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## Sierpinski’s Theorem

For any function $f:\mathbb{R} \to \mathbb{R}$ there exist two functions $f_1, \ f_2$ such that $f=f_1+f_2$ and $f_1,f_2$ have the Darboux property.
A function has the Darboux property if for any interval $I\subseteq \mathbb{R}$ we have $f(I)$ is also an interval. This is slightly different from continuity and intermediate value property. Cotinuity implies Darboux and Darboux implies Intermediate value property.

Proof: It is well known that the dimension of $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$ is $\aleph=card \mathbb{R}$. Define $x \sim y \Leftrightarrow x-y \in \mathbb{Q}$. This is obviously an equivalence relation and for any $x \in \mathbb{R}$ we will denote $[x]=\{y \in \mathbb{R} : y \sim x\}$ the equivalence class which contains $x$. It is obvious that $\mathbb{R}=\displaystyle \bigcup_{x \in \mathbb{R}}[x]$ and $y \notin [x]\Rightarrow [x]\cap [y]=\emptyset$. In the following, we will denote $\mathcal{A}=\{[x] : x \in \mathbb{R}\}$ the set of the equivalence classes, and we will find its cardinal number. We choose a basis $\mathcal{B}$ of $\mathbb{R}$ over $\mathbb{Q}$ which contains 1 (this is possible because any linearly independent set can be extended to a basis and $\{1\}$ is such set). If two different numbers, say $x$ and $y$ from $\mathcal{B}$ would be in the same equivalence class, then we have $x-y=q\cdot 1,\ q \in \mathbb{Q}$, so $1,x$ and $y$ are linear dependent. Contradiction.

We take $f:\mathbb{R} \to \mathbb{R}$ a function. We consider a bijection $g :\mathbb{R} \to \mathcal{A}$ and we denote $\mathcal{A}_1=g((-\infty,0)),\ \mathcal{A}_2=g([0,\infty))$. Thus $card \mathcal{A}_1=card \mathcal{A}_2=\aleph$, and we can define the bijections $m: \mathbb{R}\to \mathcal{A}_1$ and $n:\mathbb{R}\to \mathcal{A}_2$.

We define the functions:
$f_1(t)=\begin{cases} r&, t \in m(r) \\ f(t)-r &, t \in n(r)\end{cases}$
$f_2(t)=\begin{cases} f(t)-r & ,t \in m(r) \\ r & , t \in n(r)\end{cases}$

It is clear from their definition that $f_1$ and $f_2$ satisfy $f=f_1+f_2$. Let’s prove that $f_1$ and $f_2$ have the Darboux property. We consider $I \subset \mathbb{R}$ an interval. Because any equivalence class in $\mathcal{A}$ is dense in $\mathbb{R}$, we find that $I \cap m(r)\neq \emptyset,\ I \cap n(r)\neq \emptyset,\ \forall r \in \mathbb{R}$. Therefore, looking at the definitions of $f_1$ and $f_2$ we see that $r \in f_1(I \cap m(r))\subset f(I)$ and $r \in f_2(I\cap n(r))\subset f(I)$, for all $r \in \mathbb{R}$, yielding $f_1(I)=f_2(I)=\mathbb{R}$. We conclude that $f_1$ and $f_2$ have the Darboux property.

We saw that $f_1(I)=f_2(I)=\mathbb{R}$ for any interval $I$ so the functions $f_1$ and $f_2$ are not continuous at any point $x_0$ in $\mathbb{R}$, because $f((x_0-\delta,x_0+\delta))=\mathbb{R}$ which is not a subset of $(f(x_0)-\varepsilon ,f(x_0)+\varepsilon)$ for all $\varepsilon >0$. $\square$