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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

  1. Finsky Sergey
    December 12, 2012 at 7:28 pm

    Wow! Beautiful, at a first glance it seems that there is a drawback with defining f1, f2 (i didn’t realize that we make a bijection between A1 and R, thought, we are defining it with respect to the equivalence class) but it’s very clean! Thank you!

  1. February 10, 2012 at 12:00 pm
  2. January 26, 2014 at 4:34 pm
  3. March 19, 2016 at 5:18 pm

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