Convergence with respect to a metric in a locally convex space Suppose $X$ is a locally convex space with topology generated by a countable family of seminorms $\mathcal{P}=\{||\cdot||_{k}\}_{k\in \mathbb{N}}$. Suppose $\{x_{n}\}_{n\in \mathbb{N}}$ is a sequence on $X$ which converges to $x \in X$ in the locally convex topology. I know this is equivalent to convergence with respect to each seminorm, that is, $x_{n}\to x$ iff $||x_{n}-x||_{k} \to 0$ for every $k \in \mathbb{N}$. Now, because $\mathcal{P}$ is countable, it is actually a Fréchet space, so it is metrizable with (a possible metric) given by:
$$d(x,y) :=\sum_{k=1}^{\infty}2^{-k}\frac{||x-y||_{k}}{1+||x-y||_{k}}$$
I suppose that $x_{n}\to x$ also implies $x_{n}\to x$ with respect to the metric $d$ on $X$, since this metric defines the topology. But I'm having a hard time trying to prove it. Can someome give me any hints on how to address this problem?
 A: A remark: I believe the word Frechét spae is only used if the metric is complete.
Now suppose $x_\alpha\to x$ wrt the locally convex topology, meaning $\|x_\alpha -x\|_k\to0$ for all $k$, in particular for each $k$ you find an $a_k$ so that for all $\alpha≥a$ you have $\|x_\alpha-x\|_k≤\epsilon$. Infact  by choosing $a=\sup\{ a_1,..., a_k\}$ you actually get $\|x-x_\alpha\|_k≤\epsilon$ for all $k\in\{1,...,N\}$ for any finite $N$. Now choose $N$ so that $2^{-N}<\epsilon$ and you $a$ as before to get:
$$\sum_{k=0}^\infty 2^{-k}\frac{\|x_\alpha-x\|_k}{1+\|x_\alpha-x\|_k}=\sum_{k=0}^N2^{-k}\frac{\|x_\alpha-x\|_k}{1+\|x_\alpha-x\|_k} + \sum_{k=N+1}^\infty 2^{-k}\frac{\|x_\alpha-x\|_k}{1+\|x_\alpha-x\|_k}\\
≤\sum_{k=0}^N2^{-k}\epsilon + \sum_{k={N+1}}^\infty 2^{-k}≤\epsilon + 2^{-N}≤2\epsilon$$
for all $\alpha>a$. Now remember that $\epsilon$ was arbitrary. This step shows that the topology of the metric is coarser than the topology of the semi-norms, since any convergent net in locally convex topology is also convergent (with the same limit) in the metric topology.
For the other direction suppose that $x_\alpha\to x$ in the metric topology. The easiest way to show that $x_\alpha\to x$ for the semi-norms is to suppose that there is some semi-norm $\|\cdot\|_N$ for which $\|x_\alpha-x\|_N\not\to0$ and to get a contradiction. Well if $\|x_\alpha -x\|_N\not\to0$ then
$$\sum_{k=0}^\infty 2^{-k}\frac{\|x_\alpha-x\|_k}{1+\|x_\alpha-x\|_k} ≥ 2^{-N}\frac{\|x_{\alpha}-x\|_N}{1+\|x_{\alpha}-x\|_N}$$
and $d(x_\alpha, x)$ majorises something positive that does not converge to $0$, as such $d(x_\alpha,x)$ does not converge to zero, a contradiction.
