Proving the existence of $\emptyset$ using the Axiom of Infinity?

Question

In my Set Theory notes, one exercise is

Prove the existence of $\emptyset$ using the Axiom of Infinity.

I'm not even sure where to begin here. The Axiom of Infinity is presented to us in the form $$ \exists x \, (\exists y \quad y \in x \, \land \, \forall z \, (z \in x \to \{z\} \in x)). $$ To me this is saying there exists a non-empty set where the singleton of every element in the set is also in the set.

Now since the Axiom of Infinity only guarantees the existence of a non-empty set, how can I use it to prove the existence of an empty set?

The only thing I can think of is to fix the $x$ the Axiom guarantees exists and use Separation to obtain $\{z \in x \, | \, \{z\} \not \in x \}$. But this is using Separation, and if I were to use Separation I would just say $\emptyset= \{x \in X \, | \, x \neq x \}$ where $X$ is any old set.

If anyone could point me in the right direction, that would be very helpful! Cheers.

Once you know there is some set, then separation (as you show) lets you say that the empty set exists. But without some existence axiom, you do not know that there are any sets at all. — GEdgar, Sep 19 '17 at 00:35
I think the Existence axiom ($\exists x , (x = x)$) is assumed for every proof in my course. But I think the point of the exercise is to show $\emptyset$ exists using some clever argument about Infinity. — Tristan Batchler, Sep 19 '17 at 00:36
Of course you did not tell us the axioms you are working from, so we can only guess at how to answer your question. — GEdgar, Sep 19 '17 at 00:38
The axiom of infinity usually specified that $y$ is in fact the empty set. — Asaf Karagila, Sep 20 '17 at 05:35
This question is similar to: Axiom of Infinity implies existence of empty set. Is this circular?. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Ak3.14, Mar 08 '25 at 15:47

score 5 · Accepted Answer · answered Sep 19 '17 at 00:35

5

But this is using Separation, and if I were to use Separation I would just say $\emptyset= \{x \in X \, | \, x \neq x \}$ where $X$ is any old set.

This is the key. It works, but where are you going to get $X$? How do you know that there exists a set? That's where the exercise gives you a hint in the right direction.

answered Sep 19 '17 at 00:35

Chris Culter

27,415

Oh, I see. I think this is a little underwhelming. For example, throughout the course so far we have assumed the most basic Existence axiom ($\exists x , (x = x)$). I'm not sure why my lecturer would give us the exercise with this particular argument in mind considering we've always had that argument under out belts (there exists a set). – Tristan Batchler Sep 19 '17 at 00:41
1

I'm sure this is meant to be an exercise in "reverse" set theory. You're not investigating the theorems that you can prove using all the axioms you've been given; you're investigating which axioms you really need to prove the theorems that you want. Do you need the set existence axiom to prove the existence of the empty set? It turns out that the answer is no, provided that you have infinity and separation, as you've shown. – Chris Culter Sep 19 '17 at 00:52

Proving the existence of $\emptyset$ using the Axiom of Infinity?

1 Answers1