# About Locally Convex Frèchet Space

I need to proof the following statements, having $$\quad X = \{x_n : \mathbb{N} \rightarrow \mathbb{C}\}$$ and $$p_j = max_{k \le j}|x(k)|$$

1)$$p_j$$ is a countable family of seminorms which induces Hausdorff topology

2) $$(X,\{p_j\})$$ is a Frèchet space

For 1) I know that a countable family of seminorms induces Hausdorff topology if $$\quad \forall x \in X-0 \quad \exists p \in \{p_j\} : p(x) > 0$$. So of course any non-null sequence satisfies this condition.

But how should I prove 2)? First I should prove that it is a locally convex space, but I don't know where to start...

• A vector space with a family of seminorms is always locally convex. So to prove that $X$ is a Fréchet space, it remains only to see that the metric induced by the countable family of seminorms is complete. (see the alternative definition of a Fréchet space here) – Michh Dec 28 '18 at 2:43
• I found a statement which says that Any Frèchet Space satisfies the first numerability axiom. But Every metric space satisfies it, so should I conclude that any metric space is Frèchet?.. I don't think so.. I need completeness... – James Arten Dec 30 '18 at 17:52
• I cannot figure out the problem here. To prove that a vector space is a Fréchet space, you need to prove that: (i) the topology is induced by a countable family of seminorms, (ii) the family is separating (i.e. $\forall x \neq 0, \exists j$ such that $p_j(x) > 0$) and (iii) the metric induced by the family of seminorms is complete. In particular, there is no need to prove that the space is locally convex since this follows from point (i). – Michh Jan 1 at 10:59