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Let T_{x}and T_{y}be topologies on X and Y, respectively. Is T = { A × B : A[itex]\in[/itex]T_{x}, B[itex]\in[/itex]T_{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. (∅, ∅)[itex]\in[/itex]T and (X × Y)[itex]\in[/itex]T

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 ∅[itex]\in[/itex]A and ∅[itex]\in[/itex]B. I think this is true because the empty set is in all sets.

I'm not sure how to approach proving that X[itex]\in[/itex]A as even though A[itex]\in[/itex]T_{x}, this implies that A[itex]\in[/itex]X or A is X. I'm not sure how continue from here. Same with Y[itex]\in[/itex]B.

For ii. I think that since T_{x}and T_{y}are topologies themselves, they are closed under finite intersections, and since A[itex]\in[/itex]T_{x}and B[itex]\in[/itex]T_{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.

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# Proof: Topology of subsets on a Cartesian product

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