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!
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:
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!
By the previous theorem, for the set $A = \{1, 2, 3\}$, we have that
$$\{\ \} \subseteq A.$$