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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?

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

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    $\begingroup$ 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
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    $\begingroup$ 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

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