Cardinality of the set of all real functions which have a countable set of discontinuities

Question

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.

Answer 1

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 $\Bbb R\setminus d$.

Answer 2

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$C(\mathbb{Q^c},\mathbb{R})=c$