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Let $G_1$ be a non-compact topological space and let $G_2$ be a generic topological space. What are the compact sets in the product $G_1\times G_2$? Surely we can take the sets of the form $K_1\times K_2$ where $K_1$ and $K_2$ are respectively compact sets of $G_1$ and $G_2$. Is this right? What are the others?

gebruiker
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Richard
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    There are far, far more than that. In $\mathbb{R}^2$, every closed and bounded set is compact, and that is very different from a product of compact sets – Pedro Sánchez Terraf Feb 08 '16 at 16:10
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    To explain a bit on the previous comment: you cannot write the closed circle in R^2 as a product of compact sets in R. – DBS Feb 08 '16 at 16:14
  • @PedroSánchezTerraf but at the same time every open of a finite product spaces is the product of open sets right? – Richard Feb 08 '16 at 16:16
  • What is "open of a finite product spaces"? Do you mean "open sets in a ..."? – BigbearZzz Feb 08 '16 at 16:29
  • @BigbearZzz yes, sorry – Richard Feb 08 '16 at 16:30
  • No, the remark by DBS above also applies to open circles. They're not the product of opens. The products of open sets are a base of the product topology. – Pedro Sánchez Terraf Feb 08 '16 at 16:57
  • @PedroSánchezTerraf Last thing: so, in general, a subset of the cartesian product of two sets, say the product is $A\times B$, has not the form $A'\times B'$ where $A'\subseteq A$ and $B'\subseteq B$ right? thank you – Richard Feb 08 '16 at 17:07
  • Indeed, not even for finite $A$ and $B$. Take this with a grain of salt, but it very unlikely that you manage product spaces before clearing this up. (For instance, by drawing several pictures.) – Pedro Sánchez Terraf Feb 08 '16 at 18:52

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Supppose $Q$ is compact in $G_1\times G_2$. then $\pi_1(Q)$ is compact, $\pi_2(Q)$ is compact (the $\pi_i$ are the canonical projections), and $$Q\subseteq(\pi_1(Q)\times \pi_2(Q)).$$ It is also a closed subset of these provided the factor spaces are Hausdorff.

Conversely a closed subset of a product of two compact spaces is compact. So you have this characterizatoin. $Q$ is compact in $G_1\times G_2$ if and only if there exist $K_1$ comapact in $G_1$ and $K_2$ compact in $G_2$ so $Q\subseteq K_1\times K_2$ and $Q$ is closed therein.

John Griffin
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ncmathsadist
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    Not sure, if I'm missing something, but you explicitly state that $Q$ is closed "therein", but isn't that equivalent to $Q$ being closed in $G_1\times G_2$ (since $K_1\times K_2$ is closed (in $G_1\times G_2$))? – 0xbadf00d Jun 20 '18 at 18:46
  • @0xbadf00d What are you trying to say? – John Mars Feb 17 '21 at 17:56
  • Does this show $G_1$ and $G_2$ are compact? –  Dec 05 '21 at 17:30