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I will denote the Cantor set as $C$. We have proved earlier that every $x\in C$ can be uniquely written in a ternary representation $x=0.a_1a_2a_3...$ where all the $a_i \in \{0,2\}$.

Now we consider the function $$f:C\to C^2 \\ 0.a_1a_2a_3... \mapsto \left( 0.a_1a_3a_5... , 0.a_2a_4a_6... \right)$$ I would like to show that $f$ is continuous.

I appreciate any advice/hints.

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  • $\begingroup$ Why do you think your proof might not work if you tried to apply it instead to the case $f\colon [0,1]\to [0,1]^2$? In this case the map is not continuous. I would suggest going back to the definition of continuity. $\endgroup$ – Dan Rust Oct 21 '14 at 15:00
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Let $f=(f_1,f_2)$, with $f_1,f_2: C\to C$.

Clearly, if $\lvert x-y\rvert<\delta$, and $\delta<\dfrac{2}{3^{2n+1}}$, for some $n$, then the first $2n+1$ digits in the ternary expansions of $x$ and $y$ agree, and hence so do the first $n$ ternary digits of $f_i(x)$ and $f_i(y)$, $i=1,2$, and thus $$ \lvert\, f_1(x)-f_1(y)\rvert,\,\lvert\, f_2(x)-f_2(y)\rvert<\dfrac{2}{3^{n}}. $$ Hence, for every $\varepsilon>0$, there exists a $\delta=\dfrac{2}{3^{k}}$, where $$ k=2\lfloor\log_3 (\varepsilon/2)\rfloor+1, $$ such that...

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  • $\begingroup$ Thank you! Could you please elaborate why $$\lvert\, f_1(x)-f_1(y)\rvert,\,\lvert\, f_2(x)-f_2(y)\rvert<\dfrac{2}{3^{n}}$$? $\endgroup$ – rehband Oct 21 '14 at 16:57
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    $\begingroup$ See my edited answer. $\endgroup$ – Yiorgos S. Smyrlis Oct 21 '14 at 17:21
  • $\begingroup$ Sorry, I still don't see it :( If we write, using the ternary expansions, $$|x-y|=\left| \sum_{k=1}^{\infty} (a_k - b_k) 3^{-k} \right| < \frac{2}{3^{2n+1}}$$ why does it follow that $a_i = b_i$ for $i=1,2,...,2n+1$ ? I also have trouble seeing how the proof continues. Sorry and thank you for the patience. $\endgroup$ – rehband Oct 21 '14 at 17:55
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    $\begingroup$ If the $k-$th ternary digit of $x$ and $y$ are different, then $\lvert x-y\rvert\ge 3^{-k}$. In particular, if the ternary digits of $x$ and $y$ are 0 or 2, the their $k-$ digit is different, then $\lvert x-y\rvert\ge 2\cdot3^{-k}$ $\endgroup$ – Yiorgos S. Smyrlis Oct 21 '14 at 18:11
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    $\begingroup$ It fixed it - See new value of $k$ - It $\delta<2\cdot 3^{-k}$, and $\lvert x-y\rvert<\delta$, then the first $k$ ternary digits of $x$ and $y$ agree and hence the first $(k+1)/2=\lfloor\log_3(\varepsilon/2)\rfloor+1$ digits of $f_1(x)$ and $f_1(y)$ agree, and hence $\lvert f_1(x)-f_1(y)\rvert<2\cdot 3^{(k+1)/2}<\varepsilon$. $\endgroup$ – Yiorgos S. Smyrlis Oct 21 '14 at 18:32
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Let $h$ be the mapping between the so-called Cantor cube $K = \{0,1\}^{\mathbb{N}}$ (in the product topology, where $\{0,1\}$ is discrete) and the Cantor set (as a subset of $[0,1]$) that sends a sequence $(x_i)_{i \ge 1} \in K$ to the point in $C$ with ternary expansion $0.a_1a_2\ldots$ where $a_i = 2x_i$ for all $i$.

Key to the proof is the fact that $h$ is a homeomorphism. Once you know that, your $f$ is just the index shuffling homeomorphism between $K$ and $K \times K$, which is easily shown to be continuous using the universal property of maps into product (continuous iff all the compositions with projections are).

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