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.
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
$$A^C = \overline{A} = \{x \in \mathcal{U} \mid x \notin A\}.$$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.
Consider a universal set $\mathcal{U}$ consisting of all the whole numbers from $0$ to $10$ inclusive, meaning
$$\mathcal{U} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.$$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:
$$(x \in A^C) \Longleftrightarrow [(x \in \mathcal{U}) \land (x \notin A)]$$Reconsider the sets mentioned in the previous example:
\[ \begin{array}{rl} \mathcal{U} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\ S &= \{0, 1, 4, 9\} \end{array} \]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
\[ \begin{array}{rl} 1.5 \notin S & \\ 1.5 \notin S^C & \end{array} \]Similarly, because $-12 \notin \mathcal{U}$, we also have that
\[ \begin{array}{rl} -12 \notin S & \\ -12 \notin S^C & \end{array} \]Union
Now that we’ve talked about what’s outside a set, we can start to consider what happens when we combine two sets into one.
The first thing we can do is take all elements from two different sets, and simply mash them all up into one set.
Consider two sets $A$ and $B$ composed of elements from some universal set $\mathcal{U}$.
The union of sets $A$ and $B$, denoted $A \cup B$, is the set
$$A \cup B = \{x \in \mathcal{U} \mid (x \in A) \lor (x \in B)\}.$$As long as an element is in $A$ or $B$, or even both $A$ and $B$, it will be included in the union of $A$ and $B$.
Let’s consider some universal set, which we’ll refer to as $\mathcal{N}$, consisting of the whole numbers from $0$ to $100$ inclusive, meaning
$$\mathcal{N} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\}.$$Now consider the following sets:
\[ \begin{array}{rl} S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ C &= \{0, 1, 8, 27, 64\} \end{array} \]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
$$S \cup C = \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 0, 1, 8, 27, 64\}.$$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:
\[ \begin{array}{llll} \boldsymbol{S \cup C} & = & \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 0, 1, 8, 27, 64\} & \\ & = & \{0, 0, 1, 1, 4, 8, 9, 16, 25, 27, 36, 49, 64, 64, 81, 100\} & \text{Ordering the elements} \\ & = & \{0, 1, 4, 8, 9, 16, 25, 27, 36, 49, 64, 81, 100\} & \text{Eliminating duplicates} \end{array} \]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.
Consider the sets
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ Q &= \{0, 1, 16, 81\} \end{array} \]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
$$S \cup Q = \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\}$$and $S$.
Intersection
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.
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:
\[ \begin{array}{rl} \cup \text{ pairs with } &\lor \\ \cap \text{ pairs with } &\land \end{array} \]Going back to the example above, we have the following sets:
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ C &= \{0, 1, 8, 27, 64\} \end{array} \]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.
Reconsider the sets
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ X &= \{1, 9, 25, 49, 81\} \\ Y &= \{0, 8, 64\} \end{array} \]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.
Consider the sets
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ Q &= \{0, 1, 16, 81\} \end{array} \]We see that
$$S \cap Q = \{0, 1, 16, 81\} = Q.$$Difference
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.
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 $A - B$:
$$(x \in A - B) \Longleftrightarrow ((x \in A) \land (x \notin B)).$$Reconsider the sets
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ C &= \{0, 1, 8, 27, 64\} \end{array} \]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 $\emptyset$.
Consider the sets
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ E &= \{0, 4, 16, 36, 64, 100\} \end{array} \]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 $A$.
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:
$$(x \in A \triangle B) \Longleftrightarrow [((x \in A) \land (x \notin B)) \lor ((x \notin A) \land (x \in B))].$$Reconsider the sets
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ C &= \{0, 1, 8, 27, 64\} \end{array} \]Is $8 \in S \triangle C$? Let’s check the necessary conditions:
\[ \begin{array}{rl} 8 \in S = 0 & 8 \notin S = 1 \\ 8 \in C = 1 & 8 \notin C = 0 \\ (8 \in S) \land (8 \notin C) = 0 \land 0 &= 0 \\ (8 \notin S) \land (8 \in C) = 1 \land 1 &= 1 \end{array} \]\[ ((8 \in S) \land (8 \notin C)) \lor ((8 \notin S) \land (8 \in C)) = 0 \lor 1 = 1 \]Because this comes out to $1$, we have that $8 \in S \triangle C$ as desired.
