Suppose we have a topological vector space $X$ and $A, B\subset X$. We define A+B to be the set of the sums $a+b$ where $a\in A$ and $b\in B$. We should prove that also A+B is compact if A and B are compact. But the union of arbitrary compact sets isn't compact in generell. Thus: why is the statement wel true?
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1What do you know about $+\colon X\times X \to X$? – Daniel Fischer Sep 17 '13 at 21:16
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that this is continuous and the space is translation invariant – 17th Sep 17 '13 at 21:18
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The function $F(x,y): X \times X \to X$ defined by $F(x,y)=x+y$ is continuous, and $A \times B$ is compact in $X \times Y$.
The image of a compact set under an continuous function is compact.
N. S.
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@17th In topological vector spaces the addition is uniformly continuous, so we get the u.c. for free. But you are right, by compactness $F$ should become u.c when restricted to $A \times B$. uniform continuity is not needed, and as posted redundant, I only mentioned it originally cause we get it for free... – N. S. Sep 17 '13 at 21:38
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Is the Statement also true if we take bounded instead of compact?? Is every bounded closed set compact? Or only is several cases? – 17th Sep 17 '13 at 21:42
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@17th To speak of boundedness you typically need a norm. Now, if the topology on $X$ is given by a norm, and $A,B$ are bounded, by the triangle inequality $A+B$ is also bounded.... More exactly, if $|x| \leq C_1$ for all $x \in A$ and $|y | \leq C_2$ for all $y \in B$, then $|z| \leq C_1+C_2$ for all $z \in A+B$. – N. S. Sep 17 '13 at 21:47
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@17th But this is not a good approach, as sum of closed sets is not necessarily closed. See for example: http://math.stackexchange.com/questions/124130/sum-of-two-closed-sets-in-mathbb-r-is-closed – N. S. Sep 17 '13 at 21:52
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Could we solve it without using the fact that: The image of a compact set under an continuous function is compact. – Albanian_EAGLE Feb 09 '14 at 22:18