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Let $X$ be a TVS and $A\subseteq X$. Then it is known that for any open set $B$ in $X$, the set $A+B$ is also open. In particular, the sum of two open sets is again open. In Rudin's book in Functional Analysis, he pointed out that $A+B$ is closed whenever $A$ is compact and $B$ is closed. Of course, it is known that if $A$ and $B$ are closed then $A+B$ may fail to be closed. Am I right to say that

$A\times B$ is compact

whenever $A$ is compact and $B$ is closed? Assuming that $A\times B$ is compact, then by continuity of vector addition, it follows that the image of $A\times B$ under vector addition, which is $A+B$, is compact in $X$. It is known that every compact subset of aHausdorff topological space is necessarily closed. The fact that TVS is always Hausdorff, the result follows.

My worry right now is whether the question I raised is true or not. Any hint or solution is very much appreciated.

juniven

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What about $A = [0,1]$ and $B= \mathbb{R}$ ? Is $ A \times B$ compact in $\mathbb{R}^2$ ? –  Ahriman Oct 16 '12 at 14:16
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@Ahriman you should post this as solution –  Norbert Oct 16 '12 at 14:18

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up vote 2 down vote accepted

The fact that $A$ is closed and $B$ is compact doesn't necesarily implies that $A \times B$ is compact, as shows the particular case $A = [0,1]$ and $B= \mathbb{R}$, where $A \times B$ is a non-compact subset of $\mathbb{R}^2$. Davide Giraudo completely characterized the situation where $A \times B$ is compact.

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If $A$ is compact and non-empty, and $B\subset X$, then $A\times B$ is compact if and only if so is $B$. To see that, for a direction use the fact that the product of two compacts is compact, and for the other direction, that the range by a continuous map of a compact set is compact (here the continuous map will be the projectin with respect to the second coordinate).

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Is the empty set compact? If so, then $\varnothing \times B$ can be compact when $B$ itself is not. –  GEdgar Oct 16 '12 at 14:37
    
@GEdgar You are right, I missed this case. –  Davide Giraudo Oct 16 '12 at 14:43
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For reasons such as this, some strange mathematicians want to use the convention that $\varnothing$ is not compact. But to me that seems very strange. –  GEdgar Oct 16 '12 at 15:28
    
I agree, as the definition with covers is perfectly fitted by $\emptyset$. –  Davide Giraudo Oct 16 '12 at 15:34
    
Sort of off-topic, but interesting. Here is a consequence of compact: every chain of nonempty closed sets has nonempty intersection. Of course the intersection of the empty chain of nonempty closed sets has the whole space as its intersection. Not a problem unless the whole space is itself empty... –  GEdgar Oct 16 '12 at 15:43

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