It is well-known that a countable product of second-countable spaces is second-countable. However, as far as I know there are non-separable Fréchet spaces, at least I have seen explicit references to "separable Fréchet spaces", so it doesn't seem like every Fréchet space is separable. Now both the product topology and the topology on a Fréchet space induced by the family of semi-norms are initial topologies. And it seems to me that the proof that a countable product of second-countable spaces is second-countable doesn't use anything except the fact that the product topology is initial.
Here is a proof(?) that any initial topology induced by countably many maps into second-countable spaces is second-countable:
We start with the general setup, without assuming countability of any kind: Let $((X_\alpha, \mathcal T_\alpha))_{\alpha \in A}$ be a family of spaces, and let $X$ be a set. Now let maps $f_\alpha \colon X \to X_\alpha$ induce a topology $\mathcal T$ on $X$. That is, sets on the form $f_\alpha^{-1}(U_\alpha)$ for $U_\alpha \in \mathcal T_\alpha$ form a subbasis for $\mathcal T$. In other words, open sets in $X$ are (arbitrary) unions of finite intersections of sets $f_\alpha^{-1}(U_\alpha)$.
Now let $\mathcal B_\alpha = \{ B_\alpha^\lambda \mid \lambda \in \mathbb \Lambda_\alpha \}$ be any basis for $\mathcal T_\alpha$. I claim that sets on the form $f_\alpha^{-1}(B_\alpha^\lambda)$ constitute a subbasis for $\mathcal T$. Let $\mathcal T'$ be the topology generated by the sets $f_\alpha^{-1}(B_\alpha^\lambda)$. Clearly we have $\mathcal T' \subseteq \mathcal T$, so to prove this claim it is enough to show that every element of a subbasis for $\mathcal T$ lies in $\mathcal T'$, since $\mathcal T$ is the coarsest topology on $X$ that contains these sets.
Take such a subbasis element $f_\alpha^{-1}(U_\alpha)$ (with notation as above). Since $U_\alpha \in \mathcal T_\alpha$ and $\mathcal B_\alpha$ is a basis for $\mathcal T_\alpha$, we can write $U_\alpha = \bigcup_{\lambda \in \Lambda_\alpha'} B_\alpha^\lambda$ for some $\Lambda_\alpha' \subseteq \Lambda_\alpha$. But then
$$ f_\alpha^{-1}(U_\alpha)
= f_\alpha^{-1} \biggl( \bigcup_{\lambda \in \Lambda_\alpha'} B_\alpha^\lambda \biggr)
= \bigcup_{\lambda \in \Lambda_\alpha'} f_\alpha^{-1}(B_\alpha^\lambda), $$
which clearly lies in $\mathcal T'$. Hence $\mathcal T \subseteq \mathcal T'$, so the two topologies are equal.
Now assume that all index sets $A$ and $\Lambda_\alpha$ are countable. Then there are countably many sets $B_\alpha^\lambda$, so countably many $f_\alpha^{-1}(B_\alpha^\lambda)$. Taking finite intersections of these sets also yields a countable collection of sets, but this is a basis for $\mathcal T$, so $X$ is second-countable.
Finally, if the $f_\alpha$ are semi-norms and we let them induce a topology on $X$, then this topology is a priori induced by maps $x \mapsto f_\alpha(x – y)$ for $y \in X$ and $\alpha \in A$ (see also this question), of which there in general are uncountably many. But such a map is continuous iff $f_\alpha$ itself is continuous, so we may assume that there are only countably many such maps (given that $A$ is countable). Hence every Fréchet space is second-countable.
Can anyone help me figure out what is going on here? Thanks very much in advance!
Best Answer
I believe I have found the flaw in my argument. I claimed that if the map $f_\alpha$ were continuous, then $x \mapsto p_\alpha(x-y)$ would also be continuous. But this relies on addition being continuous, and I don't believe this is the case.
Let $p \colon \mathbb R \to \mathbb R$ be the norm $p(x) = \lvert x \rvert$, and let $\tilde{\mathbb{R}}$ be the real line equipped with the initial topology induced by $p$ (i.e. not the norm topology, at least a priori). Notice that open sets in $\tilde{\mathbb{R}}$ are symmetric around $0$ (so this clearly isn't the norm topology -- case closed).
But to give an explicit counterexample to addition being continuous: Let $A \colon \tilde{\mathbb{R}} \to \tilde{\mathbb{R}}$ be the map $x \mapsto x + 1$. I claim that this map is not continuous, say at $x = -1$. For $\epsilon > 0$ the interval $(-\epsilon,\epsilon) = p^{-1}([0,\epsilon))$ is a neighbourhood of $A(-1) = 0$. For $A$ to be continuous at $-1$ there must be a neighbourhood $U$ of $-1$ such that $A(U) \subseteq (-\epsilon,\epsilon)$. Then $U$ must contain the point $-1$, but then $1 \in U$, and $A(1) = 2 \not\in (-\epsilon,\epsilon)$ for $\epsilon$ small enough.
In case someone else has something clever to add (or something to correct me on) I will wait a bit before accepting this answer.