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I'm trying to prove that the empty set is unique.

Proof:

Let $U = \{ a \}$ be the universal set.

Assume $a \not\in \emptyset '$ and $a \not\in \emptyset$.

Without loss of generality, since $a \not\in \emptyset'$, $\emptyset '$ does not contain any elements. Since $\emptyset '$ does not contain any elements, it must by default be a subset of $\emptyset$, since the conditional statement

$$a \in \emptyset ' \Rightarrow \emptyset ' \subseteq \emptyset$$

is vacuously true.

Therefore, since $\emptyset' \subseteq \emptyset$ and $\emptyset \subseteq \emptyset '$, we have that $\emptyset ' = \emptyset$. $\tag*{$\blacksquare$}$

I would appreciate it if people could please provide feedback as to the correctness of my proof.

EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.

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    $\begingroup$ You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means $\endgroup$ – Hagen von Eitzen Jan 6 at 17:22
  • $\begingroup$ @HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect. $\endgroup$ – The Pointer Jan 6 at 17:25
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The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A \subseteq B$. Similarly, $B \subseteq A$. Hence $A=B$.

EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $a\notin \emptyset'$ and $\emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.

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Let $A$ and $B$ be two empty sets. Then the assertions $x\in A$ and $x\in B$ are logically equivalent. By the definition of equality of sets, $A=B$ iff $\forall x(x\in A\Longleftrightarrow x\in B)$, it follows that $A=B$.

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    $\begingroup$ This is not definition, this is an axiom. The definition will be closer to $\forall x(x\in A\Longleftrightarrow x\in B)\land \forall w(A\in w\iff B\in w)$ $\endgroup$ – Holo Jan 6 at 17:25
  • $\begingroup$ @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom $\endgroup$ – Holo Jan 6 at 17:45

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