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I didn't find an already existing proof for the following question:

We have two compact sets $A, B \subset \Bbb R^n$. I want to show that the addition of these is also compact.

So, I want to show $A+B:=\{a+b: a\in A, b\in B \}$ is compact.

Please help me.

bof
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  • What is the definition you know for compactness? Because if it is the finite subcover one, this shouldn't prove to be terribly hard... – Alexander Geldhof Jun 06 '20 at 12:03
  • I think it is easy with sequential compactness and harder with open cover definition. – Kavi Rama Murthy Jun 06 '20 at 12:05
  • note: I have edited the question to typeset it in MathJax. I've changed "$A, B \in \Bbb R^n$" to "$A, B \subset \Bbb R^n$" which is what I presume OP meant. – Aryaman Maithani Jun 06 '20 at 12:07
  • @KaviRamaMurthy, the sum of two compact spaces clearly has a finite subcover for every open cover, i.e. the union of two finite subcovers for each of its components. – Alexander Geldhof Jun 06 '20 at 12:08
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    @AlexanderGeldhof You are mistaking $A+B$ for $A \cup B$. Please read OP's definition of $A+B$. – Kavi Rama Murthy Jun 06 '20 at 12:10
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    But an open cover of $A+B$ isn’t necessarily even a cover of $A$ or $B$, so how do you use those to get the subcover? – Joe Jun 06 '20 at 12:11

2 Answers2

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  • The set $A\times B\subset{\mathbb R}^{2n}$ is compact.
  • The map $\quad {\rm add}: \ A\times B\to {\mathbb R}^n,\quad (a,b)\mapsto a+b\quad $ is continuous.
Christian Blatter
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Proof using sequential compactness: Suppose $a_k+b_k \to y$ where $a_k \in A$ and $b_k$ in $B$ for all $k$. There exists a subsequence $a_{k_i}$ converging to some point $a \in A$. Now look at $b_{k_i}$. This has a subsequence $b_{k_{i_l}}$ converging to some $b$ in $B$. Can you now show that $y=a+b$?

Kavi Rama Murthy
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