I'm having a trouble calculating the cardinality of the set of all functions $f:\mathbb{R} \longrightarrow \mathbb{R}$ which have at most $\aleph_0$ discontinuities (let's call the set $M$). A hint is given: Map each function $f$, with a countable set of discontinuities, $d$, to the ordered pair $(f|_d, f|_{R\setminus d})$.

I'm not entirely sure how to go on from here. I know I can create a bijective function $h: M\longrightarrow \bigcup\limits_{d\in D} A_d \times B_d$ where $D$ is the set of all countable subsets of $\mathbb{R}$, $A_d$ is the set of all real functions $f_A:d\longrightarrow \mathbb{R}$, and $B_d$ is the set of all functions $f_B:\mathbb{R} \setminus d \longrightarrow \mathbb{R}$ which do not have discontinuities (I'm not sure if the definition for $B_d$ is correct). Assuming it is correct, I know that $|A_d|=\aleph, \ \ \forall d\in D$, but I do not know how to calculate the cardinality of $B_d$.

I would appreciate any help as to how to go on from here, or hints for better methods.

  • $\begingroup$ If $d$ is a countable subset of $\mathbb{R}$, what do you know about $\mathbb{R}\setminus d$? $\endgroup$ Commented Dec 30, 2015 at 19:09
  • $\begingroup$ Every such function is the difference of two monotonic functions.See Devinatz, Advanced Calculus. $\endgroup$ Commented Dec 30, 2015 at 22:25

2 Answers 2


Hint: fix a dense countable subset of $\Bbb R\setminus d$. Then proceed as in a proof of how many continuous functions $\Bbb R\rightarrow\Bbb R$ there are.

As for constructing such a dense subset: it might be in part unsatisfactory, because I can't think of a construction not using axiom of (countable) choice. Anyways, here is the construction:

Let $U_1,U_2,...$ be the sequence of all intervals with both endpoints rational. For every $i$, choose an element $x_i\in U_i\setminus d$ (easy question: why does such an element exist?). Clearly the set of all $x_i$ is countable and subset of $\Bbb R\setminus d$. I claim it's dense. For any real numbers $a<b$, we can choose rational numbers $c,d$ with $a<c<d<b$. Now, some element $x_i$ of our set is contained in interval $(c,d)$, so $a<x_i<b$. Hence $\{x_i\}$ is dense, countable subset of $\Bbb R\setminus d$.

  • $\begingroup$ So basically, cardinality of $B_d$ is $\aleph$ right? Can you give me a hint on how to calculate the cardinality of the union when the sets of the union are not pairwise disjoint? Thanks :) $\endgroup$
    – Dylan132
    Commented Dec 30, 2015 at 21:08
  • $\begingroup$ For that, let me give you another hint: cardinality of union of sets is at most the sum of cardinalities of these sets (i.e. if the sets aren't disjoint, the sum of cardinalities won't be larger as if they were disjoint). $\endgroup$
    – Wojowu
    Commented Dec 30, 2015 at 21:20
  • $\begingroup$ Yes I figured that out thanks a lot though ! $\endgroup$
    – Dylan132
    Commented Dec 30, 2015 at 21:23
  • $\begingroup$ If I may bother you yet again, you said to fix a dense countable set in R\d, but how do I build one or prove that there exist such sets? Is there a theorem I can rely on? $\endgroup$
    – Dylan132
    Commented Dec 30, 2015 at 22:06
  • $\begingroup$ @Dylan132 I have added this to my answer. $\endgroup$
    – Wojowu
    Commented Dec 30, 2015 at 22:16

Hint to prove is to show Card $C(\mathbb{R},\mathbb{R}) $=card$C(\mathbb{Q^c},\mathbb{R})$

Step One: define function $h: C(\mathbb{R},\mathbb{R})\longrightarrow C(\mathbb{Q^c},\mathbb{R})$ such that $h$ restrict function to $Q^c$ which keeps continuity

then use density of irrational number in real number to show $h$ is injective so $C(\mathbb{R},\mathbb{R})$< card$C(\mathbb{Q^c},\mathbb{R})<c$

step 2 : on the other side consider fixed constant function which is denoted by $K(\mathbb{R},\mathbb{R})$ clearly $c=$Card $C(\mathbb{R},\mathbb{R}) $< Card $C(\mathbb{R},\mathbb{R}) $

from both step result is clear card$C(\mathbb{Q^c},\mathbb{R})=c$


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