Homeomorphism between two Cantor sets Let $K_{1}$ and $K_{2}$ two Cantor sets. Prove that they're homeomorphics.
How I begin the demonstration? I just need an idea. Thanks!
 A: There are several ways to prove this result; I’m going to suggest one that may seem a bit more abstract than some others, but it’s no harder, and it gives you an exceptionally useful way to think about Cantor sets.
In the construction of any of these Cantor sets we start with some closed interval $C_\varnothing=[a,b]$; in the case of the middle-thirds Cantor set or the fat Cantor sets described here, $[a,b]=[0,1]$, but we can use any non-degenerate closed interval. We remove an open interval $I_\varnothing$ from the interior of $C_\varnothing$ to leave two closed intervals, $C_{\langle 0\rangle}$ and $C_{\langle 1\rangle}$. Then we remove an open interval $I_{\langle 0\rangle}$ from the interior of $C_{\langle 0\rangle}$, leaving the closed intervals $C_{\langle 00\rangle}$ and $C_{\langle 01\rangle}$, and we remove an open interval $I_{\langle 1\rangle}$ from the interior of $C_{\langle 1\rangle}$, leaving the closed intervals $C_{\langle 10\rangle}$ and $C_{\langle 11\rangle}$. 
We continue in this fashion. In the end, for each finite sequence $\langle i_1,\dots,i_n\rangle$ of zeroes and ones we’ll have a closed interval $C_{\langle i_1,\dots,i_n\rangle}$, and we’ll remove an open interval $I_{\langle i_1,\dots,i_n\rangle}$ from its interior to leave two closed intervals, $C_{\langle i_1,\dots,i_n,0\rangle}$ and $C_{\langle i_1,\dots,i_n,1\rangle}$. Each of these rather odd-looking subscripts $\langle i_1,\dots,i_n\rangle$ tells us something about the location of its interval:


*

*if $i_1=0$, $C_{\langle i_1,\dots,i_n\rangle}$ is a subset of $C_{\langle 0\rangle}$, on the left side of the Cantor set, and if $i_1=1$, $C_{\langle i_1,\dots,i_n\rangle}\subseteq C_{\langle 1\rangle}$, on the right side of the Cantor set;  

*if $i_2=0$, $C_{\langle i_1,\dots,i_n\rangle}$ is a subset of $C_{\langle i_1,0\rangle}$, on the left side of $C_{\langle i_1\rangle}$, and if $i_2=1$, $C_{\langle i_1,\dots,i_n\rangle}\subseteq C_{\langle i_1,1\rangle}$, on the right side of $C_{\langle i_1\rangle}$;  

*if $i_3=0$, $C_{\langle i_1,\dots,i_n\rangle}$ is a subset of $C_{\langle i_1,i_2,0\rangle}$, on the left side of $C_{\langle i_1,i_2\rangle}$, and if $i_3=1$, $C_{\langle i_1,\dots,i_n\rangle}\subseteq C_{\langle i_1,i_2,1\rangle}$, on the right side of $C_{\langle i_1,i_2\rangle}$; and so on.


Thus, for example, $C_{\langle 0,1,0,0,1\rangle}$ is the closed interval that you reach if you start in the first lefthand interval $C_{\langle 0\rangle}$, go into its righthand subinterval $C_{\langle 0,1\rangle}$, then into the lefthand subinterval $C_{\langle 0,1,0\rangle}$ of that, the lefthand subinterval $C_{\langle 0,1,0,0\rangle}$ of that, and finally the righthand subinterval of that.
To help make the notation clearer, I’ll point out that the $2^n$ intervals $C_{\langle i_1,\dots,i_n\rangle}$ whose subscripts are sequences of length $n$ are all on the same ‘level’ of the construction; if we let $G_n$ be their union, then the Cantor set $K$ that we’re constructing is $K=\bigcap_{n\in\Bbb N}G_n$, and for each $n$, each point of $K$ belongs to exactly one of the closed intervals $C_{\langle i_1,\dots,i_n\rangle}$. 
For each $n\in\Bbb N$ and sequence $\langle i_1,\dots,i_n\rangle$ of $n$ zeroes and ones let $H_{\langle i_1,\dots,i_n\rangle}=C_{\langle i_1,\dots,i_n\rangle}\cap K$, and observe that $H_{\langle i_1,\dots,i_n\rangle}$ is both open and closed in the relative topology on $K$. 


*

*Show that the family $\mathscr{H}$ of these sets $H_{\langle i_1,\dots,i_n\rangle}$ is a base for the topology on $K$, and that for each $x\in K$ there is a unique infinite sequence $\sigma_x=\langle i_1(x),i_2(x),\dots\rangle=\langle i_k(x):k\in\Bbb Z^+\rangle$ of zeroes and ones such that $x\in H_{\langle i_1(x),\dots,i_n(x)\rangle}$ for each $n\in\Bbb Z^+$. In fact, $$\{x\}=\bigcap_{n\in\Bbb Z^+}H_{\langle i_1(x),\dots,i_n(x)\rangle}=H_{\langle i_1(x)\rangle}\cap H_{\langle i_1(x),i_2(x)\rangle}\cap H_{\langle i_1(x),i_2(x),i_3(x)\rangle}\cap\ldots\;.$$


Now for $n\in\Bbb Z^+$ let $X_n=\{0,1\}$ be a copy of the two-point space with the discrete topology, and let $X=\prod_{n\in\Bbb Z^+}X_n$ by the product of the spaces $X_n$ with the usual product topology. For each $n\in\Bbb N$ and sequence $\langle i_1,\dots,i_n\rangle$ of $n$ zeroes and ones let $B_{\langle i_1,\dots,i_n\rangle}=\{x\in X:x_k=i_k\text{ for }k=1,\dots,n\}$, and let $\mathscr{B}$ be the family of all such sets $B_{\langle i_1,\dots,i_n\rangle}$.


*

*Show that $\mathscr{B}$ is a base for the product topology on $X$, and that for each $x\in X$, $$\{x\}=\bigcap_{n\in\Bbb Z^+}B_{\langle x_1,\dots,x_n\rangle}=B_{\langle x_1\rangle}\cap B_{\langle x_1,x_2\rangle}\cap B_{\langle x_1,x_2,x_3\rangle}\cap\ldots\;.$$

*Finally, show that the map $$K\to X:x\mapsto\langle i_n(x):n\in\Bbb Z^+\rangle$$ is a homeomorphism.
When you’ve done this, you’ll have proved that every Cantor set constructed by the same general procedure as the middle-thirds Cantor set is homeomorphic to the product space $X$ and hence that all such sets are homeomorphic to one another.
