Normed Vector Spaces and Topological Vector Spaces

[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.

 

[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.

 

[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.

 

[micromass]

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

 

[andrewkirk]

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

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.

 

[FactChecker]

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.

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

 

[Bashyboy]

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

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.

 

[FactChecker]

CORRECTION: @mathwonk points out below that this link assumes F is the reals or complex numbers.

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.

Oh. The answer is that it is a topological vector space (see the Example in https://en.wikipedia.org/wiki/Topological_vector_space )

 

[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.

 

[pwsnafu]

It can't be an arbitrary field, because we need a norm on $\mathbb F$ for the norm axioms to be well-defined.

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.