# Norm on finite dimensional vector space and induced topology

Let $$V$$ be a finite dimensional vector space over $$\mathbb{R}$$, say $$V \cong \mathbb{R}^n$$ where the isomorphism is given by $$f$$ after choosing a basis $$(b_1,...,b_n).$$ Then one can define a norm on $$V$$ by $$N:V \to \mathbb{R}, v \mapsto \Vert f(v) \Vert.$$ Is this a standard/the conventional way to equip a finite dimensional real vector space with a norm?

This norm will induce a topology on $$V$$. Another way to define a topology on $$V$$ is by declaring $$U$$ to be open iff $$f(U)$$ is open in $$\mathbb{R}^n$$. As far as I know, this is the standard way to define a topology on a vector space. Are those topologies the same?

I think they are: an open $$U$$ means that $$f(U)$$ is open. Assume that $$f(U)=B_r(x)$$ for some $$x \in \mathbb{R}^n.$$ Defining $$w=f^{-1}(x)$$ we get $$f(U)=\{y \in \mathbb{R}^n \ \vert \ \Vert f(w)-y\Vert. In particular every $$u \in U$$ satisfies $$\Vert f(w)-f(u)\Vert < r$$ and every $$v \in V$$ with $$\Vert f(w)-f(v)\Vert satisfies $$f(v) \in \{y \in \mathbb{R}^n \ \vert \ \Vert f(w)-y\Vert and so $$v \in U$$. Thus $$U=\{v \in V \ \vert \ \Vert f(w)-f(v) \Vert which is in the other topology. The general case $$f(U)=\bigcup_{i \in I} B_{r_i}(x_i)$$ is then also satisfied because this gives $$U=\bigcup_i f^{-1}(B_{r_i}(x_i))$$ where any of the $$f^{-1}(B_{r_i}(x_i))$$ is in the topology. Conversely every open ball $$B=\{v \in V \ \vert \ \Vert f(w)-f(v) \Vert satisfies $$f(B)=B_r(f(w))$$ which is open.

• The topology on $\mathbb{R}^n$ is induced by its distance/metric, which is induced by its norm, so the topology you get on $V$ by using $f$ to define open spaces is the same as the topology you get by using $f$ to define a norm, and then using the norm to define a distance, and then using the distance to define a topology. Commented Jan 24, 2023 at 15:32
• @ArturoMagidin Is that obvious? Perhaps I am missing something, but I don't see it immediately. Thats why I tried to prove it in my question. Commented Jan 25, 2023 at 9:40
• Since you are using transport of structure, yes, it should be obvious given that the topology is induced by the metric. Transporting the metric and using it to define the topology yields exactly the same result as transporting the topology. Commented Jan 25, 2023 at 13:08