every subspace of a normed space of finite dimension is closed

Let $(V,\parallel \cdot \parallel )$ be a normed vector space, and $S\subset V$ a finite dimensional subspace. Then $S$ is closed.

Proof

Let $a\in \overline{S}$ and choose a sequence $\{a_n\}$ with $a_n\in S$ such that$a_n$ converges to $a$. Then $\{a_n\}$ is a Cauchy sequence in $V$ andis also a Cauchy sequence in $S$. Since a finite dimensional normedspace is a Banach space, $S$ is complete, so $\{a_n\}$ converges to anelement of $S$. Since limits in a normed space are unique, that limitmust be $a$, so $a\in S$.

Example

The result depends on the field being the real or complex numbers.Suppose the $V=Q\times R$, viewed as a vector space over $Q$ and$S=Q\times Q$ is the finite dimensional subspace. Then clearly $(1,\sqrt{2})$ is in$V$ and is a limit point of $S$ which is not in $S$. So $S$ is not closed.

Example

On the other hand, there is an example where $Q$ is the underlyingfield and we can still show a finite dimensional subspace is closed. Supposethat $V=Q^n$, the set of $n$-tuples of rational numbers, viewedas vector space over $Q$. Then if $S$ is a finite dimensional subspaceit must be that $S=\{x|Ax=0\}$ for some matrix $A$.That is, $S$ is the inverse image of the closed set $\{0\}$.Since the map $x\to Ax$ is continuous, it follows that $S$ is a closed set.