of the subsets. A
partition of a set creates these subsets while making sure that
every element is accounted for; however, the word partition also
invokes the idea of separation — no two subsets should share any
elements. The following definition of partition offers something more
precise.
PARTITION
Consider some set $A \subseteq \mathcal{U}$ along with some index set
$I$.
For each $i \in I$, let $A_i \subseteq A$ such that $A_i \neq
\emptyset$. Then the set
$$\{A_i \mid i \in I\}$$
is called a partition of $A$ if (and only if) the following two
conditions are satisfied:
Note that none of the sets in the partition can be empty. The whole
point of forming a partition is to take elements from the original set
and put them into a subset.
Example 3.10.1: Verifying a partition
Consider the following set of numbers:
$$A = \{0, 1, -1, 2, -2, 3, -3, 4, -4\}.$$
If we want to partition this set, we need to form non-empty subsets
that are all disjoint, and such that all elements in $A$ are in one of
the subsets.
Notice that we didn’t start with the index set $I = \{1, 2, 3\}$.
Instead, we formed the subsets first, and then indexed them. Of
course, we should check to see if the conditions specified in the
definition are met:
Notice that we didn’t use the letter $A$ for the names of these
subsets. The important thing is the content of the subsets themselves,
not the name given to them. As long as the two conditions are met, the
collection of sets forms a partition. We could also put every element
of $A$ into its own subset, like so:
Yet another partition we could form is just by taking the entire set
itself:
$$Z = \{0, 1, -1, 2, -2, 3, -3, 4, -4\}.$$
Since there are no other sets in this partition, it’s trivially true
that the union of all the sets forms $A$. Furthermore, since there are
no other sets in the partition, it’s trivially true that all sets in
the partition are disjoint.
Examples of Partitions
Example 3.10.3: Partitioning the integers by parity
As mentioned in the introduction, one thing we can do is split up the
integers into two subsets: one containing the even integers, and a
second subset containing the odd integers.
Here, we didn’t use the same letter for the subset names, nor did we
use indices. Again, the names of the subsets aren’t important. What’s
important is knowing that all subsets are mutually disjoint, and that
all elements from the original set are accounted for.
Example 3.10.4: Partitioning the real numbers
One way to partition the real numbers $\mathbb{R}$ is into three
subsets: the subset of all positive real numbers greater than $0$, the
subset of all negative real numbers less than $0$, and finally the
subset containing just the number $0$.
\[
\begin{array}{rl}
X &= \{x \in \mathbb{R} \mid x < 0\} \\
Y &= \{0\} \\
Z &= \{x \in \mathbb{R} \mid x > 0\}
\end{array}
\]
Some numbers that are in $X$ include $-\pi$, $-12.3372$, and
$-\sqrt{2}$.
Some numbers in $Z$ include $1.2345$, $12$, and $\pi \cdot \pi$.
Example 3.10.5: Partitioning the positive rational numbers
As one more example, we could partition the positive rational numbers
into two separate subsets.
Previously, we talked about a proof technique where we examine every
single element of a finite set. In a sense, we examined multiple
cases, and verified some result for each of those cases. An exhaustive
proof is a special kind of proof by casework.
Sometimes, we may have trouble demonstrating that some result holds
for a set of elements. However, if we partition the set into groups
that have their own special traits, and those traits make it easy to
show the desired result holds, a proof by casework can be used to
establish the desired result for the entire set.
The Underlying Argument
Suppose we were trying to prove a statement of the form
Thus, because $p = p_1 \lor p_2 \lor p_3 \lor \dots \lor p_n$, we
would also have that $p \to c$ by extension.
This proof technique is simply an extension of the Rule of Proof by
Cases:
\[
\begin{array}{l}
p \to r \\
q \to r \\
\hline
\therefore (p \lor q) \to r
\end{array}
\]
as presented in Chapter 2, Section 6. Even though we could have
discussed this proof technique in Chapter 2, we decided to hold off
until partitions were discussed, because a lot of casework is done by
considering elements from disjoint sets. However, when doing casework,
it’s not important that elements from the universe of discourse are in
only one subset of a partition.
