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Question:

Let X and Y be topological spaces. Prove that if $X \times Y$ is compact with respect to the product topology then X and Y are compact.

I've been stuck on this for a hour or so and I think what I'm missing is a subtle connection.

Attempt:

Suppose that $X \times Y$ is a compact product space. This implies that there is a collection $C_{X \times Y}=\left \{ U_{\alpha}:\alpha \in J \right \}$ of open subsets $U_{\alpha}$ that is an open cover of $X \times Y$.

In fact, every open cover of $X \times Y$ contains a finite subcollection that covers $X \times Y$.

Recall:

The product topology on the product space $X \times Y$ is generated by the basis $\bar{B}=\left \{ T \times U:T \in \tau, U \in \nu \right \}=\tau \times \nu$

This implies that any element $\left ( x,y \right )$ in the basis element B of $\bar{B} \subseteq U_{\alpha}$ for any open set $U_{\alpha}$.

To show that X is compact, we need to show that there is an open cover of X that contains X and that every open cover of X contains a finite subcollection that covers X. The same applies to Y.

Any useful help to take me further is much appreciated.

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    $\begingroup$ Aren't the projections $\pi_b : \prod_{a\in A}X_a \to X_b$ continuous surjections? $\endgroup$ – MPW Oct 13 '16 at 13:19
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    $\begingroup$ You will need to assume $X,Y$ are nonempty. The projection $\varnothing \times [0,1] \to [0,1]$ may not be surjective... $\endgroup$ – GEdgar Oct 13 '16 at 13:26
  • $\begingroup$ @GEdgar: But in that case, the assertion is false. $\varnothing \times X$ is compact (since it s empty) for any $X$, whether $X$ is compact or not. So that would have to be implicit in OP's assertion if it is to be true. $\endgroup$ – MPW Oct 13 '16 at 13:33
  • $\begingroup$ Thus, my comment is to the OP, not MPW. $\endgroup$ – GEdgar Oct 13 '16 at 13:37
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To proceed using open covers: Let $\{U_{\alpha} : \alpha \in J\}$ be an open cover of $X$ (say). The family $\{U_{\alpha} \times Y : \alpha \in J\}$ is an open cover of $X \times Y$. Since $X \times Y$ is compact, there exists a finite subcover, say $\{U_{i} \times Y : i = 1, \dots, n\}$. The collection $\{U_{i} : i = 1, \dots, n\}$ is a cover of $X$. Since $\{U_{\alpha} : \alpha \in J\}$ was an arbitrary cover, $X$ is compact. Then argue similarly for $Y$.

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  • $\begingroup$ Promising and I'm leaning towards your contribution. Is it valid to assume the statement to be true for what we want to proof? I've always understand that to show that a space X is compact, we have to show first(conventionally) that the space X has an open cover. But it seems that we could actually validly assume space X has an open cover, then, via a finite subcollection of the open cover determine if space X is compact. $\endgroup$ – Mathematicing Oct 13 '16 at 13:49
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    $\begingroup$ Every topological space $X$ admits an open cover: Take $X$ itself (a cover containing one set), or use the axioms of a topology to pick, for each $x$ in $X$, an open set $U_{x}$ containing $x$, and consider the resulting family $\{U_{x} : x \in X\}$. $\endgroup$ – Andrew D. Hwang Oct 13 '16 at 14:00
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Hint: The projection map $\pi_X\colon X\times Y\to X$ is continuous and surjective; the image of a compact space under a continuous map is…

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  • $\begingroup$ Doesn't the projection map maps a set X to a set of equivalence class X*? $\endgroup$ – Mathematicing Oct 13 '16 at 13:26
  • $\begingroup$ Different use of the word "projection". $\endgroup$ – GEdgar Oct 13 '16 at 13:28
  • $\begingroup$ Which one is it then? $\endgroup$ – Mathematicing Oct 13 '16 at 13:28
  • $\begingroup$ Well, the image of a compact space under a continuous map is compact. is this just any arbitrary 'projection' map? $\endgroup$ – Mathematicing Oct 13 '16 at 13:34
  • $\begingroup$ @Mathematicing $\pi_X(x,y)=x$ $\endgroup$ – egreg Oct 13 '16 at 14:49

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