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# every finite dimensional subspace of a normed space is closed

###### Theorem 1.

Any finite dimensional subspace of a normed vector space is closed.

Proof. Let $(V,\|\cdot\|)$ be such a normed vector space, and $S\subset V$ a finite dimensional vector subspace.

Let $x\in V$, and let $(s_{n})_{n}$ be a sequence in $S$ which converges to $x$. We want to prove that $x\in S$. Because $S$ has finite dimension, we have a basis $\{x_{{1}},...,x_{{k}}\}$ of $S$. Also, $x\in\operatorname{span}(x_{{1}},...,x_{{k}},x)$. But, as proved in the case when $V$ is finite dimensional (see this object’s parent), we have that $S$ is closed in $\operatorname{span}(x_{{1}},...,x_{{k}},x)$ (taken with the norm induced by $(V,\|\cdot\|)$) with $s_{{n}}\rightarrow x$, and then $x\in S$. QED.

# 0.0.1 Notes

The definition of a normed vector space requires the ground field to be the real or complex numbers. Indeed, consider the following counterexample if that condition doesn’t hold:

$V=\mathbb{R}$ is a $\mathbb{Q}$ - vector space, and $S=\mathbb{Q}$ is a vector subspace of $V$. It is easy to see that $\dim(S)=1$ (while $\dim(V)$ is infinite), but $S$ is not closed on $V$.

## Mathematics Subject Classification

46B99*no label found*15A03

*no label found*54E52

*no label found*

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