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Show that if $E$ is closed and $K$ is compact in $\mathbb{R}^n$, then $K+E$ is closed in $\mathbb{R}^n$, where $E+K=\{x+y\,\,\, x\in E \,\,\, , y \in K \}$ is Minkowski sum.

I tried to show it by saying that $K$ is closed and bounded but I do not know how proof goes?

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Let, $a\in\overline{E+K}$ then $\exists\{a_n\}_{n\in\mathbb{N}}\in E+K$ [Existence of such sequence is assured as $\mathbb{R}^n$ is First countable w.r.t usual norm] such that $a_n\rightarrow a$ as $n\rightarrow\infty$. Then $a_n=x_n+y_n$ for some $x_n\in E$ and $y_n\in K$, $\forall n\in\mathbb{N}$. Thus $\{x_n\}_{n\in\mathbb{N}}\in E$ and $\{y_n\}_{n\in\mathbb{N}}\in K$. Since $K$ is compact so closed and bounded and $E$ is closed so $x_n\rightarrow x\in E$ and $y_n\rightarrow y\in K$ as $n\rightarrow \infty$.

So $x+y\in E+K$. Finally $||a_n-(x+y)||=||(x_n+y_n)-(x+y)||\le||x_n-x||+||y_n-y||\rightarrow0$ as $n\rightarrow\infty$. Hence $\{a_n\}_{n\in\mathbb{N}}$ converges to $x+y\in E$ so $\overline{E+K}\subset E+K$. reverse inequality is obvious. This completes the proof.

Sujit Bhattacharyya
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  • Could you tell me why we do not use boundedness? because if we say that $K$ is closed we have all the result. –  Oct 31 '18 at 17:06
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    @SujitBhattacharyya if both are only closed and one of them is not compact, the sum might not be closed https://math.stackexchange.com/questions/124130/sum-of-two-closed-sets-in-mathbb-r-is-closed] – sonu Jan 20 '19 at 13:40
  • @Sri Krishna Sahoo Thanks for the comment. Saeed I removed my last comment because it might not true and the beautiful example is given in SK sahoo's comment. Again thanks for the comment. – Sujit Bhattacharyya Jan 22 '19 at 03:56
  • "taking a subsequence" step missing, cf. here. – Noix07 May 30 '22 at 19:24