## Sierpinski’s Theorem

For any function there exist two functions such that and have the **Darboux property**.

A function has the **Darboux property** if for any interval we have 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 considered as a vector space over is . Define . This is obviously an equivalence relation and for any we will denote the equivalence class which contains . It is obvious that and . In the following, we will denote the set of the equivalence classes, and we will find its cardinal number. We choose a basis of $\mathbb{R}$ over which contains 1 (this is possible because any linearly independent set can be extended to a basis and is such set). If two different numbers, say and from would be in the same equivalence class, then we have , so and are linear dependent. Contradiction.

We take a function. We consider a bijection and we denote . Thus , and we can define the bijections and .

We define the functions:

It is clear from their definition that and satisfy . Let’s prove that and have the Darboux property. We consider an interval. Because any equivalence class in is dense in , we find that . Therefore, looking at the definitions of and we see that and , for all , yielding . We conclude that and have the Darboux property.

We saw that for any interval so the functions and are not continuous at any point in , because which is not a subset of for all .

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!

Glad you liked it. You can take a look at the following problem also, since it uses the same method: https://mathproblems123.wordpress.com/2009/09/08/contest-problem/