# Intersection with empty index set

## Main Question or Discussion Point

If I is empty, and the collection of sets {A_i} is indexed by I, then the intersection of all the A_i is equal to the universal set.

Can someone explain why? Or better yet, give a proof?

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Here's my attempt at a proof:

Suppose I is empty. Let S be the universal set. Let x be in S. Then it is vacuously true that for all i in I (it is false that i is in I!), x belongs to A_i. Thus x belongs to the intersection of all the A_i. Hence the intersection of all the A_i equals S.

Correct?

And here's my proof attempt that U (A_i) = empty, if I is empty:

Suppose there exists x in U (A_i). Then there exists i in I such that x belongs A_i. But I is empty, so no such i exists. This contradiction means that no such x exists. Thus U (A_i) = empty. Correct? I can't find a proof anywhere.

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Here's my attempt at a proof:

Suppose I is empty. Let S be the universal set. Let x be in S. Then it is vacuously true that for all i in I (it is false that i is in I!), x belongs to A_i. Thus x belongs to the intersection of all the A_i. Hence the intersection of all the A_i equals S.

Correct?

And here's my proof attempt that U (A_i) = empty, if I is empty:

Suppose there exists x in U (A_i). Then there exists i in I such that x belongs A_i. But I is empty, so no such i exists. This contradiction means that no such x exists. Thus U (A_i) = empty. Correct? I can't find a proof anywhere.
a for all statement is not an implication statement. i think it doesn't work that way. what book is this from?

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Ok, let's start with U (A_i) = empty, where I is empty (I gave a proof above).

By DeMorgan's Laws,
intersection (A_i) = S - U (S-A_i) = S - empty = S.

Is this a better proof? I still think my first proof may be correct.

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Hurkyl
Staff Emeritus
Gold Member
Suppose I is empty. Let S be the universal set. Let x be in S. Then it is vacuously true that for all i in I (it is false that i is in I!), x belongs to A_i. Thus x belongs to the intersection of all the A_i. Hence the intersection of all the A_i equals S.
This looks right to me.

EDIT: I guess I assumed you were talking about ZFC... What theory are you talking about in particular?

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Well, if you define the intersection to be "the set of all things in all A_i", then intersection may not be well-defined (i.e., you're defining something that really can't exist). In particular, the comprehension axiom would require you to fix some A_j in the family {A_i} in order to be able to show that the set you call the intersection actually exists.

If you took as an axiom that "the set of all things in all A_i" exists, you'd immediately get a Russell paradox as the OP mentioned.

The book that I studied these topics from is Enderton's "Elements of Set Theory".

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Hurkyl
Staff Emeritus