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Prove that function is homeomorphism.

Let $$ f: \prod\limits_{1}^{\infty} ( \{0,2 \}, \mathcal{T} _{\delta}) \to ([0,1], \mathcal{T}_{e}):\{n_i \} \mapsto \sum_{i=1}^{\infty} \frac{n_i}{3^i} $$ Prove that $f$ and $f^{-1}$ are continuous.

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marked as duplicate by no identity, Davide Giraudo, Marvis, Cameron Buie, nbubis Nov 4 '12 at 20:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
@voldemort: Kindly avoid posting same question more than once and a homeomorphism by definition is bicontinuous. –  user17762 Nov 4 '12 at 18:53
    
Ok. From this time I will respect this rules. I'm new here so i hope that you will forgive me :) My problem is that I don't know how to prove that this is homeomorphism :( –  voldemort Nov 4 '12 at 19:02
    
@Marvis: This wouldn’t have happened if I’d been quicker responding to voldemort’s request for more help on the original question. At this point I think that it makes better sense to leave this question open, since the more extensive answer is here; there are two pointers to the earlier question that allow access to anything there that isn’t here. –  Brian M. Scott Nov 4 '12 at 19:08
    
@BrianM.Scott, you don't have to be quicker! You're helping me and I'm not allowed to demand from you anything. I can just be greatfull that you want to help me :) –  voldemort Nov 4 '12 at 19:16

2 Answers 2

Let $X$ be the product that is the domain of $f$, and let $C=f[X]$.

You already know that $f$ is a bijection. To show that $f^{-1}$ is continuous, it suffices to show that $f$ is an open map, one that takes open sets to open sets.

For $n\in\Bbb Z^+$ let $\Phi_n$ be the set of functions from $\{1,\dots,n\}$ to $\{0,2\}$. For each $\varphi\in\Phi_n$ let $$B(\varphi)=\{x\in X:x_k=\varphi(k)\text{ for }k=0,\dots,n\}\;.$$ Let $\Phi=\bigcup_{n\in\Bbb Z^+}\Phi_n$, and let $\mathscr{B}=\{B(\varphi):\varphi\in\Phi\}$; then $\mathscr{B}$ is a base for the product topology on $X$.

The proof of the following lemma is very straightforward, and I’ll leave it to you.

Lemma. If $f[B]$ is open in $C$ for all $B\in\mathscr{B}$, then $f$ is an open map, and $f^{-1}$ is continuous.

To show that each $f[B]$ is open, you need to figure out what $f[B]$ is. Suppose that $\varphi\in\Phi_n$. By definition

$$\begin{align*} f[B(\varphi)]&=\left\{\sum_{k\ge 1}\frac{x_k}{3^k}:x\in B(\varphi)\right\}\\ &=\left\{\sum_{k=1}^n\frac{\varphi(k)}{3^k}+\sum_{k\ge n+1}\frac{x_k}{3^k}:x\in B(\varphi)\right\}\;; \end{align*}$$

show that this set is equal to one of the $2^n$ ‘blocks’ of the set $C_n$ in this construction of the middle-thirds Cantor set, and explain why each of those blocks is an open subset of $C$.

To show that $f$ is continuous, show that those blocks are a base for the topology of $C$, and use what I’ve done above to show that their inverse images under $f$ are open inn $X$.

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Thank you :) Now I have to study this and I hope I will understand this :) –  voldemort Nov 4 '12 at 19:13

It is a basic problem in fractal ,actually, the cantor set is satified the strong open set condition that directly leads to your result.

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As it stands, this is not an answer. Please provide more details. –  robjohn Nov 4 '12 at 22:18

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