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From The Cantor set is homeomorphic to infinite product of $\{0,1\}$ with itself - cylinder basis - and it topology and the excerpt: For the continuity you may want to use the fact that the product topology of $\{0,1\}^\mathbb{N}$ is generated by the sets of the form $U(N,a)=\{(a_n)_{n=1}^\infty\in\{0,1\}^\mathbb{N}:a_N=a\}$ where $N\in\mathbb{N}$ and $a\in\{0,1\}$, and hence it suffices to show that the preimages of these sets $U(N,a)$ are open in the Cantor set.
I don't know how it suffices to show the preimage is open, what is a more precise proof of this part?

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user3987
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  • If you show the preimages of these sets are open, then so are the preimages of finite intersections of them, and unions of those, because taking the inverse image of unions and intersections "commutes" with the set operations. – BrianO Nov 07 '15 at 23:33
  • Thanks. I'm still not sure how to show the preimages of these sets are open. Are sets of the for U(N,a) open in this case? – user3987 Nov 07 '15 at 23:45
  • First step: for any finite sequence $s$ of $0,1$ define $V(s) = {a\in {0,1}^{\mathbb{N}} \mid a\restriction length(s) = s }$. This set in the range corresponds to one finite path in the Cantor set down to a subtree from which middle thirds have continued to be removed. If $s=(s_0,\ldots,s_N)$ then the corresponding "rational" in base 3 is $(2s_0, \ldots, 2s_N)$. Now, each $U(N,a)$ is a finite union of $V(s)$'s -- of $2_{N-1}$ of them, in fact. – BrianO Nov 07 '15 at 23:54
  • Yes, the $U(N,a)$ are all sub-basic open sets in the product topology. – BrianO Nov 07 '15 at 23:55

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For any $x$ in the Cantor set $C$ there is a unique $f(x)= (x_n)_{n\in N}\in \{0,1\}^N$ such that $\sum_{n=1}^{\infty} 2 x_n 3^{-n}=x.$ For $j\in N$ let $H(j)$ be the set of functions from $\{1,...,j\}$ to $\{0,1\}$. For $a\in \{0,1\}$ and $j\in N$ let $H(j,a)=\{h\in H(j):h(j)=a\}$. We have $$f^{-1}U(j,a)=\{x\in C: x_j=a\}=\cup_{h\in H(j,a)}\{x\in C: \forall i\leq j (x_i=h(i))\}.$$ Observe that for $a=0$ and $j\in N$ and $h\in H(j,0)$ we have $$\{x\in C:\forall i\leq j\, (\,x_i=h(i)\,)\}= C\cap [h^*,h^*+2.3^{-j})$$ $$\text {where } h^*=\sum_{i=1}^{i=j} 2 h(i) 3^{-i}.$$ Now either $h^*=0$ or $h^*$ is the upper endpoint of the open interval $(h^*-3^{-(j-1)},h^*)$ that was removed in the construction of $C$. Therefore, when $a=0$, $$\{x\in C:\forall i\leq j (x=h(i)\}=C\cap (h^*-3^{j-1},h^*+2.3^{-j})$$ which is open in $C$. The case $a=1$ is handled similarly.

DanielWainfleet
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  • can you also show how to prove that the inverse is continuous? –  Apr 11 '16 at 20:52
  • please :) I would appreciate it very much. –  Apr 11 '16 at 21:40
  • The Cantor set is a compact Hausdorff space .If $X$ is a compact Hausdorff space and $Y$ is a Hausdorff space and $f:X\to Y$ is a continuous bijection then $ f$ is a homeomorphism – DanielWainfleet Apr 11 '16 at 21:44
  • @user265665, is there a name for that theorem? Or anywhere I could find a proof for it? –  Apr 11 '16 at 21:50
  • I dk if there is a name. But I would look under "continuous images of compact (Hausdorff) spaces" , I think. – DanielWainfleet Apr 11 '16 at 22:08
  • one more thing. About your notation, the last two times you have ${ x\in C: \forall i \leq j(x=h(i))}$, those are supposed to be $x=h(i)$ and not $x_{i}=h(i)$, right? –  Apr 12 '16 at 14:37
  • It should be $x_i=h(i)$ as $h(i)\in {0,1}$ and $2 x_i$ is the $i$th digit of $x$ in the base-$3$ representation of $x$. – DanielWainfleet Apr 13 '16 at 01:03
  • @DanielWainfleet where do you get $2.3^{-j}$ in the sixth line? –  Apr 02 '19 at 20:32