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I know that probably this question has already been answered, but I'd like to present my attempt of solution.

Let $(E,\|\cdot\|)$ be a normed vector space and let $F\subseteq E$ be a finite-dimensional subspace. Suppose that $\dim_{\mathbb{R}} F=n$.

First of all I proved the following lemma.

Lemma 1 Every finite-dimensional normed vector space is complete.

Then, I noticed that $(F,\|\cdot\|_{F})$ is a finite-dimensional normed vector space, where $\|\cdot\|_{F}$ is the induced norm. So, $F$ is complete by Lemma 1.

Lemma 2 If $(E,\|\cdot\|)$ is a normed vector space such that $\dim_{\mathbb{R}}{E}=n$, then $E$ is algebraically and topologically isomorphic to $\mathbb{R}^{n}$.

By Lemma 2, I can view $F$ as a complete subspace of $\mathbb{R}^n$ and, since $\mathbb{R}^n$ is itself complete, I can conclude that $F$ is closed. Here I used the fact that every complete subspace of a complete space is closed.

Is there something wrong?

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    $\begingroup$ doesn't completeness from Lemma 1 imply closedness? $\endgroup$
    – user66081
    Commented Jul 2, 2014 at 12:37

1 Answer 1

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The part with lemma 2 isn't necessary. Any complete subspace of a metric space is closed. So the proposition follows directly from lemma 1 and you don't need to assume that $E$ is finite-dimensional. But can you prove lemma 1?

Actually even if you use just “complete subspace of complete space is closed” you don't need lemma 2 since both $E$, $F$ are complete by lemma 1.

On the other hand lemma 1 is a consequence of lemma 2. That's because a homeomorphism between normed linear spaces which is also linear isomorphism preserves completeness. It is so because the mapping is then bilipschitz which goes from the following fact:

For a linear map between normed linear spaces the following are equivalent:

  1. it is bounded
  2. it is lipschitz
  3. it is continuous
  4. it is continuous at zero

The only non-trivial implication is $(4) \implies (1)$. By continuity at zero, there is $δ > 0$ such that $f[B(0, δ)] ⊆ B(f(0), 1)$. But then $f[B(0, 1)] ⊆ B(f(0), \frac{1}{δ})$ and $\lVert f(x)\rVert ≤ \frac{1}{δ}\lVert x \rVert$ for every $x$ by linearity.

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  • $\begingroup$ Why is its consequence? Does any linear and topological isomorphism preserve completeness? $\endgroup$
    – avati91
    Commented Jul 2, 2014 at 12:44
  • $\begingroup$ @avati91 Yes, you are right. It preserves just complete metrizability. But maybe it can be strengthened. However there is no point in proving completeness of $F$ by lemma 1 while proving completeness of $E$ by lemma 2, which is incorrect as you said. $\endgroup$
    – user87690
    Commented Jul 2, 2014 at 12:49
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    $\begingroup$ @avati91 Actually a homeomorphism between two normed spaces which is linear isomorphism preserves completeness, because it is both lipschitz and inverse lipschitz. $\endgroup$
    – user87690
    Commented Jul 2, 2014 at 12:55
  • $\begingroup$ @avati91 Yes, it holds. $\endgroup$
    – user87690
    Commented Jul 2, 2014 at 13:01
  • $\begingroup$ If $T:(E,\|\cdot\|_E)\to (F,\|\cdot\|_F)$ is a linear and topological isomorphism, then it is also a bi-Lipschitz map. But...why? Which is the Lipschitz constant? $\endgroup$
    – avati91
    Commented Jul 2, 2014 at 13:03

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