3

I defined a function $f$ as follows:

Let $I_{n,k}=\left(x_1,x_2\right)$ the removed open interval in step $n$ at position $k$ of the $\frac{1}{3}$-cantor set and $J_{n,k}=\left(y_1,y_2\right)$ the removed open interval in step $n$ at position $k$ of a arbitrarily fat cantor set. Map the closure of $I_{n,k}$ onto the closure of $J_{n,k}$ like:

$$f:clo(I_{n,k}) \rightarrow clo(J_{n,k}): x\mapsto \begin{cases}f(x):=y_1&x=x_1\\f(x):=y_2&x=x_2\\f(x):=\frac{y_2-y_1}{x_2-x_1}\cdot x & x\in\left(x_1,x_2\right)\end{cases}$$

Due to the construction $f$ is bijective and continuous at $\left(x_1,x_2\right)$ for every $I_{n,k}$ but I struggle with the continouity at the endpoint of the intervals $I_{n,k}$. The fact that I want to map onto the fat cantor set it's not possible to argue with $\left(\frac{1}{2}\right)^n$. What am I missing?

For example:

$clo\left(I_{1,1}\right)=clo\left(\left(\frac{1}{3},\frac{2}{3}\right)\right)=\left[\frac{1}{3},\frac{2}{3}\right]$ maps to $clo\left(J_{1,1}\right)=\left[y_1,y_2\right]$ with the definition of $f$, repeating for every $n \in \mathbb{N}$ with the inherent intervals.

That's what I am thinking:

Sketch of homeomorphism

So $f\left( \frac{1}{3}\right)=y_1$, $f\left(\frac{2}{3}\right)=y_2$ and for all $x \in I_{1,1}$ is $f\left(x\right)=\frac{y_2-y_1}{\frac{2}{3}-\frac{1}{3}}=3(y_2-y_1) \cdot x$.

monoid
  • 608
  • Your mapping goes between removed intervals and not between the discontinua. – user87690 Jul 06 '14 at 08:59
  • But with the first two definitions I think to map the endpoints of the closure of $I_{n,k}$. But with those I get just countable but infinite members of $\mathcal{C}$. The linear function in between should take the uncountable rest of $\mathcal{C}$ and map them bijective. – monoid Jul 06 '14 at 09:04
  • But you you have defined those linear parts on excluded open intervals which are not part of $\mathcal{C}$. – user87690 Jul 06 '14 at 10:49
  • In other words: you have defined bunch of maps $f_{n, k}: \overline{I_{n, k}} \to \overline{J_{n, n}}$. But how do you want to get the homeomorphism between Cantor sets from them? – user87690 Jul 06 '14 at 10:51
  • See http://math.stackexchange.com/questions/429674/homeomorphism-between-two-cantor-sets?rq=1. – Moishe Kohan Jul 06 '14 at 12:08
  • Thanks but your link uses topology, base of a topology, etc. which I am not allowed to use. – monoid Jul 06 '14 at 12:21
  • 2
    That's like being told not to use a hammer when you have to put in a nail. – Lee Mosher Jul 06 '14 at 14:22
  • Oh, I think the function must be $f(x) = \frac{y_2-y_1}{x_2-x_1}\cdot \left(x-x_1\right) + y_1$ with $x \in I_{n,k}$? – monoid Jul 06 '14 at 14:44
  • @LeeMosher Flawed comparison. It's reasonable to demonstrate the existence of a homeomorphism between two concrete subsets of $\mathbb R$ without invoking an abstract infinite product of topological spaces. –  Jul 06 '14 at 14:52
  • @Thisismuchhealthier: My comment refers not to any particular link, but to the OP's comment that he is not allowed to use the basic concepts of topology. – Lee Mosher Jul 06 '14 at 15:21

1 Answers1

4

Your idea is good. We already know that a continuous strictly increasing function is a homeomorphism. If we can find such a function $f:[0,1]\to [0,1]$ which carries the complement of one set onto the complement of another, this function will also carry one Cantor set onto the other.

But the task of piecing an infinite collection of linear functions together, checking the continuity of the whole thing, is rather tedious. I suggest building $f$ as a uniform limit of continuous maps $f_n$, following the process of creation of Cantor sets themselves. That is, let $C_n$ and $F_n$ be pre-Cantor sets of $n$th generation; with $C_0=F_0=[0,1]$. Let $f_0(x)=x$. Having defined $f_n$, proceed to $f_{n+1}=f_n$ on $[0,1]\setminus C_n$ and $f_{n+1}$ being an appropriate three-chain broken line on each component of $C_n$.

Alternatively, define $f_n$ as the unique increasing piecewise linear function on $[0,1]$ that maps the boundary points of $C_n$ bijectively onto the boundary points of $F_n$, and only changes the slope at these points.

This is essentially what you do already with linear pieces, with the difference that each $f_n$ is defined on all $[0,1]$.

By construction, $\sup |f_{n+1}-f_{n}| \le \max_k |J_{n,k}|\le 2^{-n}$. (I use $|\cdot|$ to denote length, and assume $J_{n,k}|$ is the same for all $k$, as on the picture.) Hence $\sup |f_{n+1}-f_{n }|\le 2^{-n}$, which implies that the sequence $f_n$ converges uniformly, by the Weierstrass $M$-test applied to $\sum (f_{n+1}-f_n)$. Let $f$ be its limit. Since $f_n(C_m)\subset F_m$ for all $n\ge m$, we have $f(C_m)\subset F_m$. Hence $f(C)\subset F$.

Everything in the preceding paragraph applies to the inverse maps $f_{n}^{-1}$: namely, $\sup |f_{n+1}^{-1}-f_{n}^{-1}| \le \max_k |I_{n,k}|$, and so on. Let $g= \lim_{n\to\infty} f_n^{-1}$. Then both $g\circ f = \lim_n g_n\circ f_n$ and $f\circ g = \lim_n f_n\circ g_n$ are identity maps. So, $g$ is the inverse of $f$.