Whether we form a partition of the universe, or we split the universe
up into overlapping subsets, isn’t important. What’s important is
that, once we have our desired subsets, every element is represented
by at least one of those subsets.
Picking Arbitrary Elements from a Partition
When we partition a set, we split it into disjoint subsets where every
element is accounted for in one of the subsets. We don’t have to form
these subsets arbitrarily. Instead, we could take all elements from
the original set that have some common property, and stick those
elements into a subset. We can repeat this process until all elements
are accounted for.
Then, from each of these subsets, we choose an arbitrary element.
Then, whatever is true of that element will be true for all elements
within that subset. In essence, we’re using the principles of
Universal Specification and Universal Generalization, along with a
partition, to prove a theorem.
Our first theorem will be a result about absolute values of real
numbers. As a reminder, here’s the definition of absolute value of a
real number.
ABSOLUTE VALUE
For any real number $x$, the absolute value of $x$, commonly
denoted $|x|$, is the non-negative real number given by the following:
\[
\begin{array}{rl}
|x| = x & \text{when } x \geq 0 \\
|x| = -x & \text{when } x < 0
\end{array}
\]
Another way to think of the absolute value of a number is as its
distance from $0$.
Theorem 3.11.1: The absolute value of a product
For any two real numbers $x$ and $y$, we have that
$$|xy| = |x||y|.$$
Proof 3.11.1
General Strategy: Because the absolute value changes based on
whether the given number is negative or non-negative, we can consider
combinations of negative and non-negative numbers for $x$ and $y$
respectively.
We can start by partitioning the real numbers into two subsets: one
subset will be the real numbers greater than or equal to $0$, and the
other subset will be the real numbers less than $0$.
\[
\begin{array}{rl}
\mathbb{R}^+ &= \{x \mid x \geq 0\} \\
\mathbb{R}^- &= \{x \mid x < 0\}
\end{array}
\]
Notice that we have two variables, $x$ and $y$. Because of how we
partitioned the real numbers, we can pick values for $x$ and $y$ from
either $\mathbb{R}^+$ or $\mathbb{R}^-$. This gives us four cases to
check.
Case 1: $x \in \mathbb{R}^+$, $y \in \mathbb{R}^+$.
Because both $x \geq 0$ and $y \geq 0$, we have that $xy \geq 0$. We
also have that $|x| = x$ and $|y| = y$. This tells us that
$$|xy| = xy = |x||y|$$
which is the desired result.
Case 2: $x \in \mathbb{R}^+$, $y \in \mathbb{R}^-$.
Because $y < 0$, we have that $|y| = -y$. We still have that
$|x| = x$. Additionally, we know that $xy < 0$, and so $|xy| = -xy$.
This tells us that
$$|xy| = -xy = x(-y) = |x||y|.$$
We’ve reached the same desired result.
Case 3: $x \in \mathbb{R}^-$, $y \in \mathbb{R}^+$.
This case is just a mirrored version of Case 2, and so we get the same
result.
Case 4: $x \in \mathbb{R}^-$, $y \in \mathbb{R}^-$.
Since both $x$ and $y$ are less than $0$, we have that $|x| = -x$ and
$|y| = -y$. However, notice that since both $x$ and $y$ are negative,
their product $xy$ will be positive, and so $|xy| = xy$. This gives us
the following result:
Here, we used the fact that $1 = (-1)(-1)$. We also used the
commutative property of multiplication to move around the numbers.
Conclusion. In all four cases, we achieved the result as desired.
An Example with Lots of Cases
Here, we’re going to discuss perfect squares and what kinds of numbers
can be the units digits of perfect squares. Before we discuss the next
theorem, let’s make some observations about the non-negative integers.
Example 3.11.1: Every non-negative integer has the form 10a + b
What we see is that any non-negative integer can be written as
$$10a + b$$
where $a$ is any non-negative integer and $b$ is one of $0, 1, 2, 3,
4, 5, 6, 7, 8$, or $9$.
