Variants of Dirichlet-type function as a pointwise limit of continuous functions

Question

Problem

Suppose $f$ is a function from a complete metric space $X$ to a metric space $Y$, and suppose $Y$ has points $y_{0}$, $y_{1}$ such that the subsets $f^{-1}(y_{0})$ and $f^{-1}(y_{1})$ are both dense in $Y$.

Show that in the above situation, if $f$ maps every point of $X$ either to $y_{0}$ or to $y_{1}$ and if $f^{-1}(y_{1})$ is countable, then $f$ cannot be expressed as a pointwise limit of functions $X \rightarrow Y$ that are continuous at all points of $f^{-1}(y_{1})$; but that it can be expressed as a pointwise limit of functions $X \rightarrow Y$ that are continuous at all points of $f^{-1}(y_{0})$.

Question

It is well known by the Baire category theorem that $f$ can not be a pointwise limit of continuous functions $X \to Y$. However, I have no clue how to prove $f$ can not be a pointwise limit of functions being continuous at a dense set $f^{-1}(y_{1})$. Moreover, I think the "countable" condition is redundant for the first part since it has been proved that if $f$ is continuous on a dense set, then it is continuous on an uncountable dense set (e.g., see Theorem 1 in 1).

Also, I believe the "countable" condition is only helpful for the 2nd part as follows,

Assume $f^{-1}(y_{1}) = \{x_{1},x_{2},\ldots\}$.

Construct \begin{equation*} f_{n}(x) = \begin{cases} y_{0}, & x \in f^{-1}(y_{0}) \cup \{x_{n+1}, x_{n+2},\ldots\} \\ y_{1}, & x \in \{x_{1},x_{2},\ldots,x_{n}\} \end{cases} \end{equation*} It is easy to check $(f_{n})$ is continuous at $f^{-1}(y_{0})$.

Answer 1

Suppose for a contradiction that the function $f$ is a pointwise limit of a sequence $(f_n)_{n\in\mathbb N}$ of functions from $X$ to $Y$ that are continuous at all points of $f^{-1}(y_{1})$. For each natrual $n$ let $C_n$ be the set of continuity points of the function $f_n$. It is well-known (see, for instance, Proposition 3.6 from [Kech]) that $C_n$ is a $G_\delta$ subset of $X$. Then $C=\bigcap_{n\in\mathbb N} C_n$ is a (dense) $G_\delta$ subset of $X$, and so Baire. That is the restriction $f|C$ is a pointwise limit of the sequence $(f_n|C)_{n\in\mathbb N}$ of continuous restrictions $f_n|C$, $n\in\mathbb N$.

For each natural $n$ put $C_n=\{x\in C:|f_m(x)-f(x)|\le 1/n$ for all natural $m\ge n\}$. Then $C$ is a union $\bigcup_{n\in\mathbb N} C_n$ of a nondecreasing family $\{C_n:n\in\mathbb N\}$. Since $C=\bigcup_{n\in\mathbb N} \overline{C_n}^C$ and $C$ is a Baire space, there exist $n\in\mathbb N$ and a nonempty open subset $U$ of $C$ such that $U\subset \overline{C_n}^C$, that is $C_n$ is dense in $U$. Let $d$ be the metric of the space $Y$ and $\varepsilon=d(y_0,y_1)>0$. Increasing $n$, if needed, we can ensure that $n\ge\tfrac 3\varepsilon$. Since $f^{-1}(y_1)$ is dense in $C$, there exists a point $x\in U\cap f^{-1}(y_1)$. Since the function $f_n$ is continuous at $x$, there exists an open neighborhood $V\subset U$ of $x$ such that $d(f_n(x),f_n(x'))<\tfrac{\varepsilon}{3}$ for each $x'\in V$. Since $f^{-1}(y_0)$ and $C$ are dense $G_\delta$ subsets of Baire space $X$, their intersection $f^{-1}(y_0)\cap C$ is a dense subset of $X$ and so of $C$. Pick any point $x'\in V\cap f^{-1}(y_0)$. Then by the triangle inequality $$d(f(x),f(x'))\le d(f(x),f_n(x))+d(f_n(x),f_n(x'))+(f_n(x'),f(x'))<$$ $$\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}= \varepsilon.$$ Therefore $y_1=f(x)=f(x')=y_0$, a contradiction.

In the proof we used that the set $f^{-1}(y_0)$ is nonmeager (and dense) in $X$ to conclude that $f^{-1}(y_0)\cap C$ is dense in $C$. If we relax this condition then there is a counterexample: in fact, even in this example $f$ is a pointwise limit of functions from $X$ to $Y$ that are continuous at all points of $f^{-1}(y_{0})$.

References

[Kech] Alexander S. Kechris. Classical Descriptive Set Theory , Springer, 1995.