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

How do I prove that the subspace 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?


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 $\mathbb{Q} \subseteq \mathbb{Q}[\sqrt{2}]$.

  • 1
    $\begingroup$ why is K being a complete normed field causing that the finite dimensional normed space U is complete? $\endgroup$ – Matan L Jan 11 '16 at 12:19
  • 2
    $\begingroup$ 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 $K$. $\endgroup$ – Jendrik Stelzner Jan 12 '16 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.