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 $(V,\parallel \cdot \parallel )$ 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 $\{x_1,\mathrm{\dots },x_k\}$ of $S$. Also, $x\in \mathrm{span}(x_1,\mathrm{\dots },x_k,x)$. 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 $\mathrm{span}(x_1,\mathrm{\dots },x_k,x)$ (taken with the norm induced by $(V,\parallel \cdot \parallel )$) with $s_n\to x$, 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:

$V=ℝ$ 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 $V$.