Forming a homeomorphism between perfect closed set of measure $\alpha; 0<\alpha<1$ and the Cantor set. Lets call this perfect closed set of measure $\alpha$ F. It was made quite similar to the Cantor set. I will include the making of each here.
$$C=\bigcap_{n=1}^{\infty}C_n$$ where $C_1=[0,1]\setminus (\frac{1}{3}, \frac{2}{3}); C_2=C_1\setminus ((\frac{1}{9},\frac{2}{9})\cup (\frac{7}{9},\frac{7}{9})); C_3=C_2\setminus(....)$
and $$F=\bigcap_{n=1}^{\infty}F_n$$ 
First let $\beta=1-\alpha $ then $$\vartriangle_{F11}= (\frac{1}{2}-\frac{\beta}{4}, \frac{1}
{2}+\frac{\beta}{4})$$
then $x_1, x_2$ are chosen each from the two remaining intervals from $[0,1]$ outside of $\vartriangle_{F11}$
$$\vartriangle_{F21}= (x_1-\frac{\beta}{4^2}, x_1+\frac{\beta}{4^2})$$
$$\vartriangle_{F22}= (x_2-\frac{\beta}{4^2}, x_2+\frac{\beta}{4^2})$$
Then again same procedure with $x_3,x_4,x_5,x_5$ getting intervals $$\vartriangle_{F31},\vartriangle_{F32},\vartriangle_{F33},\vartriangle_{F34}$$
Then we finally have $$F_1=[0,1]\setminus\vartriangle_{F11} \\ 
                       F_2=[0,1]\setminus(\vartriangle_{F21}\cup \vartriangle_{F22}) \\ 
  F_3=[0,1]\setminus(\vartriangle_{F31}\cup \vartriangle_{F32} \cup \vartriangle_{F33}\cup \vartriangle_{F34})\\ ..... \\ F=\bigcap_{n=1}^{\infty}F_n$$
I think to make this homeomorphism I would somehow need to make a mapping between these $\vartriangle_{Fij}$ with $\vartriangle_{Cij}$-these being the excluded sets in the Cantor set. But I honestly don't know how.
 A: HINT: Can you find a homeomorphism $f_n$ from $C_n$ to $F_n$, for each $n$?
This does not solve the problem, obviously, but it's a first step. We then want to define a homeomorphism $f$ from $C$ to $F$, which is the "limit" of the $f_n$s. You'll need to argue why such a limit in fact exists (subhint: show that for each $x\in C$, the sequence $f_n(x)$ converges . . .).
A: HINT: I would go about it in a slightly indirect fashion. Show that each point of $C$ and each point of $F$ is uniquely determined by an infinite sequence $\langle d_n:n\in\Bbb N\rangle$, where each $d_n$ is $0$ or $1$. Specifically, $d_n$ is $0$ if the point is in the left subinterval at stage $n$ and is $1$ if the point is in the right subinterval at stage $n$. Then show that $C$ and $F$ are both homeomorphic to the product $\{0,1\}^{\Bbb N}$, where $\{0,1\}$ has the discrete topology. The homeomorphism between $F$ and $C$ is then the map that takes each point of $F$ to the point of $C$ with the same associated sequence $\langle d_n:n\in\Bbb N\rangle$.
