Question

Suppose B is a set, {Ai | i ∈ I } is a familiy of sets, and I ≠ ∅?

Answer 1

@hartnn

Answer 2

If \(x\in B\) and \(x\in A_0\), then certainly \(x\in B\cap A_0\) for at least one \(A_0\) in the union of \(A\)'s.

Regarding your last question: why wouldn't those two statements be equivalent?

Answer 3

well because existential quantifier can not be distributed over conjunction. I said that my textbook

Answer 4

Well, I can't seem to think of an example that holds true to that, but I suppose in general, the equivalence might not hold.

Whatever the case may be, we're using the definition of set intersection, which is \(x\in A\cap B\iff (x\in A)\wedge(x\in B)\).

Answer 5

It's not the intersection that's confusing me. It's the fact that they moved existential quantifier to the front without any justification that is confusing me.

I was thinking perhaps because the set B does not depend on index i. If whether or not x is in Ai does not change the fact that x is in B.

Answer 6

The proof (which I'm assuming has been provided for you) only says that there's at least some set \(A_0\) in the union that contains \(x\). That's all we can say, that \(\exists~A_0\) for which \(x\in A_0\).

The proof does not say that \(\exists~A_0\) such that \(x\in B\); the fact that \(x\in B\) is already given (by the directional assumption).

Besides, \(\exists~A_0\) such that \((x\in B)\wedge(x\in A_0)\) doesn't really make much sense.

Answer 7

I think you're introducing the quantifier where it doesn't need to introduced; instead, you need only consider the intersection definition.

Answer 8

huhm... I does makes sense if I only consider the intersection.

Thank you

Answer 9

np!