In Wikipedia's definition of Fréchet space, it is stated that a Fréchet space is a topological vector space that satisfies the following:
- It is locally convex
- Its topology can be induced by a translation-invariant metric
- Any (hence every) translation-invariant metric inducing the topology is complete
1) My first question lies with the "hence every" clause. It is known that completeness is a property of the metric, not topology, the standard example being $(0,1)$ under the absolute value and arctan metrics. As such how is it that we can conclude completeness here for every translation-invariant metric inducing the topology?
2) Wikipedia gives yet another equivalent condition for the completeness property - namely, let $\{p_k\}_{k\in\mathbb{N}}$ denote a countable family of seminorms defining the topology of $X$; then the space is complete with respect to the family of seminorms (i.e. if $(x_n)$ is a sequence in $X$ which is Cauchy with respect to each seminorm $p_k$, then there exists $x\in X$ such that $(x_n)$ converges to $x$ with respect to each seminorm $p_k$). How does the completeness with respect to any translation-invariant metric imply the completeness with respect to any countable family of seminorms and vice versa?
Remark: I suppose an answer to Q2 will answer Q1.
Let $F$ be your Fréchet space.
Suppose that $d$ is a translation invariant metric that induced the topology of $F$ and that $F$ is complete under $d$. Let $\mathcal P=\{p_k\}_{k\in\mathbb N}$ be any family of seminorms inducing the topology of $F$. Let $$d'(x,y)=\sum_{k=1}^\infty 2^{-k}\,\frac {p_k(x-y)}{1+p_k (x-y)}$$ be the metric induced by $\mathcal P$. It is easy to check that this metric induces the same topology that $\mathcal P$, that is the topology of $F$.
Looking at the respective unit balls, we can find (since both metrics induce the same topology) $\alpha,\beta>0$ such that $$\alpha\,d(x,0)\leq d'(x,0)\leq \beta\,d(x,0).$$ Because the metrics are translation invariant, for any $x,y\in F$ we have $d(x,y)=d(x-y,0)$ and the same for $d'$ so $$\tag1\alpha\,d(x,y)\leq d'(x,y)\leq \beta\,d(x,y)$$for all $x,y\in F$. Thus a sequence will be complete for $d$ if and only if it is complete for $d'$.
As $(1)$ is an equivalence relation, we have that completeness is the same for any translation invariant metric that induces the topology for $F$, and also agrees with completeness with respect to any sequence of seminorms that induces the topology.