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Suppose X and Y are topological spaces. If the product topology on X × Y is the discrete topology, prove that the topologies on X and Y are both discrete.

I am really struggling to get on the right track here. I know that it is sufficient to show that every singleton is in X and Y, since there is a theorem saying that a topology is discrete iff the topology contains every singleton. However, I am having trouble showing that this is true

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Take any singleton in the product topology. It‘s open in the discrete topology. What does that say about the singleton using the definition of the product topology? (How is it produced and why does this help?)

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  • $\begingroup$ Is the following sufficient? Let (x,y) be a point in XxY. Then {(x,y)} is open in XxY on the discrete topology. By def of product topology, {(x,y)} must be the product of open sets in X and Y. Since {(x,y)} = {x} x {y}, {x} is open in X and {y} is open in y. Thus, all singleton sets are open in X and Y. Therefore, X and Y have the discrete topology. $\endgroup$ – user710492 Oct 8 '19 at 3:23
  • $\begingroup$ Great! Only the step “by def of the product topology...” could be a bit more precise, though I don‘t know your definition of the product topology. (And I‘ll trust you that you know why that step is true.) Other than that, it‘s perfect, well done. $\endgroup$ – Qi Zhu Oct 8 '19 at 3:26
  • $\begingroup$ We defined the product topology as a basis. What might be a better way to say this? $\endgroup$ – user710492 Oct 8 '19 at 3:27
  • $\begingroup$ I think you mean you defined the topology using the basis. Basically just that the singleton is the union of these basis elements from which one can see that the components must be open as well. $\endgroup$ – Qi Zhu Oct 8 '19 at 3:32
  • $\begingroup$ Thank you for your help $\endgroup$ – user710492 Oct 8 '19 at 3:41
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One way: the projections $\pi_X: X \times Y \to X$ and $\pi_Y: X \times Y \to Y$ are open and continuous maps in the product topology.

If $A \subseteq X$ then $A$ is open as $A=\pi_X[A \times Y]$ and $A \times Y$ is open in $X \times Y$ as the space is assumed to be discrete. So $X$ is discrete. Use $\pi_Y$ to show $Y$ is discrete too.

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  • $\begingroup$ I feel like the main step in this problem is to actually prove that the projection maps are open (which is of course not hard to do). $\endgroup$ – Qi Zhu Oct 8 '19 at 13:35
  • $\begingroup$ @QiZhu It's open on basic sets, so open. It's a one-liner, essentially. It's in most text books as an important example of open maps. $\endgroup$ – Henno Brandsma Oct 8 '19 at 20:16
  • $\begingroup$ Yes, I agree with you that it‘s immediate. Though it‘s also the key statement for this problem. But you‘re right, of course, a student in Topology should know this. $\endgroup$ – Qi Zhu Oct 9 '19 at 0:48

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