When topological vector space metrizable

functional-analysisgeneral-topologyreal-analysistopological-vector-spaces

This question comes from Chapter 5 of Folland's Real Analysis.It's a proposition (5.16)without proof.

Let $X$ be a vector space equipped with the topology defined by a family $\left\{ p_{\alpha} \right\} _{\alpha \in A}$ of seminorms.
a, $X$ is Hausdorff iff for each $x\ne0 $ there exists $\alpha \in A$ such that $p_\alpha(x)\ne 0$.
b.If $X$ is Hausdorff and $A$ is countable,then $X$ is metrizable with a translation invariant metric (i.e.,$\rho(x,y)=\rho(x+z,y+z)$ for all $x,y,z\in X$).

The so-called topology defined by seminorms comes from the theorem:

Let $\left\{ p_{\alpha} \right\} _{\alpha \in A}$ be a family of seminorm on the vector space $X$.If $x\in X,\alpha\in A$,and $\varepsilon>0$,let
$$
U_{x\alpha \varepsilon}=\left\{ y\in X:p_{\alpha}\left( y-x \right) <\varepsilon \right\} ,
$$

and let $\mathcal{T}$ be the topology generated by the sets $U_{x\alpha\varepsilon}$.Then $(X,\mathcal{T})$ is a locally convex topological vector space.

I have solved the problem $a$,but I cannot make any progress on b.

Best Answer

We consider the seminorms $\{p_\alpha\}_{\alpha\in A}$ which define the topology on your vector space. In part (b), we assume that $X$ is a Hausdorff space and that $A$ is countable. Thus we may as well assume that $A=\mathbb{N}$. Now I claim that the function $$ \rho:X\times X\to [0,\infty) $$ given by $$ \rho(x,y) = \sum_{k=1}^\infty \frac{1}{2^k} \frac{p_k(x-y)}{1+p_k(x-y)} $$ is a metric which is translation invariant.

It is clear from the formulae and the definition of a seminorm that $\rho$ is translation invariant and symmetric. I will leave it to you to prove that $\rho$ is finite valued, positive definite, and satisfies the triangle inequality. (In checking that $\rho$ is positive definite, you will need the assumption that $X$ is Hausdorff and the result of part (a).)

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