Let $(X,\| \cdot \|)$ be a finite dimensional normed linear space. Can we say that $(X,\| \cdot \|)$ is a locally compact normed linear space?
Please help me in understanding this concept.
Thank you very much.
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$\begingroup$ In what? I know that it is true for locally compact normed linear spaces. $\endgroup$ – Dbchatto67 Aug 19 '18 at 7:31
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$\begingroup$ These are just the $\Bbb R^n$ with the usual topology. $\endgroup$ – Lord Shark the Unknown Aug 19 '18 at 7:38
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If $X$ is a normed vector space on $\mathbb{R}$ (or $\mathbb{C}$) then $X$ is locally compact iff it is finite dimensionnal.
$\mathbb{Q}$, as seen as a $1$-dimensionnal normed vector space on $\mathbb{Q}$, is not locally compact since no neighbourhood of the origin is compact.
Proving this is the same as proving that the closed unit ball of $X$ is compact iff $X$ is finite dimensionnal. You can find a proof or hints here.
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$\begingroup$ Can you please prove what you have stated @nicomezi? Let $X$ be a linear space over $\Bbb R$ and $\dim X=n$ then it is clear that $X \simeq \Bbb R^n$. In $\Bbb R^n$ the closed unit ball is clearly compact by the Heine-Borel theorem. How does that guarantee the compactness of the closed unit ball in $X$? Would you please elaborate this fact? Let $T$ be a linear isomorphism from $\Bbb R^n$ onto $X$. Can we say that the image of a compact subset of $\Bbb R^n$ under $T$ is compact in $X$ too? $\endgroup$ – Dbchatto67 Aug 19 '18 at 7:50
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$\begingroup$ @DebabrataChattopadhyay. This is a classical theorem. See for example:mathonline.wikidot.com/…. $\endgroup$ – mathcounterexamples.net Aug 19 '18 at 8:03
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$\begingroup$ Yeah I got it. $T$ is actually becoming a homeomorphism here. Right? $\endgroup$ – Dbchatto67 Aug 19 '18 at 8:18
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If $X$ is a finite dimensional vector space, then there exists a unique linear topology on $X$. It follows that if you choose a linear isomorphism $T \ : \ X \to \mathbb R^n$, then $T$ is a homeomorphism. Since $\mathbb R^n$ is locally compact, so is $X$.
