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$A, B$ 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

real analysis - Continuous on rationals, discontinuous on irrationals - Mathematics Stack Exchange:

There is no function that is continuous on the rationals and discontinuous on the irrationals.

real analysis - Does there exists a function $f:\Bbb{R}\to \Bbb{R}$ s.t $f$ is continuous on a dense $G\_{\delta}$ set and discontinuous a.e on $\Bbb{R}$? - Mathematics Stack Exchange

Y.D.X.
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    Did you want such a function for every partition of $\mathbb R$ into two dense sets, or just for some partition into two dense sets? The former is impossible, as noted in comment by @geetha290krm. I could give you an example of the latter but I don't know if that would answer your question. – bof Oct 23 '23 at 11:35
  • @bof I mean the latter. Could you show the example? Thanks! – Y.D.X. Oct 23 '23 at 11:40

1 Answers1

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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.

bof
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