# Compact sets in TVS

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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No. Think about $\mathbb{Q}_p.$ Also please post this level of question on Math.StackExchange.com –  JSpecter Jun 12 '11 at 17:33
@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. –  Qfwfq Jun 12 '11 at 19:44
It really looks like a homework in a functional analysis course... –  Alain Valette Jun 12 '11 at 20:21
@Alain: maybe you're right. –  Todd Trimble Jun 12 '11 at 21:57

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.