# Finite dimensional subspace if a normed vector space is closed using equivalence of norms

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!

• Think of completeness. Aug 20, 2020 at 18:19
• @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? Aug 20, 2020 at 20:21
• 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. Aug 20, 2020 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.

• Thank you! However I was trying to deduce directly from equivalency of norms instead rather than proving something new but thank you nonetheless. Aug 20, 2020 at 16:49
• I think you need result $[1]$ to prove the equivalence of the norms. In that case both results that I mentioned are colloraries of $[1]$.
– user798113
Aug 20, 2020 at 16:58
• @JustWandering How do you showed that two norms in finite-$\mathbb{R}$-vector space dimensional are equivalent?
– user798113
Aug 20, 2020 at 17:03
• 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. Aug 20, 2020 at 17:07