Compactness implies total boundedness in any topological vector space?

Question

I know that in a complete metric space, compactness is equivalent to closedness and total boundedness. However, in an arbitrary topological vector space, I think compactness implies total boundedness. (The definition of total boundedness in any tvs is as in Rudin's functional analysis definition 3.19.) Is my judgement correct?

Answer 1

This follows immediately from the definition. Let $E$ be compact and $V$ be a neighborhood of $0$. Then $(x+V)_{x \in E}$ is an open cover of $E$ so there is a finite subcover $(x_i+V)_{1\leq i \leq n}$. Let $F=\{x_1,x_2,...,x_n\}$ . Then $E \subset F+V$.