I have shown any norms on a finite dimensional real vector space are equivalent then the question asks why would this imply every finite-dimensional subspace of normed vector space is closed. (Closed in the sense that it is toplogically closed, its complement is an open subset.)

I understand that equivalent norms yield the same notion of convergence however I have very few ideas on where to start. I have seen a few posts showing the subspace is complete instead, but I do not think it is in the spirit of this problem.

How should I proceed? Many thanks in advance!

  • $\begingroup$ Think of completeness. $\endgroup$ Aug 20 '20 at 18:19
  • $\begingroup$ @DanielFischer Thank you! Just confirming, are you referring to showing the normed vector space (the Ambient space) is complete first and then using complete subset of a complete space is closed? $\endgroup$ Aug 20 '20 at 20:21
  • $\begingroup$ The ambient space need not be complete. But a complete subset of a metric space is closed, whether the ambient space is complete or not. $\endgroup$ Aug 20 '20 at 20:23

I know that if $X$ is normed space over some field $\mathbb{F}$ and finite-dimensional with dimension $n$, so you can prove $X$ is isomorphic to $\mathbb{F}^{n}$ with the euclidean norm. $[1]$

A collorary of above result is that if $X$ be a finite dimensional vector space with norms $||\cdot||_{1}$ and $||\cdot||_{2}$. Then $||\cdot||_{1}$ and $||\cdot||_{2}$ are equivalent.

Now, if you can prove that result $[1]$ then you have that any finite dimensional subspace of normed linear space is closed.

  • $\begingroup$ Thank you! However I was trying to deduce directly from equivalency of norms instead rather than proving something new but thank you nonetheless. $\endgroup$ Aug 20 '20 at 16:49
  • $\begingroup$ I think you need result $[1]$ to prove the equivalence of the norms. In that case both results that I mentioned are colloraries of $[1]$. $\endgroup$
    – mathproof
    Aug 20 '20 at 16:58
  • $\begingroup$ @JustWandering How do you showed that two norms in finite-$\mathbb{R}$-vector space dimensional are equivalent? $\endgroup$
    – mathproof
    Aug 20 '20 at 17:03
  • $\begingroup$ I used the fact that they are both equivalent to $d_1$ norm and that required some tricks in compactness and topological properties such that a continuous function on a compact set attains its bounds etc. $\endgroup$ Aug 20 '20 at 17:07

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