# Does we have separation theorem for closed subsets in a topological vector space?

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 $$(K+V) \cap (C+V) = \emptyset?$$

• What if in $\mathbb{R}$ you have $K = \bigcup_{n=1}^\infty [\sqrt{4n}, \sqrt{4n+1}]$ and $C = \bigcup_{n=1}^\infty [\sqrt{4n+2}, \sqrt{4n+3}]$? – Daniel Schepler Nov 14 '18 at 0:32

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.
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].