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Suppose $A_i$ is a compact subset of a topological space $X_i$ for all $i\in I$.

Tychonoff’s theorem says the product of compact spaces is compact. Since $A_i\subseteq X_i$ is compact, this means $A_i$ is compact as a subspace of $X_i$ and thus the product space $\prod A_i$ (with product topology) is compact.

However, I’m not sure if what I have shown is enough to say $\prod A_i$ is a compact subset of $\prod X_i$ (with product topology). Is there a better way to show this?

Partey5
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1 Answers1

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That is correct. All it still needs is the observation that the product topology on $\prod A_i$ is equal to the topology induced on $\prod A_i$ by $\prod X_i$ (and it is easy to check that this holds indeed).

José Carlos Santos
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  • I’m not sure how to show this. – Partey5 May 12 '21 at 14:52
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    A basis of the topology on $\prod X_i$ are the sets $\prod Y_i$ in which each $Y_i$ is an open subset of $X_i$ and that for all but finitely many $i$'s you have $Y_i=X_i$. So, a basis of the topology of $\prod A_i$ are the sets of the form $(\prod Y_i)\cap(\prod A_i)$ where the $Y_i$'s are as before. But these are procisely the sets of the form $\prod B_i$ in which each $B_i$ is an open subset of $A_i$ and that for all but finitely many $i$'s you have $B_i=A_i$. – José Carlos Santos May 12 '21 at 15:00
  • It's a general consequence of the transitive law of initial topologies that I discuss here. Taking subspaces, then products is just the same as the direct subspace topology on that product subset. – Henno Brandsma May 13 '21 at 13:44