Compactness implies closedness and boundedness? I know this is true for a metric space. But how about a topological vector space? Is there a reference explicitly stating this result?
 A: In a topological vector space $\mathbf{X}$, every compact set $K$ is bounded in the sense that for every open neighbourhood $V$ of $0$ there exists a $t \in \mathbb{R}$ with $K \subseteq tV$.
To see this, we first observe that for any $x \in \mathbf{X}$ and any open neighbourhood $V$ of $0$ there exists some $\varepsilon > 0$ such that $\varepsilon' x \in V$ for all $0 \leq \varepsilon' < \varepsilon$. It follows that if $V$ is an open neighbourhood of $0$, then $\mathbf{X} = \bigcup_{n \in \mathbb{N}} nV$, which yields the desired claim.
In a topological vector space $\mathbf{X}$, being compact implies being closed iff $\mathbf{X}$ is Hausdorff - which some people include in being a topological vector space, and others don't. That Hausdorff suffices is true for topological spaces in general. Conversely, a topological vector space that fails to be Hausdorff also fails to be $T_1$, so there exists some $x \in \mathbf{X}$ such that $\{x\}$ is not closed. But any singleton is trivially compact.
