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In Rudin's Functional Analysis, page $10,$ he stated the following separation theorem for topological vector space.

Theorem $1.10:$ Suppose $K$ and $C$ are subsets of a topological vector space $X,$ $K$ is compact, $C$ is closed, and $K\cap C = \emptyset.$ Then $0$ has a neighbourhood $V$ such that $$(K+V) \cap (K+C) = \emptyset.$$

Note that $$K+V = \bigcup_{x\in K}(x+V).$$

I am interested whether one would obtain the same conclusion if we assume that $K$ is closed instead of compact. More precisely,

Question: If $K$ and $C$ are closed subsets of a topological vector space $X$ such that $K\cap C = \emptyset,$ does there exist a neighbourhood $V$ of $0$ such that $$(K+V) \cap (C+V) = \emptyset?$$

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    $\begingroup$ What if in $\mathbb{R}$ you have $K = \bigcup_{n=1}^\infty [\sqrt{4n}, \sqrt{4n+1}]$ and $C = \bigcup_{n=1}^\infty [\sqrt{4n+2}, \sqrt{4n+3}]$? $\endgroup$ – Daniel Schepler Nov 14 '18 at 0:32
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No. Consider $\mathbb{R}^2$ with two closed subsets $$ K=\{y=0\},\quad C=\{xy=1\} $$ then $d(\overbrace{(n,0)}^{\in K},\overbrace{(n,n^{-1})}^{\in C})=n^{-1}\to 0$ as $n\to\infty$, so there cannot be any $\varepsilon>0$ such that $K+B_\varepsilon$ and $C+B_\varepsilon$ are disjoint.

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Let $K=\{2,3,\cdots\}$ and $C=\{n+\frac 1 n:n\geq 2\}$. Then there is no such neighborhood. [If $\epsilon >0$ then $n \in (K+(-\epsilon, \epsilon)) \cap (C+(-\epsilon, \epsilon))$ for all $n$ sufficiently large].

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