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# every subspace of a normed space of finite dimension is closed

Let $(V,\|\cdot\|)$ 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 $V$ and is also a Cauchy sequence in $S$. Since a finite dimensional normed space is a Banach space, $S$ is complete, so $\{a_{n}\}$ converges to an element of $S$. Since limits in a normed space are unique, that limit must 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 underlying field and we can still show a finite dimensional subspace is closed. Suppose that $V=Q^{n}$, the set of $n$-tuples of rational numbers, viewed as vector space over $Q$. Then if $S$ is a finite dimensional subspace it 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.

## Mathematics Subject Classification

54E52*no label found*15A03

*no label found*46B99

*no label found*

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## Corrections

ugh! by Mathprof ✓

ugh! by Mathprof ✓

More details in the proof by romeubraz ✓

contains own proof by CWoo ✓