# Are all compact subsets of a topological vector space bounded?

If we have a compact subset $A$ of a topological vector space, is it necessary for $A$ to be bounded? If not, can we put any simple conditions on the topological vector space to make boundedness a necessary condition for compactness?

By boundedness, I mean the following definition: We say that a subset $A$ of a topological vector space $V$ is bounded iff for every open neighbourhood U of $0$, there is a scalar $x$ such that for every $y > x$, $A \subset yU$.

Any compact metric space is bounded as the product $A\times A$ is compact and the metric is a real-valued function on $A\times A$, which must have a maximum and a minimum. For another possible intepretation of bounded, the distance function from the origin is a real valued function on $A$ and therefore has a maximum and a minimum.
For your edit, note that for any open neighborhood $U$ of $0$, the neighborhoods $yU$ for $y\in \mathbb{R}$ cover $A$. Thus $A$ is covered by finitely many such neighborhoods by compactness. In particular there is a $x$ such that $A\subseteq xU$, and therefore $A\subseteq yU$ for all $y\geq x$.
• @MattSamuel don't you need to argue that $U$ may be assumed balanced w.l.o.g.? Otherwise, how do you deduce $A \subseteq yU$ from $A \subseteq xU$? – Tom Hirschowitz Aug 14 '18 at 20:43