We can ask the same thing of $64$. Let’s run through the calculations again, this time using $x = 64$:
\[ \begin{array}{rl} 64 \in S = 1 & 64 \notin S = 0 \\ 64 \in C = 1 & 64 \notin C = 0 \\ (64 \in S) \land (64 \notin C) = 1 \land 0 &= 0 \\ (64 \notin S) \land (64 \in C) = 0 \land 1 &= 0 \end{array} \]\[ ((64 \in S) \land (64 \notin C)) \lor ((64 \notin S) \land (64 \in C)) = 0 \lor 0 = 0. \]Because this comes out to $0$, we have that $64 \notin S \triangle C$.
Examining all elements in $\mathcal{N}$, we see that
$$S \triangle C = \{4, 8, 9, 16, 25, 27, 36, 49, 81, 100\}.$$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.
Consider the sets
\[ \begin{array}{rl} \mathcal{N} &= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots, 98, 99, 100\} \\ S &= \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \\ E &= \{0, 4, 16, 36, 64, 100\} \end{array} \]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.
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).
Consider the universe $\mathbb{Z}$ of all whole numbers, including $0$ and all negative whole numbers:
$$\mathbb{Z} = \{0, 1, -1, 2, -2, 3, -3, 4, -4, \dots\}.$$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:
\[ \begin{array}{rl} A_1 &= \{2, 4, 6, 8, 10\} \\ A_2 &= \{-1, 0, 1, 2, 3, 4, 5\} \\ A_3 &= \{1, 2, 7, 8\} \\ A_4 &= \{1, 3, 4\} \\ A_5 &= \{-5, -12, -13\} \end{array} \]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.
Reconsider the set $\mathbb{Z} = \{0, 1, -1, 2, -2, 3, -3, 4, -4, \dots\}$. Now consider the index set
$$I = \{a, b, c, d, e\}.$$Then we could have the following sets:
\[ \begin{array}{rl} A_a &= \{2, 4, 6, 8, 10\} \\ A_b &= \{-1, 0, 1, 2, 3, 4, 5\} \\ A_c &= \{1, 2, 7, 8\} \\ A_d &= \{1, 3, 4\} \\ A_e &= \{-5, -12, -13\} \end{array} \]where $A_a, A_b, A_c, A_d, A_e \subseteq \mathbb{Z}$, but this time, $I \not\subseteq \mathbb{Z}$.
Once again, reconsider the set $\mathbb{Z} = \{0, 1, -1, 2, -2, 3, -3, 4, -4, \dots\}$. We can make the index set anything we want, such as
$$I = \{1, a, x, \boxminus, \star\}.$$Then we have
\[ \begin{array}{rl} A_1 &= \{2, 4, 6, 8, 10\} \\ A_a &= \{-1, 0, 1, 2, 3, 4, 5\} \\ A_x &= \{1, 2, 7, 8\} \\ A_\boxminus &= \{1, 3, 4\} \\ A_\star &= \{-5, -12, -13\} \end{array} \]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.
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:
$$(x \in \bigcup_{i \in I} A_i) \Longleftrightarrow (\exists i \in I\ [x \in A_i]).$$Of course, if the index set uses whole numbers starting from $1$ and ending at $n$, we can instead write this logical equivalency as
$$(x \in \bigcup_{i=1}^{n} A_i) \Longleftrightarrow (\exists i \in \{1, \dots, n\}\ [x \in A_i]).$$Once again, reconsider the set $\mathbb{Z} = \{0, 1, -1, 2, -2, 3, -3, 4, -4, \dots\}$, with the following index set:
$$I = \{1, 2, 3, 4\},$$along with the following sets:
\[ \begin{array}{rl} A_1 &= \{2, 4, 6, 8, 10\} \\ A_2 &= \{-1, 0, 1, 2, 3, 4, 5\} \\ A_3 &= \{1, 2, 4, 7, 8\} \\ A_4 &= \{1, 3, 4\} \end{array} \]Is $8$ in the multiple union of $A_1$, $A_2$, $A_3$, and $A_4$? We can check to see if $8$ is in at least one of them:
$$(8 \in A_1) \lor (8 \in A_2) \lor (8 \in A_3) \lor (8 \in A_4) = 1 \lor 0 \lor 1 \lor 0 = 1$$As such, we see that $8$ is indeed in the multiple union of $A_1$, $A_2$, $A_3$, and $A_4$.
