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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 $.

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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 $y\geq x$.

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  • $\begingroup$ I've edited my question to clarify what I mean by boundedness. This is a notion independent of the metric on the vector space, even if such a metric exists. $\endgroup$ – Ormi Mar 22 '15 at 16:22
  • $\begingroup$ @Ormi Please see my edit. $\endgroup$ – Matt Samuel Mar 22 '15 at 16:25
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    $\begingroup$ @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 $A \subseteq xU$? $\endgroup$ – Tom Hirschowitz Aug 14 '18 at 20:43

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