Proving a set exists or not

Question

I am struggling to comprehend how to prove the difference between a set existing but containing no elements (the empty set) and a set not existing. I understand the proof that the set of all sets does not exist, but when it comes to proving whether a specific set exists I do not know how to show the difference. For example how do I prove whether the set { x : $\forall$ y (x $\in$ y) } where x is a set exists or not?

Answer 1

The difference comes down to the axioms of set theory. Any mathematical system will start from a certain list of axioms, from which everything else is derived. The Peano axioms, for instance, define the natural numbers as used in number theory. And the most common (in my experience) axioms for set theory are called ZFC: the axioms proposed by Zermelo and Fraenkel, including the Axiom of Choice.

These axioms define precisely what is a set. Anything which the axioms don't cover, isn't a set. For example, the ZFC axioms don't say anything about negative numbers, so negative numbers aren't sets. Similarly, they don't say anything about penguins, so penguins also aren't sets.

More subtly, one of the ZFC axioms is the Axiom of Regularity: every non-empty set $\alpha$ contains an element $\beta$ such that $\alpha \cap \beta = \varnothing$. This is the axiom which kills Russell's Paradox, since it means no set can contain itself.

Proving that a given set "exists" (can be formed by following the axioms of ZFC) is a rather complicated process. You can either start from the axioms and build your set from there, in which case you've proven it exists. Or you can show that your set violates one of the axioms, in which case you've proven that it doesn't exist. But in most cases you'll encounter in everyday life, you can assume that all the sets you come across must exist. "Sets that don't exist" (that is, things that look like sets but can't be constructed within ZFC) are the rare exception, not the rule.

EDIT: I was asked how regularity implies that no set can contain itself, and the proof is short and sweet so I'm including it here. Let $S$ be an arbitrary set. By the axioms of ZFC, $\{S\}$ is a non-empty set. By regularity, $\{S\} \cap S = \varnothing$. Thus $S \notin S$.