Proof: Topology of subsets on a Cartesian product

[Colossus91]

Homework Statement

Let T_x and T_y be topologies on X and Y, respectively. Is T = { A × B : A\inT_x, B\inT_y } a topology on X × Y?

The attempt at a solution

I know that in order to prove T is a topology on X × Y I need to prove:
i. (∅, ∅)\inT and (X × Y)\inT
ii. T is closed under finite intersections
iii. T is closed under arbitrary unions

In order to prove (i) I would have to prove that ∅\inA and ∅\inB. I think this is true because the empty set is in all sets.
I'm not sure how to approach proving that X\inA as even though A\inT_x, this implies that A\inX or A is X. I'm not sure how continue from here. Same with Y\inB.

For ii. I think that since T_x and T_y are topologies themselves, they are closed under finite intersections, and since A\inT_x and B\inT_y then A and B are also closed under finite intersections, thus T is closed under finite intersections. I have to go more into detail with this but I just want to make sure if this is the right idea.

I think iii. could also be proved with a similar argument to the one used to prove ii.

 

[Slats18]

For (i), your topology T is the set of all open sets A x B such that A is an element of T_x, and B is an element of T_y. X is an element of T_x, as it is required to be one by the same rules we a re trying to prove, as well as the empty set, and vice versa for Y being an element on T_y. Use this fact to show that X x Y is in your topology T.

 

[Fredrik]

i. (∅, ∅)\inT and (X × Y)\inT

This one should say $\emptyset\in T$ and $X\times Y\in T$.

In order to prove (i) I would have to prove that ∅\inA and ∅\inB. I think this is true because the empty set is in all sets.

It's not. It's a subset of all sets, but most sets don't have it as a member.

I'm not sure how to approach proving that X\inA as even though A\inT_x, this implies that A\inX or A is X. I'm not sure how continue from here. Same with Y\inB.

It doesn't make sense to try to prove that X is a member of A when you haven't specified what A is.

$A\in T_x$ doesn't imply what you say it implies. It just means that A is an open subset of X.

Note that the definition of T says that T consists of of all cartesian products of two open subsets of X and Y, such that the first set is a subset of X and the second a subset of Y. To check if $\emptyset\in T$, you should ask yourself if ∅ can be expressed as a cartesian product at all. X×Y is obviously a cartesian product, so to prove that X×Y is in T, you only have to prove...what?

For ii. I think that since T_x and T_y are topologies themselves, they are closed under finite intersections, and since A\inT_x and B\inT_y then A and B are also closed under finite intersections, thus T is closed under finite intersections. I have to go more into detail with this but I just want to make sure if this is the right idea.

I don't see what A and B being closed under finite intersections have to do with anything. You need to start with a statement like "Let n be an arbitrary positive integer, and let $E_1,\dots,E_n$ be arbitrary members of T". Then you prove that $\bigcap_{k=1}^n E_k\in T$.