# Thread: Cantor Set Homeomorphic to Product Space

1. For each integer $n$ let $X_n$ denote the two point topological space ${0, 1}$. Then the product space $X = \prod_{n=1}^{\infty} X_i$ is homeomorphic to the Cantor set. (here I'm looking at the usual Cantor Set... the middle third Cantor set)

Here the two point set would have the discrete topology.
The product has the product topology.
And the Cantor set has the subspace topology.

I defined my function $f$ to be,
$f: C \rightarrow X$

$\sum_{n=1}^{\infty} \frac{a_n}{2^n} \rightarrow (\frac{a_1}{2}, \frac{a_2}{2}, \frac{a_3}{2}, ...)$, where $a_n \in \left\{0, 2\right\}$

Then the inverse is...
$f^{-1}: X \rightarrow C$

$x \rightarrow \sum_{n=1}^{\infty} \frac{2 \cdot x_n}{2^n}$, where $x = (x_1, x_2, x_3, ...), x_n \in \left\{0, 1\right\}$

Checking they are inverses:

$f(f^{-1}(x)) = f(\sum_{n=1}^{\infty} \frac{2 \cdot x_n}{2^n}) = (\frac{2 \cdot x_1}{2}, \frac{2 \cdot x_2}{2}, \frac{2 \cdot x_3}{2}, ...) = (x_1, x_2, x_3, ...) = x$

$f^{-1}(f(\sum_{n=1}^{\infty} \frac{a_n}{2^n})) = f^{-1}(\frac{a_1}{2}, \frac{a_2}{2}, \frac{a_3}{2}, ...) = \sum_{n=1}^{\infty} \frac{2 \cdot a_n/2}{2^n} = \sum_{n=1}^{\infty} \frac{a_n}{2^n}$

$\implies$ they are inverses.

$\implies$ f is bijective.

All that is left is to show $f$ and $f^{-1}$ are continuous.
I'm having some trouble showing that $f^{-1}$ is continuous.

Consider an open set in $U \subset C$. Then, $U = C \cap (a, b), (a, b) \subset R$.
It seems straight forward that $f(U) = f(C \cap (a, b)$ would be in $X$. How do I show it would be open in $X$?

2.

3. If a bijective map between compact Hausdorff spaces is continuous then its inverse is automatically continuous. (Proof .)

Originally Posted by Opalg
If a bijective map between compact Hausdorff spaces is continuous then its inverse is automatically continuous. (Proof .)
Yeah, that'd do it! But I guess I'm just wondering what an open set would look like in $X$.

5. Originally Posted by joypav
For each integer $n$ let $X_n$ denote the two point topological space ${0, 1}$. Then the product space $X = \prod_{n=1}^{\infty} X_i$ is homeomorphic to the Cantor set. (here I'm looking at the usual Cantor Set... the middle third Cantor set)

Here the two point set would have the discrete topology.
The product has the product topology.
And the Cantor set has the subspace topology.

I defined my function $f$ to be,
$f: C \rightarrow X$

$\sum_{n=1}^{\infty} \frac{a_n}{2^n} \rightarrow (\frac{a_1}{2}, \frac{a_2}{2}, \frac{a_3}{2}, ...)$, where $a_n \in \left\{0, 2\right\}$

Then the inverse is...
$f^{-1}: X \rightarrow C$

$x \rightarrow \sum_{n=1}^{\infty} \frac{2 \cdot x_n}{2^n}$, where $x = (x_1, x_2, x_3, ...), x_n \in \left\{0, 1\right\}$

Checking they are inverses:

$f(f^{-1}(x)) = f(\sum_{n=1}^{\infty} \frac{2 \cdot x_n}{2^n}) = (\frac{2 \cdot x_1}{2}, \frac{2 \cdot x_2}{2}, \frac{2 \cdot x_3}{2}, ...) = (x_1, x_2, x_3, ...) = x$

$f^{-1}(f(\sum_{n=1}^{\infty} \frac{a_n}{2^n})) = f^{-1}(\frac{a_1}{2}, \frac{a_2}{2}, \frac{a_3}{2}, ...) = \sum_{n=1}^{\infty} \frac{2 \cdot a_n/2}{2^n} = \sum_{n=1}^{\infty} \frac{a_n}{2^n}$

$\implies$ they are inverses.

$\implies$ f is bijective.

All that is left is to show $f$ and $f^{-1}$ are continuous.
I'm having some trouble showing that $f^{-1}$ is continuous.

Consider an open set in $U \subset C$. Then, $U = C \cap (a, b), (a, b) \subset R$.
It seems straight forward that $f(U) = f(C \cap (a, b)$ would be in $X$. How do I show it would be open in $X$?
I have not checked all of your working, but once you have checked that $f$ is bijective and continuous, then $f^{-1}$ is automatically continuous by the following general result, which is easy to prove:

Let $f:X\to Y$ be a bijective continuous map from a compact space to a Hausdorff space. Then $f^{-1}$ is also continuous.

Basically what you need to know to prove the above is that continuous image of a compact set is also compact, and that compact subspaces of a Hausdorff space is a closed set in the Hausdorff space.

(PS: I somehow missed opalg's response).