This will be important to use in our next theorem, but for now we can
organize this observation using the following table, where each entry
gives the value of $10a + b$ for the row’s $a$ and the column’s $b$:
From the observations presented in the previous example, we see that
any non-negative integer can be written as $10a + b$. This observation
will help us in the upcoming theorem.
Theorem 3.11.2: The units digits of perfect squares
The only numbers that can occur in the units digit of a perfect square
are $0$, $1$, $4$, $5$, $6$, and $9$.
Proof 3.11.2
General Strategy: We use the observation above that any
non-negative integer can be written as $10a + b$. We square that
quantity to see what determines the units digit of any perfect square.
Any non-negative integer can be written as $10a + b$ where $a$ is any
non-negative integer ($0, 1, 2, 3, 4, 5, \dots$) and $b$ can only be
one of $0, 1, 2, 3, 4, 5, 6, 7, 8$, or $9$.
Because the quantity $10a^2 + 2ab$ is being multiplied by $10$, the
units digit of $(10a + b)^2$ will be the same as the units digit of
$b^2$. Remember that $b$ can only take on one of ten values. Thus we
could simply check each of those values and see if some pattern
emerges.
Case 1: $b = 0$. When $b = 0$, we have that $b^2 = 0$. The units
digit of $0$ is of course $0$, so the units digit of $(10a + b)^2$
will also be $0$.
Case 2: $b = 1$. When $b = 1$, we have that $b^2 = 1$. The units
digit of $1$ is of course $1$, so the units digit of $(10a + b)^2$
will also be $1$.
Case 3: $b = 2$. When $b = 2$, we have that $b^2 = 4$. The units
digit of $4$ is of course $4$, so the units digit of $(10a + b)^2$
will also be $4$.
Case 4: $b = 3$. When $b = 3$, we have that $b^2 = 9$. The units
digit of $9$ is of course $9$, so the units digit of $(10a + b)^2$
will also be $9$.
Case 5: $b = 4$. When $b = 4$, we have that $b^2 = 16$. The units
digit of $16$ is $6$, so the units digit of $(10a + b)^2$ will also be
$6$.
Case 6: $b = 5$. When $b = 5$, we have that $b^2 = 25$. The units
digit of $25$ is $5$, so the units digit of $(10a + b)^2$ will also be
$5$.
Case 7: $b = 6$. When $b = 6$, we have that $b^2 = 36$. The units
digit of $36$ is $6$, so the units digit of $(10a + b)^2$ will also be
$6$.
Case 8: $b = 7$. When $b = 7$, we have that $b^2 = 49$. The units
digit of $49$ is $9$, so the units digit of $(10a + b)^2$ will also be
$9$.
Case 9: $b = 8$. When $b = 8$, we have that $b^2 = 64$. The units
digit of $64$ is of course $4$, so the units digit of $(10a + b)^2$
will also be $4$.
Case 10: $b = 9$. When $b = 9$, we have that $b^2 = 81$. The units
digit of $81$ is $1$, so the units digit of $(10a + b)^2$ will also be
$1$.
Conclusion. After examining all ten cases, we see that the only
units digits encountered were $0, 1, 4, 5, 6$, and $9$. This
establishes the desired result.
A Common Mistake
The most common error that can occur when doing a proof by cases is
not accounting for every element in the universe being considered. It
is essential that all the cases you consider account for every
possible element.
Example 3.11.2: An incomplete case split
Consider the universe of all integers.
Whether we have a negative or positive integer, squaring that integer
yields a number that is either larger than, or equal to, the original
integer.
For example, when we start with a positive integer $n \geq 1$, we have
that
$$(n \geq 1) \Longrightarrow (n^2 \geq n).$$
Likewise, squaring any negative integer always yields a positive
integer, like so:
It may seem like the previous example demonstrates that $n^2 \geq n$
for all integers. After all, we showed it was true for positive
integers and negative integers. However, there’s one integer that’s
neither positive nor negative, namely $0$. In this situation, since
it’s only one integer left unaccounted for, we could simply compute
$0^2$ and compare it to $0$ to determine that the above statement is
indeed true for all integers. We formalize this proof now.
