# every finite dimensional subspace of a normed space is closed

###### Theorem 1

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 parent (http://planetmath.org/EverySubspaceOfANormedSpaceOfFiniteDimensionIsClosed)), 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 $V$.

Title every finite dimensional subspace of a normed space is closed EveryFiniteDimensionalSubspaceOfANormedSpaceIsClosed 2013-03-22 14:58:56 2013-03-22 14:58:56 Mathprof (13753) Mathprof (13753) 14 Mathprof (13753) Corollary msc 46B99 msc 15A03 msc 54E52