# Show that the Cantor space is homeomorphic to the Cantor set, thought of as a subspace of $\left[0,1\right]$.

Let $$X$$ be the Cantor space and let $$Y$$ be the Cantor middle thirds set. Define the function $$f:X\rightarrow Y$$ such that:

$$f(x) = f((x_0, x_1, ...)) = 0.y_0y_1...\quad\text{where }\begin{cases} y_i = x_i&\text{if }x_i = 0 \\ y_i = 2&\text{if }x_i = 1 \end{cases}$$

where the output is a decimal expansion in ternary base. Clearly this is a bijective function since the Cantor set does not contain any numbers with decimal expansion in ternary base containing something other than 0 or 2. Let $$U$$ be open in $$Y$$, then under the product topology all but finitely many coordinates of any $$x\in U$$ are the same for each $$x$$. So there exists a $$n$$ such that for all $$x\in U$$, and for all $$N\geq n$$ $$x_N$$'s are equal. Observe that each element in $$f^{-1}(U)$$ is then of the form $$0.y_0y_1...y_ny_\alpha y_\beta...$$, where $$y_0,...,y_n$$ are fixed but $$y_\alpha, y_\beta,...$$ can be either 0 or 2. So $$f^{-1}(U)$$ is some $$\epsilon/\left(3^n\right)$$-ball intersected with the Cantor set, generating an open set in the subspace topology inherited from $$\left[0,1\right]$$. Hence $$f$$ is continuous. Let $$V$$ be open in $$X$$, then $$V$$ is some $$\epsilon'$$-ball intersecting with the Cantor set. Note then that $$V = \epsilon/\left(3^n\right)$$, where $$n = \min\left\lbrace i \right\rbrace$$ such that $$\epsilon/\left(3^i\right)\leq\epsilon'$$. Therefore elements in $$V$$ are of the form $$0.y_0y_1...y_ny_\alpha y_\beta...$$, where $$y_0,...,y_n$$ are fixed but $$y_\alpha, y_\beta,...$$ can be either 0 or 2. So $$f(V) = (f^{-1})^{-1}$$ consists of elements $$x$$ such that for all $$N\geq n$$ $$x_N$$'s are equal. Hence $$f(V)$$ is open in the product topology on the Cantor space. Hence $$f^{-1}$$ is continuous. This shows that $$f$$ is a homeomorphism.

I would like some advice on how to clean up the proof or if I forgot anything in the proof because it looks somewhat messy to me.

• You are confusing X and Y in your post. Otherwise yes it is the proof. Jan 29, 2020 at 6:18
• You should tell us what the the Cantor space is. It seems to be the infinite product of copies of $\{0,1\}$. Jan 29, 2020 at 10:58

Unfortunately your proof is not correct.

As reuns observed in his comment, in your arguments you have to exchange $$X$$ and $$Y$$ ("Let $$U$$ be open in $$Y$$, then under the product topology ..." does not make sense). But that is easily repaired.

A more serious mistake is this:

Let $$U$$ be open in $$X$$, then under the product topology all but finitely many coordinates of any $$x \in U$$ are the same for each $$x$$. So there exists an $$n$$ such that for all $$x \in U$$ and for all $$N \ge n$$ the $$x_N$$'s are equal.

This is not true. In fact, if $$x \in U$$, there exists a basic open $$V = \prod_{i=1}^\infty V_i$$ (i.e. $$V_i \subset \{0,1\}$$ open such that $$V_i = \{0,1\}$$ for almost all $$i$$) with $$x \in V$$. This means that for $$x \in U$$ the coordinates $$x_N$$ are arbitrary for $$N \ge n$$.

On the other hand, it seems that in the rest of your proof you do not use that for all $$N \ge n$$ the $$x_N$$'s are equal, but that they are arbitrary!

To correct your proof, I recommend to write for $$x = (x_1,x_2,x_3\ldots)$$ [note that I start the sequence with index $$1$$] $$f(x) = \sum_{i=1}^\infty \frac{2x_i}{3^i} .$$ Let us moreover define $$f_N(x) = \sum_{i=N}^\infty \frac{2x_i}{3^i} .$$ Then for all $$x, y$$ we have $$\lvert f_N(x) - f_N(y) \rvert = \left\lvert \sum_{i=N}^\infty \frac{2x_i - 2y_i}{3^i} \right\lvert \le \sum_{i=N}^\infty \frac{2}{3^i} = \frac{1}{3^{N-1}} .$$ Given $$x$$, let $$V_i = \{ x_i\}$$ for $$i < N$$ and $$V_i = \{ 0,1\}$$ for $$i \ge N$$. Then $$V_N(x) = \prod_{i=0}^\infty V_i$$ is an open neigborhood of $$x$$ such that for $$y \in V_N(x)$$ we have $$y_i = x_i$$ for $$i < N$$.

You can easily see that $$f$$ is continuous because for $$y \in V_N(x)$$ $$\lvert f(x) - f(y) \rvert = \lvert f_N(x) - f_N(y) \rvert \le \frac{1}{3^{N-1}} .$$ To see that $$f^{-1}$$ is continuous it suffices to show that for each $$x$$ and each $$N$$ there exists $$\epsilon > 0$$ such that $$\lvert f(x) - f(y) \rvert < \epsilon$$ implies $$y \in V_N(x)$$.

So let $$\epsilon = \frac{1}{3^{N-1}}$$. We have $$\left\lvert\frac{2x_1 - 2y_1}{3} \right\lvert = \left\lvert f(x) - f_2(x) - f(y) + f_2(y) \right\rvert \le \left\lvert f(x) - f(y) \right\rvert + \left\lvert f_2(x) - f_2(y) \right\rvert < \frac{1}{3} + \frac{1}{3} = \frac{2}{3}.$$ This implies $$x_1 = y_1$$ because otherwise $$\left\lvert\frac{2x_1 - 2y_1}{3} \right\lvert = \frac{2}{3}$$ which is impossible.

Proceed inductively to show that $$\left\lvert\frac{2x_i - 2y_i}{3} \right\lvert < \frac{2}{3^i}$$ for $$i < N$$ (using that $$x_j = y_j$$ for $$j=1,\ldots,i-1$$). Thus $$y_i = x_i$$ for $$i < N$$. This shows $$y \in V _N(x)$$.