# Empty set

The empty set is the set having no members. It is usually denoted as \(\emptyset\). Whatever object is considered, it can’t be a member of \(\emptyset\). It might be useful in the beginning to think about the empty set as an empty box. It has nothing inside it, but it still does exist.

Formally, the existence of the empty set is asserted by the **Empty Set Axiom**:

The empty set axiom itself does not postulate the uniqueness of \(\emptyset\). However, this fact is easy to prove using the axiom of extensionality. Consider sets \(A\) and \(B\) such that both \(\forall x : x∉A\) and \(\forall x: x∉B\). noteThat is, suppose we had two empty sets. Remember that the extensionality axiom tells us that if we can show \(\forall x : (x ∈ A \Leftrightarrow x ∈ B)\), then we may deduce that \(A=B\). In this case, for every \(x\), both parts of the statement \((x ∈ A \Leftrightarrow x ∈ B)\) are false: we have \(x \not \in A\) and \(x \not \in B\). Therefore the iff relation is true.

The existence of the empty set can be derived from the existence of any other set using the axiom schema of bounded comprehension, which states that for any formula \(\phi\) in the language of set theory, \(\forall a \exists b \forall x : x \in b \Leftrightarrow (x \in a \wedge \phi(x))\). In particular, taking \(\phi\) to be \(\bot\), the always-false formula, we have that \(\forall a \exists b \forall x : x \in b \Leftrightarrow (x \in a \wedge \bot)\). Since \(x \in b \Leftrightarrow (x \in a \wedge \bot)\) is logically equivalent to \(x \in b \Leftrightarrow \bot\) and hence to \(x \notin b\), the quantified statement is logically equivalent to \(\forall a \exists b \forall x : x \notin b\), and as soon as we have the existence of at least one set to use as \(a\), we obtain the Empty Set Axiom above.

It is worth noting that the empty set is itself a single object. One can construct a set *containing* the empty set: \(\{\emptyset\}\). \(\{\emptyset\} \not= \emptyset\), because \(\emptyset ∈ \{\emptyset\}\) but \(\emptyset ∉ \emptyset\); so the two sets have different elements and therefore cannot be equal by extensionality.
noteIn terms of the box metaphor above, \(\{\emptyset\}\) is a box, containing an empty box, whilst \(\emptyset\) is just an empty box

Another way to think about this is using cardinality. Indeed, \(|\{\emptyset\}| = 1\) (as this set contains a single element - \(\emptyset\)) and \(|\emptyset| = 0\) (as it contains no elements at all). Consequently, the two sets have different amounts of members and can not be equal.

the empty set is often used to represent the ordinal 0

Punctuation can be weird in this edit, as the author is not a native English speaker. Might need to be improved

@5hc Thanks for the edit! I made a couple of linguistic changes, and made the “uniqueness of \(\emptyset\)” a bit less compact.