Prove that function is continuous. Possible Duplicate:
  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.
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

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

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