Is $\{0,1\}^{\Bbb N}$ homeomorphic to the Cantor Set?

[Chris L T521]

This will be my last Graduate POTW submission here on MHB; I'm running out of ideas (after doing this for 124 weeks), and I think it's time to get someone fresh in here to do things from now on. It's been a pleasure doing this for roughly 2.5 years now, and I hope you guys give the person who will be taking my place the same kind of support I've received from you during my time running the Graduate POTW.

Anyway, here's this week's problem!

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Problem: Let $X$ denote the discrete topological space with two elements. Show that $X^{\mathbb{N}}$ is homeomorphic to the Cantor Set.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge. You can find his solution below.

[sp]It suffices to show that $\{0,1\}^{\Bbb N}$ is homeomorphic to the Cantor set, $C$. Define a map $f : \{0,1\}^{\Bbb N} \to C$ by setting

$$f((a_n)) = \sum_{n = 1}^\infty \frac{2a_n}{3^n}.$$

It has an inverse map $g : C \to \{0,1\}^{\Bbb N}$ given by

$$g\left(\sum_{n = 1}^\infty \frac{a_n}{3^n}\right) = \left(\frac{a_n}{2}\right).$$

Thus $f$ is bijective. Furthermore, $f$ is continuous. For consider the metric $d$ on $\{0,1\}^{\Bbb N}$ given by

$$d((a_n), (b_n)) = \sum_{n = 1}^\infty \frac{|a_n - b_n|}{2^n}.$$

Given $\varepsilon > 0$, set $\delta = \frac{\varepsilon}{2}$. For all $(a_n), (b_n)\in \{0,1\}^{\Bbb N}$, $d((a_n), (b_n)) < \delta$ implies

$$|f(a_n) - f(b_n)| \le 2\sum_{n = 1}^\infty \frac{|a_n - b_n|}{3^n} \le 2\, d((a_n), (b_n)) < \varepsilon.$$

Since $\varepsilon$ was arbitrary, continuity of $f$ follows. As $\{0,1\}^{\Bbb N}$ is compact (by Tychonoff's theorem), $f$ is a bijective continuous map of a compact space onto the Hausdorff space $C$, so $f$ is a homeomorphism.[/sp]