Locus of Continuity of a Function between Polish Spaces I think the question is maybe a bit trivial, but right now my head doesn't work that much well so I'm asking y'all to have some clarification.
I know that, for a subset $A\subseteq \mathbb R$, the following are equivalente:
• there exists a function $f:\mathbb R\to\mathbb R$ such that the locus of continuity of $f$ is exactly $A$;
• $A$ is a G-delta subset of $\mathbb R$ (so, equivalently, $A$ is Polish).
My question is: how do I generalize this to a generic Polish space $X$? So, which of the following ones are equivalent to $A\subseteq X$ is a G-delta subset of $X$?
a) there exists a function $f:X\to X$ such that the locus of continuity of $f$ is exactly $A$;
b) there exists a Polish space $Y$ and a function $f:X\to Y$ such that the locus of continuity of $f$ is exactly $A$;
c) for all Polish spaces $Y$ there exists a function $f:X\to Y$ such that the locus of continuity of $f$ is exactly $A$.
Thank you in advance and sorry, this is one of my first questions here and I'm not English so my language is bad
 A: Let $X, Y$ be topological spaces.
First a few definitions:

a) if $f: X \rightarrow Y$ is a function,
$S(f) := \{x \in X: f \text{ is continuous at } x\}$
(= locus of continuity of $f$).

b) $I(X) := \{x \in X: \{x\} \text{ is open}\}$ (= set of isolated points of $X$).

c) $X$ is resolvable, if there exist two dense, disjoint subsets $A, B$ of $X$.

d) $X$ is quasi-resolvable, if
$X \setminus \overline{I(X)}$ is resolvable.
Part a) of the following lemma is well-known (and not very difficult to see) and b) is an immediate consequence:  
Lemma 1: Let $X$ be T0 and first countable.

a) If $I(X) = \emptyset$, then X is resolvable. 
b) $X$ is quasi-resolvable.
The following is easy to see: 
Lemma 2: Let $f: X \rightarrow Y$ be a function. 
a) $I(X) \subset S(f)$.  
b) If $Y$ is discrete, then $S(f)$ is open in $X$.
Theorem 3: Let $f: X \rightarrow Y$ be a function, $Y$ metrizable. Then $S(f)$ is a $G_\delta$ subset of $X$.
Proof.
Let $d: Y \times Y \rightarrow \mathbb{R}^{\ge 0}$ be a metric on Y, which induces the given topology.
Let $S := S(f)$.  
Let $k \in \mathbb N, x \in S$.
Then there exists an open subset $U_{x, k}$ of $X$ such that $d(f(x), f(u)) < \frac{1}{k}$ for all $u \in U_{x,k}$.  
It is $S \subset G_k := \bigcup_{x \in S} U_{x,k}$ open in $X$. Hence $S \subset \bigcap_{k \in \mathbb N} G_k$ is a $G_\delta$ subset of $X$.
It is not difficult to see that
$\bigcap_{k \in \mathbb N} G_k \subset S$.
Note: In case of $X = Y = \mathbb R$ this is the well-known Young's theorem. The above proof follows its standard proof. Sorry, I only found a reference to German Wikipedia.
Theorem 4: Let X be quasi-resolvable and Y first countable, T2, not discrete. Let S be a $G_\delta$ subset of $X$, $I(X) \subset S$. Then there exists a function $f: X \rightarrow Y$ such that $S(f) = S$.
Proof.
Choose $y \in Y \setminus I(Y)$.
Since $Y$ is first countable and T2, there exist sequences
$(a_n)_{n \in \mathbb N}, (b_n)_{n \in \mathbb N}$ in
$Y \setminus\{y\}$ such that
$(a_n)_n, (b_n)_n \rightarrow y$ and
$\{a_n: n \in \mathbb N\} \cap \{b_n: n \in \mathbb N\} = \emptyset$.

Since $X$ is quasi-resolvable, there exists a subset $T$ of $X$, such that for every open set $U$ of $X$ which is not contained in $I(X)$, we have
$U \cap T \neq \emptyset$ and $U \cap (X \setminus T) \neq \emptyset$.

Let $X \setminus S = \bigcup_{n \in \mathbb N} F_n$ with
$F_n$ closed in $X$ for each $n \in \mathbb N$.
For $x \in X \setminus S$ define
$n(x) := \text{min} \{n \in \mathbb N: x \in F_n \}$.
Now define
$f: X \rightarrow Y, 
f(x) = \begin{cases} 
  a_{n(x)}, & x \in (X \setminus S) \cap T \\ 
  b_{n(x)}, & x \in (X \setminus S) \setminus T \\
  y, & x \in S
\end{cases}$ 
It is not too difficult to show that $S(f) = S$.
Note: the above function is a modification of the Thomae's function.
Corollary: Let X be metrizable, $A$ a subset of $X$. The following are equivalent:

*

*$A$ is a $G_\delta$ subset of $X$ and $I(X) \subset 
   A$.

*There exists a function $f: X \rightarrow X$ such
that $S(f) = A$.

*For every first countable T2 topological space $Y$,
which is not discrete, there exists
a function $f: X \rightarrow Y$ such that
$S(f) = A$.

*There exists a function $f: X \rightarrow \mathbb R$
such that $S(f) = A$.

*There exists a function
$f: X \rightarrow \{\frac{1}{n}: n \in \mathbb N \} \cup \{0\}$
such that $S(f) = A$.

