# I Normed Vector Spaces and Topological Vector Spaces

1. Aug 16, 2016

### Bashyboy

Let $(V, ||\cdot||)$ be some finite-dimensional vector space over field $\mathbb{F}$ with $\dim V = n$. Endowing this vector space with the metric topology, where the metric is induced by the norm, will $V$ become a topological vector space? It seems that this might be true, given that finite-dimensional vector spaces are nice e.g., all norms on a finite dimensional vector space are equivalent and $V$ is isomorphic $\mathbb{F}^n$. In short, my question is, can all normed finite-dimensional vector spaces be made into a topological vector space by considering the metric topology?

Does anyone know if this is true? I would like to know before I attempt at proving it.

2. Aug 16, 2016

### andrewkirk

What do we know about the field $\mathbb F$? It must have a topology for us to even be able to ask the question of whether $V$ is a topological VS. Are you prepared to assume that the topology on $\mathbb F$ is a metric topology? If so, I would be confident that $V$ is a topological VS, as we can then use $\epsilon-\delta$ arguments throughout to prove the continuity of the two functions:
• $f:V\times V\to V$ given by $f(u,v)=u+v$; and
• $g:F\times V\to V$ given by $g(k,v)=kv$
If the topology on $\mathbb F$ is not metric it is not immediately clear (to me at least) whether a proof would succeed. If it did, I imagine it would be more difficult than with a metric topology.

3. Aug 17, 2016

### Bashyboy

Further stipulating that $\mathbb{F}$ is endowed with a metric topology would be perfectly fine, although I am interested in knowing whether the more general theorem applies. I will try proving this special case with the metric topology.

4. Aug 17, 2016

### micromass

If $\mathbb{F}$ is an arbitrary field, then how do you define a normed vector space?

5. Aug 17, 2016

### andrewkirk

It can't be an arbitrary field, because we need a norm on $\mathbb F$ for the norm axioms to be well-defined. My question was about whether the topology on $\mathbb F$ is assumed equal to the metric topology induced by the norm.

6. Aug 18, 2016

### FactChecker

Yes. A normed vector space always has an induced topology. It doesn't have to be finite-dimensional.

7. Aug 19, 2016

### Bashyboy

I wasn't questioning whether it could have a topology induced by the norm, but rather if the induced topology would yield a topological vector space.

8. Aug 19, 2016

### FactChecker

CORRECTION: @mathwonk points out below that this link assumes F is the reals or complex numbers.
Oh. The answer is that it is a topological vector space (see the Example in https://en.wikipedia.org/wiki/Topological_vector_space )

Last edited: Aug 19, 2016
9. Aug 19, 2016

### mathwonk

the wiki article you linked assumes, as is natural, that the field of scalars is the reals or complexes. the present confusion stems from trying to ask the question about a case so general it is not what the definitions and assumptions are. presumably the first case to consider to clarify this question would be the one dimensional case.

10. Sep 2, 2016

### pwsnafu

We don't want a norm on the field rather an absolute value. Yes this pedantic to the point of ridiculousness.
It's worth pointing out if $\mathbb{F} = \mathbb{Q}$ then you have the ordinary absolute value and the p-adic one. The latter is...different.