Theorem 3.11.3: Every integer is at most its own square
For any integer $n$, we have that
$$n \leq n^2.$$
Proof 3.11.3
General Strategy: Here, we’ll partition the integers into three
subsets. We’ll consider negative integers, positive integers, and $0$.
Case 1: $n < 0$.
All square numbers are greater than or equal to $0$, meaning
$n^2 \geq 0$.
Since $n < 0$ and $n^2 \geq 0$, we necessarily have that
$n \leq n^2$ as desired.
Case 2: $n = 0$.
Here, since $0^2 = 0$, we also have that $0 \leq 0^2$. This gives us
that $n \leq n^2$ as desired.
Case 3: $n > 0$.
Since $n$ is an integer $> 0$, we have that $n \geq 1$. Multiplying
both sides by $n$ gives us that $n^2 \geq n$ as desired.
Conclusion. All integers are accounted for with the three cases
presented above. Hence we have the desired result.
Russell's Paradox and a Formal Resolution
Sets are a powerful abstraction that allows us to collect a wide
variety of objects into one structure. Typically, all objects within a
set share some common characteristic other than mere inclusion in the
set. They may be points on the plane, equilateral triangles, even
numbers, or even fruits.
As powerful as sets are, we haven’t given a formal definition of what
a set is. We’ve been relying on an intuitive definition, and though it
works well enough for our purposes, sooner or later the cracks start
to show. Sometime around 1901, a mathematician named Bertrand Russell
formulated his now-infamous paradox that seemed to dismantle the
entire theory of sets. The scary thing about the paradox is that a lot
of mathematical research and important results rested on the
foundations of Set Theory. If Set Theory is wrong, are all results
depending on sets wrong as well?
In this section, we’ll eke out what this paradox is, and hint at a
solution that solves the problem.
Sets That Are Members of Other Sets
Over the course of this chapter, we’ve defined sets that mostly have
numbers, though we’ve constructed other more intricate sets. Sometimes
we’ve constructed sets that contain other sets.
Example 3.12.1: A set containing another set
Let’s consider two different sets:
\[
\begin{array}{rl}
A &= \{1, 2, 3\} \\
B &= \{1, 2, 3, \{1, 2, 3\}\}
\end{array}
\]
Here, we can see that $B$ contains the set $\{1, 2, 3\}$ as an
element, which is something we’ve seen before, but we’re going to
rewrite set $B$ to make it more obvious that it contains a different
set:
$$B = \{1, 2, 3, A\}.$$
We’ve got a handle on this situation, but let’s ask a more interesting
question:
Can a Set Contain Itself?
What would it look like for a set to contain itself? Let’s see if we
can make such a set.
Example 3.12.2: Trying to build a set that contains itself
Let’s consider the set
$$\boxtimes = \{1\}.$$
Does $\boxtimes$ contain the set $\{1\}$ as an element? Of course not!
Right now, $\boxtimes$ only contains a single element, which is a
number.
Let’s redefine $\boxtimes$ to contain the set $\{1\}$ as well:
$$\boxtimes = \{1, \{1\}\}.$$
Does $\boxtimes$ contain itself now? Not quite, because now
$\boxtimes$ contains two elements, one of which is a number, and the
other one being a set. Instead the set $\{1, \{1\}, \{1, \{1\}\}\}$
contains $1$, $\{1\}$, and $\boxtimes$, but this set isn’t the same
thing as $\{1, \{1\}\}$, which is what we’ve defined $\boxtimes$ to
be.
Let’s try again, this time including the set $\{1, \{1\}\}$:
aren’t the same. We see that the set on the right could be written as
$\{1, \{1\}, \{1, \{1\}\}, \boxtimes\}$.
