# 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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## 1 Answer

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.

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Ahh perfect! So then the axiom of schema of separation would apply here because we know for sure there is one set that contains all possible candidates, but not in russel's situation because there is no set of all sets. –  Nick Alger May 31 '12 at 0:48
*Typo above, what I wrote should read schema of specification not separation. –  Nick Alger May 31 '12 at 0:55
@ThomasE.: My answer said "specification", Nick's comment said "separation". –  Robert Israel May 31 '12 at 4:20
Robert Israel got it right in his answer, I typed it wrong in my comment but couldn't edit in time, thus the correction. –  Nick Alger May 31 '12 at 4:30
@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
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