# Existence of infinite intersection

The intersection of an infinite family of sets $S_\alpha$ is defined by membership: $$(x \in \cap_{\alpha \in A} S_\alpha) \text{ if and only if } (x \in S_\alpha \text{ for all } \alpha \in A).$$

This is probably a silly question, but how do we know that this legitimately defines a set? It doesn't actually construct a single element of the intersection, it just provides a yes-or-no question that can be asked about any potential candidate for membership.

This style - definition of a set by properties of it's members - recalls Russel's classic "set of all sets that don't contain themselves", $$(x \in R) \text{ if and only if } (x \notin x)$$

Russel's set $R$ doesn't exist, though the "definition" follows the same technique.

What's different here? How do we know that the infinite intersection is a well defined concept?

I think I can show that an $\cap_{\alpha \in A} S_\alpha$ exists by transfinite induction, but well-ordering the index set $A$ requires the well ordering principle which is equivalent to the axiom of choice, whereas I've heard that intersection does not require the axiom of choice. Also, at this level its so easy to make a subtle mistake about what things even mean, so I don't really trust my proof.

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If $A = \emptyset$ you do have a problem. Otherwise, pick some $\alpha_0 \in A$ and the intersection is $\{x \in S_{\alpha_0}: x \in S_\alpha \ \text{for all}\ \alpha \in A\}$, which is OK by the axiom schema of specification.
@Nick, Robert: Actually an important remark is that $\bigcap\varnothing$ is not well-defined. The $\bigcap A$ (intersection of all the members of $A$) can be written in two equivalent-for-non-empty-families ways, but one would result in $\bigcap\varnothing=\varnothing$ and the other would result in $\bigcap\varnothing=V$ (where $V$ is the universe). – Asaf Karagila May 31 '12 at 5:44