Theorem: Let $A$ be a compact set and $B$ be a non-compact set. Then $A\times B$ is non-compact.

I know that if $B$ is non-compact, then there exists an open cover $O$ of $B$ that does not have a finite sub-collection that also covers $B$. How can I use this to construct a cover of $A\times B$ that does not have a finite sub-collection?


  • $\begingroup$ Trivial remark: you need of course to assume that $A$ is nonempty. $\endgroup$ – J.-E. Pin Mar 20 '16 at 0:12
  • $\begingroup$ @J.-E.Pin more specifically, that it has an open cover? $\endgroup$ – stochman Mar 20 '16 at 0:18
  • $\begingroup$ @stochman, huh? No. If A is empty, then the product is compact. $\endgroup$ – Mariano Suárez-Álvarez Mar 20 '16 at 0:36

HINT: If $\mathscr{U}$ is an open cover of $B$ with no finite subcover, consider $\{A\times U:U\in\mathscr{U}\}$.

  • $\begingroup$ OK, so $\mathcal{O}$ ={$A$ $\times$ $U$ : $U$ $\in$ $\mathcal{U}$} is clearly an open cover of $A \times B$. I now need to show that there is not a finite sub-collection. It seems obvious, but I'm not quite sure how to rigorously prove it. $\endgroup$ – stochman Mar 19 '16 at 23:48
  • $\begingroup$ @stochman: Suppose that $\{A\times U_1,\ldots,A\times U_n\}$ is a finite subset of $\mathcal{O}$. Show that its union is $A\times\bigcup_{k=1}^nU_k$. Then explain why this is not all of $A\times B$. $\endgroup$ – Brian M. Scott Mar 19 '16 at 23:51

The (surjective) projection map $A\times B\to B$ is continuous. So if $A\times B$ is compact, then also $B$ is compact.

More generally, a product space $\prod_{i}X_i$ is compact if and only if every factor $X_i$ is compact.

The “if” part is Tychonov's theorem, the “only if” part is the argument above.

  • $\begingroup$ contraposition was a clever idea. I wouldn't have thought of that +1 $\endgroup$ – Andres Mejia Mar 19 '16 at 23:41

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