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How can I show that a finite dimensional linear subspace F of an arbitrary normed space X is complete, hence closed?

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  • $\begingroup$ Can you do it for $\mathbb{R}^n$? $\endgroup$ – sinbadh Jan 26 '16 at 18:58
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Let $(F, ||\cdot||)$ be a finite dimensional normed space over $\mathbb{F}$ (where $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$). Choose some linear isomorphism $T \colon \mathbb{F}^n \rightarrow F$ and define a norm $||\cdot||_1$ on $\mathbb{F}^n$ by $||v||_1 := ||Tv||$. Then $T \colon (\mathbb{F}^n, ||\cdot||_1) \rightarrow (F,||\cdot||)$ becomes an isometry. Since all the norms on $\mathbb{F}^n$ are equivalent and $(\mathbb{F}^n, ||\cdot||_2)$ is complete (here, $||\cdot||_2$ is the standard norm on $\mathbb{F}^n$), this implies that $(\mathbb{F}^n, ||\cdot||_1)$ is complete which implies in turn that $(F, ||\cdot||)$ is complete.

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