Proving that the Empty Set $\emptyset$ is Unique Verification

Question

I know that this question has been asked before, but I am presenting my proof for verification. I do not want anyone to give me a proof or suggest a better one; I only want to know if the following proof is legitimate.

Let $A$ and $B$ be empty sets. Let $U$ be the universe in which $A$ and $B$ abide. Clearly, $U = \overline{A}$ and $U = \overline{B}$. However, this implies that $\overline{U} = \bar{\bar{A}} = A$ and $\overline{U} = \bar{\bar{B}} = B$. Since $\overline{U} = \overline{U}$, we conclude that $A=B$, that is, the empty set is unique.

Is the overline supposed to denote complement? That notation is very nonstandard in my experience. If so, I think this is fine. — Cameron L. Williams, Sep 25 '15 at 22:37
Yes it is. I'm guessing $A^c$ is better notation for set complementation? — J. Dunivin, Sep 25 '15 at 22:40
@Benedict Voltaire $A^c$ is a good notation when it is known what the universe is. $U\backslash A$ is even better. — Guest, Sep 25 '15 at 22:41
Guest's notation is best by far. There's no ambiguity, but $A^c$ is fair game. — Cameron L. Williams, Sep 25 '15 at 22:42

score 1 · Answer 1 · answered Sep 25 '15 at 22:56

1

Your proof is not legitimate in ZF (in which there is no "universe" in which $A$ and $B$ abide). (I would say more but you explicitly asked me not to.)

answered Sep 25 '15 at 22:56

Rob Arthan

51,538
4
53
105

I'm thinking $U$ is meant to be a fixed universe set (e.g. the reals), not the universe set. – Cameron L. Williams Sep 25 '15 at 23:07
1

@CameronWilliams: maybe, but the OP apparently wants a yes-no answer about the legitimacy of the proposed proof and no more. – Rob Arthan Sep 25 '15 at 23:13
Fair enough. I definitely didn't like the tone, myself, but that's immaterial. – Cameron L. Williams Sep 25 '15 at 23:14

score 1 · Accepted Answer · answered Sep 25 '15 at 23:06

1

As Rob Arthan alluded to in his answer, you have not really proved uniqueness of the empty set. You have shown that for any set $U$, there is at most one empty subset. But you cannot take $U$ to be "all sets" because this is not a set in ZF theory. You would be best to try to prove this without referring to any such $U$, and in particular avoiding use of complements (which require a base space to define).

answered Sep 25 '15 at 23:06

Jason

15,726

1

But the proof begins with two empty sets and every empty set is a subset of any set. So why does taking any set $U$ and concluding that $A=B$ only shows that "for any set $U$, there is at most one empty subset" instead of showing that "any two empty sets are equal"? – Guest Sep 25 '15 at 23:23
That argument is a little circular. The fact that "every empty set is a subset of any set" requires proof and indeed is the usual way to show that any two empty sets are equal. You could of course tweak the OP's argument to create a correct proof using the union axiom, but the most efficient proof of uniqueness of the empty set should not require any other axioms than the axiom of extensionality. – Jason Sep 26 '15 at 02:21

Proving that the Empty Set $\emptyset$ is Unique Verification

2 Answers2