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.