Essentially discontinuous on a dense subset of $\mathbb{R}$, but continuous on the complement which is also dense

Question

$A, B$ is a partition of $\mathbb{R}$, and both $A$ and $B$ are dense. Is there a function $f: \mathbb{R} \to \mathbb{R}$, that is essentially discontinuous ($\lim f$ does not exist) at everywhere in $B$, and continuous at everywhere in $A$?

(I mean if there exists $A,B,f$ as a combination that satisfies the condition. In other words, I want either a proof about “for any partition such f does not exists”, or a positive example.)

The popcorn function (aka. Thomae's function) is discontinuous on $\mathbb{Q}$ and continuous on $\mathbb{R} \setminus \mathbb{Q}$, but not a positive example: Those are all removable discontinuities, because $\forall x_0\in \mathbb{R},\, \lim_{x \to x_0} f(x) \equiv 0$.

My attempt

I guess such a function does not exist, so I am trying to prove “continuous on $A$ $\implies \lim f$ exists on $B$”.

There exists a list of sequences $\{\beta_{ab}\}$, where

• $\beta_{ab} \in B$,
• $\lim_{b \to +\infty} \beta_{ab} = \alpha_a \in A$,
• $\lim_{a \to+\infty} \alpha_a = \beta \in B$.

I try to construct another sequence $\{\beta_{a_n,b_n}\}_{n}$ whose limit is $\beta$. $$ |\beta_{a_n,b_n} - \beta| \leq |\beta_{a_n,b_n} - \alpha_{a_n}| + |\alpha_{a_n} - \beta| \leq \varepsilon' + \varepsilon. $$ I get stuck here…

Relevant questions

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Answer 1

Let $B_1,B_2,B_3,\dots$ be an infinite sequence of pairwise disjoint nowhere dense perfect sets (closed sets without isolated points) whose union $B=\bigcup_{n=1}^\infty B_n$ is dense in $\mathbb R$.

(To construct such a sequence, let $I_1,I_2,I_3,\dots$ be an enumeration of the open intervals with rational endpoints, and recursively construct a Cantor set $B_n\subseteq I_n\setminus(B_1\cup\cdots\cup B_{n-1})$.)

The set $A=\mathbb R\setminus B$ is also dense in $\mathbb R$ by the Baire category theorem. Define $f(x)=0$ if $X\in A$ and $f(x)=\frac1n$ if $x\in B_n$. The function $f$ is continuous at each point of $A$ because the sets $B_n$ are closed; $f$ does not have a limit at any point of $B$ because $A$ is dense and $B_n$ has no isolated points.