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Proof: Topology of subsets on a Cartesian product

by Colossus91
Tags: cartesian, product, proof, subsets, topology
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Colossus91
#1
Jan10-12, 08:22 PM
P: 1
The problem statement, all variables and given/known data

Let Tx and Ty be topologies on X and Y, respectively. Is T = { A B : A[itex]\in[/itex]Tx, B[itex]\in[/itex]Ty } 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]Tx, 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 Tx and Ty are topologies themselves, they are closed under finite intersections, and since A[itex]\in[/itex]Tx and B[itex]\in[/itex]Ty 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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Slats18
#2
Jan10-12, 11:25 PM
P: 47
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
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Jan11-12, 12:34 AM
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Quote Quote by Colossus91 View Post
i. (∅, ∅)[itex]\in[/itex]T and (X Y)[itex]\in[/itex]T
This one should say ##\emptyset\in T## and ##X\times Y\in T##.

Quote Quote by Colossus91 View Post
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.
It's not. It's a subset of all sets, but most sets don't have it as a member.

Quote Quote by Colossus91 View Post
I'm not sure how to approach proving that X[itex]\in[/itex]A as even though A[itex]\in[/itex]Tx, 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.
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. XY is obviously a cartesian product, so to prove that XY is in T, you only have to prove...what?

Quote Quote by Colossus91 View Post
For ii. I think that since Tx and Ty are topologies themselves, they are closed under finite intersections, and since A[itex]\in[/itex]Tx and B[itex]\in[/itex]Ty 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##.


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