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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.

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

1 Answer 1

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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 $y \in V _N(x)$.

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