While wondering about why locally compact vector space are important, I was told that in a locally compact vector space if one scales a compact set by some number, than we still have a compact set, while this might fail in a vector space that is not locally compact. I was trying to see this with the space $\ell^2$ (space of real sequences with finite sum of squares) which is according to this post not locally compact. But then I realized that I don't even know what compact space look like in this vector space.
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Since scaling is a continuous map of any Topological Vector Space to itself, the image of any compact subset of any TVS under scaling will also be compact.
uniquesolution
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I see. I don't remember what property it was then... – edamondo Mar 04 '21 at 18:01
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Still out of curiosity, what do the compact sets of $\ell^2$ look like? Is there a characterization like for finite $\mathbb{R}^n$? – edamondo Mar 04 '21 at 19:33
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Have a look here: https://math.stackexchange.com/questions/941897/examples-of-compact-sets-that-are-infinite-dimensional-and-not-bounded – uniquesolution Mar 04 '21 at 19:37
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thank you for the link. The motivation behind the original question was a proof in which I have a sequence of compact sets $K_n\subset X$ and it is written : "Since $X$ is locally compact, up to enlarging inductively each compact set, we may assume $K_n\subset \mathring{K}_{n+1}$ for any $n\in\mathbb{N}$". – edamondo Mar 05 '21 at 10:51
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Locally compact topological vector spaces are finite dimensional (see, e.g., Rudin's book) and thus not very interesting for functional analysis.
Jochen
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