If $X$ is a topological vector space and $K \subset X$ a compact set, can we say that $K$ is also bounded?
By Rudin's functional analysis any topological vector space is Haussdorf, I read that in any Haussdorf space intersection between closed and compact set is compact, is there a similar property for boundness instead?
I cannot manage to find a proof and neither to work out one.
Can you provide or point out a proof?
How do you define “bounded” in a topological vector space? If you say that $A$ is bounded if there is a neighborhood $N$ of $0$ and a number $r>0$ such that $A\subset rN$, then your statement can be proved as follows: fix an open neighborhood $N$ of $0$ consider the set $\{rN\,|\,r>0\}$. Since $\bigcup_{r>0}rN=X$, that set is an open cover of $K$. Therefore, there is a finite set $F\subset(0,\infty)$ such that $K\subset\bigcup_{r\in F}rN$. But then$$K\subset\left(\max F\right)N.$$