Proving that the empty set is unique

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.

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Elementary Set Theory

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