It seems that no matter how many iterations we perform, we never quite
get a set that contains itself. The only thing we may be able to do is
to think about what would happen if we continue this process
indefinitely:
Does this version of $\boxtimes$ contain itself as a member? This is a
harder question to ask, because we’re now dealing with an infinite
set.
Determining if a set is contained in another set when both sets
contain infinitely many elements takes some care, so we’ll sidestep
that issue by considering the following set:
$$\boxtimes = \{S \mid (S \text{ is a set}) \land (S \notin S)\}.$$
What kind of set is $\boxtimes$? It’s a set whose elements are sets.
But these are special sets. In the previous example, we were concerned
with what a set that contains itself would look like. Here, we don’t
scrutinize whether a given set contains itself. Instead we’re just
assuming that we’ve somehow curated every set, and are only including
sets that don’t contain themselves.
Now consider the following question: is $\boxtimes$ contained within
$\boxtimes$? Remember that for propositions such as
$\boxtimes \in \boxtimes$, there are only two possibilities: either
$\boxtimes \in \boxtimes = 0$ or $\boxtimes \in \boxtimes = 1$. Let’s
consider both possibilities.
Case 1: $\boxtimes \in \boxtimes = 0$.
Here, we’re considering the possibility where $\boxtimes$ doesn’t
contain itself. Well, since $\boxtimes \notin \boxtimes$, we know that
\[
\begin{array}{lll}
& \boldsymbol{\boxtimes \notin \boxtimes} & \textbf{Reason} \\
\Longrightarrow & \neg((\boxtimes \text{ is a set}) \land (\boxtimes \notin \boxtimes)) & \text{Definition of } \boxtimes \\
\Longrightarrow & \neg(\boxtimes \text{ is a set}) \lor \neg(\boxtimes \notin \boxtimes) & \text{DeMorgan's Laws} \\
\Longrightarrow & (\boxtimes \text{ is not a set}) \lor (\boxtimes \in \boxtimes) & \text{Law of Double Negation}
\end{array}
\]
However, notice what the logical implications are if
$\boxtimes \notin \boxtimes$: either $\boxtimes$ is not a set (it is,
so this would be a contradiction) or $\boxtimes \in \boxtimes$ (which
contradicts the supposition that $\boxtimes \notin \boxtimes$). Either
way, the proposition $\boxtimes \notin \boxtimes$ yields
contradictions. As such, we may suspect that
$\boxtimes \in \boxtimes = 1$. Let’s check.
Case 2: $\boxtimes \in \boxtimes = 1$.
Suppose we somehow knew that $\boxtimes \in \boxtimes$.
So, by assuming $\boxtimes \in \boxtimes$, the logical implication is
that $\boxtimes \notin \boxtimes$? This is a clear contradiction.
At this point, both cases yield contradictions. This could be a
problem, but solutions do exist.
Resolving the Paradox
How come we haven’t had this problem dealing with any of the
previously defined sets in this chapter? It has to do with the fact
that we were constructing sets with particular objects. Even numbers
are easy to grasp, quadrilaterals are easy to grasp, but abstract set
requirements aren’t so easy to grasp. We even had trouble trying to
come up with a set that contained itself in the second example above.
While not a resolution to the paradox, defining sets using
easy-to-grasp elements does at least avoid the issue.
Perhaps the most commonly used formal system of Set Theory is ZFC
Set Theory. ZFC is short for Zermelo–Fraenkel–Choice, where Zermelo
and Fraenkel are the two mathematicians who formulated this type of
Set Theory, and the word choice refers to the Axiom of Choice, which
is something that will be examined later in this book.
ZFC Set Theory came around in the early 20th century, and is free of
paradoxes like Russell’s Paradox. ZFC Set Theory is an intricate
system, and requires some more advanced mathematics to understand.
While it isn’t discussed in this book, further reading on ZFC Set
Theory and on the history of Russell’s Paradox is easy to find, for
anyone curious enough to look into how it inspired the search for more
axiomatic systems of Set Theory.