However, when it comes to a number like $13$, we see that
$$(13 \in A_1) \lor (13 \in A_2) \lor (13 \in A_3) \lor (13 \in A_4) = 0 \lor 0 \lor 0 \lor 0 = 0$$and so $13$ is not in the multiple union of $A_1$, $A_2$, $A_3$, and $A_4$.
We can just go ahead and determine the entire multiple union:
\[ \begin{array}{lll} \boldsymbol{A_1 \cup A_2 \cup A_3 \cup A_4} & = & \{2, 4, 6, 8, 10\} \cup \{-1, 0, 1, 2, 3, 4, 5\} \cup \{1, 2, 4, 7, 8\} \cup \{1, 3, 4\} \\ = & \{2, 4, 6, 8, 10, -1, 0, 1, 2, 3, 4, 5, 1, 2, 4, 7, 8, 1, 3, 4\} & \text{Combining all elements} \\ = & \{-1, 0, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8, 8, 10\} & \text{Ordering the elements} \\ = & \{-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 10\} & \text{Eliminating duplicates} \end{array} \]Another way we can write the set representing the multiple union is
$$\{x \in \mathcal{U} \mid \exists i \in I\ [x \in A_i]\}.$$Naturally, there’s an intersection counterpart to the 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 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:
$$(x \in \bigcap_{i \in I} A_i) \Longleftrightarrow (\forall i \in I\ [x \in A_i]).$$Of course, if the index set uses whole numbers starting from $1$ and ending at $n$, we can instead write this logical equivalency as
$$(x \in \bigcap_{i=1}^{n} A_i) \Longleftrightarrow (\forall i \in \{1, \dots, n\}\ [x \in A_i]).$$Once again, reconsider the set $\mathbb{Z} = \{0, 1, -1, 2, -2, 3, -3, 4, -4, \dots\}$, with the following index set:
$$I = \{1, 2, 3, 4\},$$along with the following sets:
\[ \begin{array}{rl} A_1 &= \{2, 4, 6, 8, 10\} \\ A_2 &= \{-1, 0, 1, 2, 3, 4, 5\} \\ A_3 &= \{1, 2, 4, 7, 8\} \\ A_4 &= \{1, 3, 4\} \end{array} \]Is $4$ in the multiple intersection of $A_1$, $A_2$, $A_3$, and $A_4$? We can check to see if $4$ is in all of them:
$$(4 \in A_1) \land (4 \in A_2) \land (4 \in A_3) \land (4 \in A_4) = 1 \land 1 \land 1 \land 1 = 1$$As such, we see that $4$ is indeed in the multiple intersection of $A_1$, $A_2$, $A_3$, and $A_4$.
However, when it comes to a number like $8$, we see that
$$(8 \in A_1) \land (8 \in A_2) \land (8 \in A_3) \land (8 \in A_4) = 1 \land 0 \land 1 \land 0 = 0$$and so $8$ is not in the multiple intersection of $A_1$, $A_2$, $A_3$, and $A_4$.
We can just go ahead and determine the entire multiple intersection:
\[ \begin{array}{llll} \boldsymbol{A_1 \cap A_2 \cap A_3 \cap A_4} & = & \{2, 4, 6, 8, 10\} \cap \{-1, 0, 1, 2, 3, 4, 5\} \cap \{1, 2, 4, 7, 8\} \cap \{1, 3, 4\} & \\ & = & \{2, 4\} \cap \{1, 2, 4, 7, 8\} \cap \{1, 3, 4\} & \text{Intersecting } A_1 \text{ and } A_2 \\ & = & \{2, 4\} \cap \{1, 3, 4\} & \text{Intersecting the result with } A_3 \\ & = & \{4\} & \text{Intersecting the result with } A_4 \end{array} \]Another way of writing the set representing the multiple intersection is
$$\{x \in \mathcal{U} \mid \forall i \in I\ [x \in A_i]\}.$$