The following question is from my assignment in operator theory and I am not able to prove the assertion that is asked.
Let $X$ be a topological vector space. Show that if $F\subseteq X$ is closed and $K\subseteq X$ is compact, then the set $F+K$ is closed in $X$.
Well in my class notes, not much is said regarding the compact subsets of topological vector space. So, I am not able to think which proposition I should be using to ground my proof on.
Please help me in proving this question.
To show that $F+K$ is closed in $X$, we need to show that its complement, $(F+K)^c$, is open in $X$. Let $x\in (F+K)^c$. Then, $x\not\in F+K$, so there exists a neighborhood $U$ of $x$ such that $U\cap (F+K)=\varnothing$. This means that for all $y\in U$ and all $z\in K$, we have $y+z\not\in F$.
Since $K$ is compact, it is covered by finitely many neighborhoods, say $U_1,\dots, U_n$. Then for each $i$, we have $U_i\cap K\neq \varnothing$, and hence there exists $z_i\in K$ such that $z_i\in U_i$. It follows that $$[U\cap (F+K) \subseteq \bigcup_{i=1}^n (U\cap U_i) - z_i\subseteq \bigcup_{i=1}^n (U\cap U_i) - K.]$$ Since $U\cap (F+K)=\varnothing$, it follows that $\bigcup_{i=1}^n (U\cap U_i) - K=\varnothing$. But each set $U\cap U_i$ is open, so $\bigcup_{i=1}^n (U\cap U_i)$ is also open. Therefore, $\bigcup_{i=1}^n (U\cap U_i) - K=\varnothing$ implies that $K=\varnothing$, which is a contradiction. We conclude that $x\in (F+K)^c$ cannot be in the interior of $(F+K)^c$, so $(F+K)^c$ is open and $F+K$ is closed.
