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My question doesn't concern the standard-proof of the uniqueness of the empty set, which uses the lemma that the empty set is a subset of every set. My question concerns an alternate proof that I made before reading the standard one, and which now seams wrong, but I can't find my mistake. Can someone point me in the right direction?

My proof goes as follows:

Let us assume for contradiction that the empty set is not unique. This means that there is another empty set, let's call it $\varnothing'$, for which $ \varnothing' \neq \varnothing $. By the axiom of extensionality we have the following $\varnothing = \varnothing' :\iff \forall x: x \in \varnothing' \leftrightarrow x \in \varnothing $, which implies $\varnothing \neq \varnothing' \iff \exists x: \neg (x \in \varnothing' \leftrightarrow x \in \varnothing)$. This is equivalent to $ (x \in \varnothing' \land x \notin \varnothing) \lor (x \notin \varnothing' \land x \in \varnothing) $ . But this is never true, because the empty set (by definition) doesn’t contain any elements. Hence, the empty set is unique.

I have read the following question. My question doesn't only concern the correctness of my proof but also the difference between my proof and the standard one. If my proof is right, then I don't see why one should use the "standard" proof, which uses another lemma, instead of just applying the axiom of extensionality.

Marwan
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  • How does it follow from the equality of sets that there exists $x\in\emptyset'$ with $x\notin\emptyset$? – Luke Collins Jul 11 '19 at 12:13
  • Correct........ The proof relies on the Extensionality axiom. – Mauro ALLEGRANZA Jul 11 '19 at 12:13
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    Phrasing the proof as a proof by contradiction is a detour. You can just apply the axiom of extensionality directly, since if $A$ and $B$ are empty sets, then $x\in A$ and $x\in B$ are both always false and therefore surely equivalent. Since $\forall x(x\in A \leftrightarrow x \in B)$ the axiom of extensionality explicitly tells you $A=B$. – hmakholm left over Monica Jul 11 '19 at 12:16
  • @MauroALLEGRANZA you mean my proof is correct? Then why isn't this the "standard" proof. It seams way easier since it doesn't rely on other lemmas or anything. – Marwan Jul 11 '19 at 12:17
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    Why do you think that the "standard proof through the lemma that the empty set is a subset of every set" is different ? The standard proof is $A=B \text { iff } A \subseteq B \text { and } B \subseteq A$ that means (unpacking the def) : $A=B \text { iff } \forall x (x \in A \leftrightarrow x \in B)$ – Mauro ALLEGRANZA Jul 11 '19 at 12:20

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The proof is correct, but it is not immediately clear that

$$\emptyset\neq\emptyset' \implies \exists x\in\emptyset':x\notin\emptyset.$$

I would expand on why that is the case. Indeed, $\emptyset\neq\emptyset'$ means $\neg(\emptyset\subseteq\emptyset'\land\emptyset'\subseteq\emptyset)$, i.e., there exists $x\in\emptyset$ such that $x\notin\emptyset'$ or there exists $x\in\emptyset'$ such that $x\notin\emptyset$. The former is a contradiction ($x\in\emptyset$), so the latter must be the case.

Then the proof continues as you did.

It is not the "standard" proof because the statement follows immediately by the definition of equality of sets (the so-called extensionality axiom). Two sets are equal if they have the same elements, which is vacuously true if we have two empty sets $\emptyset$ and $\emptyset'$.

Luke Collins
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