Let $K$ be a compact subset of a Hausdorff topological vector space. Is it true that $\bigcap_{n\in \mathbb{N}}\frac{1}{n}K$ is either empty of or is a set consisting of the origin only?
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1$\begingroup$ No. Think about $\mathbb{Q}_p.$ Also please post this level of question on Math.StackExchange.com $\endgroup$– JSpecterCommented Jun 12, 2011 at 17:33
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1$\begingroup$ @JSpecter: very often by "TVS" one understands "TVS over $\mathbb{R}$ or $\mathbb{C}$". If the OP meant "TVS over a local field", he would have probably specified it. $\endgroup$– QfwfqCommented Jun 12, 2011 at 19:44
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$\begingroup$ It really looks like a homework in a functional analysis course... $\endgroup$– Alain ValetteCommented Jun 12, 2011 at 20:21
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$\begingroup$ @Alain: maybe you're right. $\endgroup$– Todd TrimbleCommented Jun 12, 2011 at 21:57
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1 Answer
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I'll assume this is a TVS over $\mathbb{R}$ or $\mathbb{C}$, and not over some other local field.
If $x$ is nonzero and belongs to $\frac1{n}K$ for every $n \geq 1$, then $nx \in K$ for each $n$. If $L$ is the line through $x$, then the subspace topology coincides with the standard topology on the ground field by the TVS axioms, and $L \cap K$ is a compact subset of the line which is unbounded since it contains all the $nx$. This gives a contradiction.
answered Jun 12, 2011 at 17:35