Can a Frechet space be embedded into a power of another Frechet space?

Question

I know that such embedding is possible for the Kolmogorov space since a topological space $X$ is Kolmogorov if and only if $X$ can be embedded into a power of $2$ space where $2 = \{0,1\}$ equipped with the topology $\{ \emptyset, 2,\{1\}\}$.

My question is that if the same thing can be done for the Frechet space. Given a topological space $X$, is there a Frechet space $Y$ such that $X$ is Frechet if and only if $X$ can be embedded into a power of $Y$?

I heard that there was no such $Y$, but have been struggling to prove it. I think that the proof involves the following, given a Frechet space $Y^{*}$, we construct a Frechet space $(X^{*},\tau)$, where $\tau$ is cofinite, such that $|X^{*}| > |Y^{*}||\mathbb{N}|$ and every continuous map $f:Y^{*} \to X^{*}$ is constant. I have been struggling to come up with such $X^{*}$ as well as how this would imply that $X^{*}$ cannot be embedded into a power of $Y^{*}$. Any help will be greatly appreciated.

Answer 1

Given $X$ there certainly is such a $Y$. Let $Y = X\cup \{p\}$ where $p \notin X$ and give $Y$ the cofinite topology (which is Fréchet).

Then if $X$ embeds into a power of $Y$ it certainly is Fréchet as any subspace of a power of Fréchet spaces is still Fréchet by standard preservation facts.

If $X$ is itself Fréchet, it embeds in a power of $Y$, using the standard Tychonoff's embedding theorem:

Let $i: X \to Y$ be the inclusion map, and for every closed set $C$ of $X$ let $f_C: X \to Y$ be defined by $f(x)=p$ for $x \in C$, $f(x)=x$ for $x \notin C$. Note that all $f_C$ are continuous when $X$ is Fréchet, as all inverse images of singletons are closed in $X$.

Then the family $$\mathcal{F}=\{i\} \cup \{f_C: C \text{ closed in } X\}$$ separates points and points and closed sets in $X$ and so $X$ embeds into $Y^{\mathcal{F}}$ in the standard way.

Of course every Fréchet space embeds into the power of a Fréchet space (itself, trivially), but the above argument shows we can find one such space for every countable Fréchet space (and for every cardinality class, in fact).

The second part seems to refer to a classic fact due to Herrlich that for any Fréchet space $X$ there is a non-trivial $T_3$ space $Z(X)$ (so certainly $T_1$) (having more than 1 point) such that all continuous maps from $Z(X)$ to $X$ are constant. If you read German, you can read it here (click access to full text).

This implies that we cannot have a universal $T_1$ space $Y$ such that every Fréchet space $X$ embeds into a power of $Y$ (the problem is that one $Y$ has to work for all $X$, instead of being able to choose $Y$ based on $X$):

Suppose we had such a $Y$ and consider the $Z(Y)$ from above and suppose we had an embedding $e: Z(Y) \to Y^I$ for some index set $I$. Then for each $i: p_i \circ e: Z(Y) \to Y$ is continuous (where $p_i$ is the continuous projection onto the $i$-th coordinate) and hence constant. Which makes $e$ a constant map, contradicting injectivity and non-triviality of $Z(Y)$.

Note that it is well-known that two point Sierpinski space is "power-universal" for $T_0$ spaces and $[0,1]$ for completeley regular $T_1$ spaces, so the $T_1$ (and the same argument applies to $T_2$ and $T_3$ classes) case is unfortunate..

Maybe we can find a co-finite $Z(Y)$ for a Fréchet $Y$? I'm not quite sure about that yet.