# Are finite dimensional normed linear spaces locally compact?

Let $(X,\| \cdot \|)$ be a finite dimensional normed linear space. Can we say that $(X,\| \cdot \|)$ is a locally compact normed linear space?

Thank you very much.

• In what? I know that it is true for locally compact normed linear spaces. – Dbchatto67 Aug 19 '18 at 7:31
• These are just the $\Bbb R^n$ with the usual topology. – Lord Shark the Unknown Aug 19 '18 at 7:38

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.
• 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? – Dbchatto67 Aug 19 '18 at 7:50
• Yeah I got it. $T$ is actually becoming a homeomorphism here. Right? – Dbchatto67 Aug 19 '18 at 8:18
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 $X$.