# My proof that the empty set is unique

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.

• You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means – Hagen von Eitzen Jan 6 at 17:22
• @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. – The Pointer Jan 6 at 17:25

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.

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 $$A=B$$.

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