Compact set in an TVS Hausdorff space

Question

We know that given $X$ a topological vector Hausdorff space, a compact set $K$ in $X$ is closed. I wonder if there exists a TVS Hausdorff space in which $K$ is compact, therefore closed but not bounded. I apologise if the questions is already asked or there is a theorem which prove that space not exists. I searched previous question and on web with no result. To summarise : $K\subset X$ compact and $ X $ TVS Hausdorff $\implies K=\overline K$. Does it implies also K bounded, in term of topological vector space? Thanks

Answer 1

The answer is no. Every compact set in a TVS must be bounded (and more generally, this is the case for every uniform space. More precisely, a subset of a uniform space is compact iff it is complete and totally bounded). To see it, take $U$ an disc neighbourhood of the origin and notice that $\{nU\}_{n\in \mathbb N}$ is an open cover of your compact set $K$, so it must have a finite subcover, $\{U,\dots, mU\}$. This obviously implies $K\subset \frac 1m U$ and we are done.

The more interesting question is whether every closed and bounded subset of a TVS $X$ is compact: this is not always the case (for example, in every infinite dimensional Banach space the closed unit ball is not compact) but there's an interesting class of TVS satisfying this property: those are known as Montel spaces.

Answer 2

"Bounded" means nothing without a metric (you phrased your question in terms of general topology).

But in the presence of a metric, every compact set is bounded. That's quite easy to see by trialling the open cover $\{B(x,1):x\in K\}$ of the compact set $K$.