# Cross product of a set is compact implies constituent set is compact

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.

• Aren't the projections $\pi_b : \prod_{a\in A}X_a \to X_b$ continuous surjections? – MPW Oct 13 '16 at 13:19
• You will need to assume $X,Y$ are nonempty. The projection $\varnothing \times [0,1] \to [0,1]$ may not be surjective... – GEdgar Oct 13 '16 at 13:26
• @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. – MPW Oct 13 '16 at 13:33
• Thus, my comment is to the OP, not MPW. – GEdgar Oct 13 '16 at 13:37

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 $Y$.
• 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 $\{U_{x} : x \in X\}$. – Andrew D. Hwang Oct 13 '16 at 14:00
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…
• @Mathematicing $\pi_X(x,y)=x$ – egreg Oct 13 '16 at 14:49