Proving that the empty set is unique

I haven't been able to figure out if the following reasoning is correct, so I'd like to have the opinion of other people on that.

The goal is to prove that the empty set is unique.

In order to do that, let $E$ be a set, $A$ be an empty set and $B$ be an empty set. I want to prove that $A = B$.

First, I can try proving that $A \subset B$.

I know that $\forall x \in E, x \notin A$. Now I can consider this proposition : $x \in A \Rightarrow x \in B$

Since $x \notin A$, I can safely say that that the proposition above is true.

Hence, I can conclude that $A \subset B$

Am I right in my conclusion?

Your conclusion is correct. You can also resort directly to the Axiom of Extension

$A=B\iff \forall z\colon(z\in A\leftrightarrow z\in B)$

For the given conditions give us $z\notin A$ and (equivalently) $z\notin B$ for arbitrary $z$, hence the right hand side in the axiom is true.

I think you might be overlooking the importance (or unaware) of the Axiom of Extensionality, which simply says two sets are equal if they have the same members. Thus two empty sets are equal by this axiom.

Added: In some variations of set theory, we may want to have more than one thing without members, and these are called ur-elements. The Axiom of Extensionality can then be modified to allow for more than one "empty set", while still allowing us to compare and identify things that have members belonging to them.

• The OP apparently uses a "lemma" ($A\subseteq B\land B\subseteq A\implies A=B$). The Axiom of Extensionality of course hides in the proof of this lemma. – Hagen von Eitzen Oct 13 '14 at 20:31

Here's an approach that's probably cleaner:

Per Axiom of Existence, there exists a set which has no elements. Let $A$ and $B$ each be a set which has no elements. Since $x \in A \iff x \in B$ is vacuously true, by Axiom of Extensionality $A=B$.

Equivalently, that $x \in A \iff x \in B$ is true can be established using the truth table of the biconditional, by recognizing that $x \in A$ and $x \in B$ are in each case false (ex falso quodlibet).