Interpreting a problem on Frechet spaces (topology)

[Mathmos6]

Homework Statement

I've been given the following problem:

"Suppose that U is a finite-dimensional subspace of a Fréchet space $(V,\tau)$. Show that the subspace topology on U is the usual topology (given for example by a Euclidean norm) and that U is a closed linear subspace of V."

I feel a bit confused: I'm not sure what precisely is meant by 'the usual topology (given for example by a Euclidean norm)', but IIRC any 2 norms on a finite dimensional vector space are equivalent (to e.g. the Euclidean norm) so am I right in thinking that the question just requires me to show that the subspace topology on U is a topology induced by some norm (and therefore equivalent to any norm on U)?

I suppose to show closure is fairly simple, just play around with bases and limits, and it's a subspace of a vector space so linearity is also trivial, right? So then it's only the first part which I need help with, if I understand everything else correctly - thanks!

 

[lippyka]

Homework EquationsN/AThe Attempt at a SolutionI think the answer to the first part of the question is that, yes, the subspace topology on U is a topology induced by some norm (and therefore equivalent to any norm on U). To show this, let us consider a basis {u1, u2,..., uk} of U and let x be a point in U. Then, we can express x as a linear combination of the basis vectors:x = c1u1 + c2u2 + ... + ckuKwhere the ci are real numbers. Let ||·|| be a norm on U. We can then define a topology on U by taking all sets of the form N_ε(x) = {y ∈ U : ||x - y|| < ε}It is easy to see that these sets form a topology on U, and that this topology is induced by the norm ||·||. Thus, the subspace topology on U is a topology induced by some norm and is thus equivalent to any norm on U.