The countable product of Fréchet spaces is a Fréchet space

Question

Let $\{E_n \; ; \; n \in \mathbb{N}\}$ be a family of Fréchet spaces. I want to prove that the product $$E:= \prod_{n=1}^{\infty} E_n$$ is a Fréchet space, that is, $E$ is metrizable (Hausdorff space and admits a countable basis of neighborhoods of $0\in E$), complete and locally convex (admits a basis of neighborhoods of $0\in E$ consisting of convex sets).

I already know that $ E $ is Hausdorff, locally convex and complete space. I don't know how to prove that $ E $ is metrizable. How to proceed?

How do you show the completeness without the metric, I wonder? — Henno Brandsma, Jun 29 '20 at 21:34
Using the concepts of filters and Cauchy filters on topological vector spaces. For instante, see Exercise 5.5 for this book. — Guilherme, Jun 29 '20 at 21:52

score 1 · Accepted Answer · answered Jun 29 '20 at 18:32

1

Let $d_n$ be one metric in each $E_n$. Then $$d(\tilde{x}, \tilde{y}):= \sum_{n=1}^{+\infty} \frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}\frac{1}{2^n} $$ where $\tilde{x}, \tilde{y}\in E$, is a metric in $E$.

answered Jun 29 '20 at 18:32

alphaomega

2,216

And this metric define the topology of $ E $ ? – Guilherme Jun 29 '20 at 18:41
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Isn't this in general the case? I mean if $X_n$ are metric spaces and $X$ is the countable product of them with the product topology then the function $$d(\tilde{x}, \tilde{y}):= \sum_{n=1}^{+\infty} \frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}\frac{1}{2^n}$$ is a metric in $X$ and the induced topology is equivalent to the product topology. – alphaomega Jun 29 '20 at 19:21

score 1 · Answer 2 · answered Jun 29 '20 at 21:33

In this answer I deal with the metrisability dierctly, as a general result in metric spaces. Of course, if $\mathcal{U}_n$ is a countable local base of convex neighbourhoods of $0$ for $E_n$, (such that their intersection is $\{0\}$, which is equivalent to Hausdorffness in a TVS), the standard product local base for the $(0)$ product point

$$\{\prod_n U_n \mid \exists F \subseteq \Bbb N \text{ finite }: \forall n: ((n \in F) \to (U_n \in \mathcal{U_n})) \land (n \notin F ) \to (U_n = E_n)\}$$

consists of convex sets, is countable (as there are only countably many finite subsets of $\Bbb N$ and all $\mathcal{U}_n$ are countable), and intersects to $\{(0)\}$, so $\prod_n E_n$ is Fréchet.

The first proof I referred to has the added bonus that if all $d_n$ are complete, then so is the product sum-metric I define there. So that'll give completeness more easily, which I think is not yet apparent from the local base alone.

The countable product of Fréchet spaces is a Fréchet space

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