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?

  • 4
    $\begingroup$ doesn't completeness from Lemma 1 imply closedness? $\endgroup$ – user66081 Jul 2 '14 at 12:37

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.

  • $\begingroup$ Why is its consequence? Does any linear and topological isomorphism preserve completeness? $\endgroup$ – avati91 Jul 2 '14 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 Jul 2 '14 at 12:49
  • 1
    $\begingroup$ @avati91 Actually a homeomorphism between two normed spaces which is linear isomorphism preserves completeness, because it is both lipschitz and inverse lipschitz. $\endgroup$ – user87690 Jul 2 '14 at 12:55
  • $\begingroup$ @avati91 Yes, it holds. $\endgroup$ – user87690 Jul 2 '14 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 Jul 2 '14 at 13:03

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.