Intersection with empty index set

[mathboy]

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?

 

[mathboy]

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.

 

[ice109]

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?

 

[mathboy]

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.

 

[Hurkyl]

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.

 

[dans595]

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

OLD:

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".

 

[Hurkyl]

The opening poster sounded like he was talking about working in a fixed universe of discourse.


Incidentally, even in ZF, you can take an empty intersection; it would simply be the (proper) class of all sets.

 

[dans595]

That's interesting. How do you define the intersection then, in ZF? The resources that I've learned from define the intersection of a set to be the set given by the comprehension axiom. Are you specifically defining the empty intersection to be the class of all sets?