How to prove that a finite-dimensional linear subspace is a closed set

Question

Given a linear space $V$, a field $F$, a norm $\|\cdot\|$ on $V$ and a Base $B$.

How do I prove that the subspace $\text{span}\{b_1,b_2,\ldots,b_n\}$ where $b_i \in B$ is a closed set under the topology that is created from the metric space that the norm creates?

Is it generally true? Do I need $V$ to be a Banach space? Or do I need $F$ to be the real numbers?

score 33 · Accepted Answer · answered Jan 10 '16 at 08:09

33

You only need that the ground field $K$ is a complete normed field, e.g. $K \in \{\mathbb{R},\mathbb{C}\}$.

If $(x^{(k)})_k$ is a sequence in $U := \langle b_1, \dotsc, b_n \rangle$ which converges to some $x \in V$ then $(x^{(k)})_k$ is a Cauchy sequence. Because $K$ is a complete normed field the finite dimensional normed space $U$ is complete. Therefore $(x^{(k)})_k$ converges to some $y \in U$. By the uniqueness of limits in $V$ we already have $x = y \in U$. So $U$ is closed.

Notice that the statement does not necessarily hold for normed spaces over non-complete normed fields , even if all occuring vector spaces are finite-dimensional. Take for example the $\mathbb{Q}$-vector spaces $\mathbb{Q} \subseteq \mathbb{Q}[\sqrt{2}]$.

answered Jan 10 '16 at 08:09

Jendrik Stelzner

16,401

3

why is K being a complete normed field causing that the finite dimensional normed space U is complete? – Matan L Jan 11 '16 at 12:19
4

A generalization of the fact that all norms on $\mathbb{R}^n$ are equivalent is that all norms on $K^n$ are equivalent (page 15). The $|\cdot|1$ norm on $K^n$, i.e. $|(x_1, \dotsc, x_n)|_1 = \sum{i=1}^n |x_i|$, is complete (because $x^{(k)} \in K^n$ is a Cauchy sequence if and only if each coordinate $x^{(k)}_i$ is a Cauchy sequence). Hence all norms on $K^n$ are complete. So the same goes for every finite dimensional normed space over $K$. – Jendrik Stelzner Jan 12 '16 at 09:16
1

Why does sequential closedness imply closedness in this case? – ThorbenK Jan 18 '23 at 21:42
@ThorbenK because we are working with a normed space! – Omer Feb 11 '23 at 21:24

How to prove that a finite-dimensional linear subspace is a closed set

1 Answers1

Linked