Topology induced by the Euclidean metric

Question

Let $V$ be a finite dimensional real vector space, then $V$ is isomorphic to $\mathbb{R}^n$.

So, is the Euclidean metric defined by $d(x,y)= ||(\lambda_i)-(\gamma_i) ||$ where $\lambda_i$ and $\gamma_i$ are the coefficients?

In this case, if $(V,\tau)$ is a topological vector space then the linear map $id: (V,\tau_d) \rightarrow (V,\tau)$ is continuous because $(V,\tau_d)$ is Hausdorff and finite dimensional and finally the topology induced by the Euclidean metris $ \tau_d$ is at least as fine as $\tau$ ?

Am I right?

Thanks!

Answer 1

Let V be a normed vector space with norm n:V -> R.
The norm induces a metric for V, d(u,v) = n(u - v).
The topology Td, induced by the norm metric cannot be
compared to other topologies making V a TVS.

That is because V with the discrete topology
Tdis and V with the indiscrete are both TVS's.

Notice that the map id:(V,Td) -> (V,Tdis) is not continuous.

In the event, as is common, a TVS is defined to also
be Hausdorff, then for the TVS V with topology T
if id:(V,T) -> (V,Td) is continuous, then T is finer than Td.