$C$ be a closed subset of the Cantor set $\Delta$. Show the existence of a continuous function $f:\Delta\to C$ s.t. $f(x)=x$, $x\in C$ Question: Let $C$ be a closed subset of the Cantor set $\Delta$. Prove there is a continuous function $f$ from $\Delta$ onto $C$ s.t. for every $x \in C$ we have $f(x)=x$.
Context: Advanced Undergraduate Analysis. I am familiar with Rudin and Carothers. This was a fact posed by a professor that has been on my mind for awhile now. 
I was considering the Cantor function $\Delta \rightarrow [0,1]$ but I don't know how I could show that it is continuous on $\Delta$. 
Any insight or help would be appreciated. 
 A: Let $\Delta$ be the middle-thirds Cantor set, and let $F$ be a non-empty closed subset of $\Delta$. Then $[0,1]\setminus F$ is an open set in $[0,1]$, so it is the union of a countable family $\mathscr{I}$ of pairwise disjoint open intervals in $[0,1]$. (Note that intervals of the forms $[0,a)$ and $(a,1]$ are open in $[0,1]$.) Let $I=(a_I,b_I)\in\mathscr{I}$; clearly $a_I,b_I\in F$. $\Delta$ does not contain any non-empty open interval, so there is an $x_I\in(a_I,b_I)\setminus\Delta$. Now define
$$f:\Delta\to F:x\mapsto\begin{cases}
x,&\text{if }x\in F\\
a_I,&\text{if }x\in(a_I,x_I)\text{ for some }I\in\mathscr{I}\\
b_I,&\text{if }x\in(x_I,b_I)\text{ for some }I\in\mathscr{I}\;.
\end{cases}$$
Can you prove now that $f$ is continuous?
Added: If $0\notin F$, there will be an $I\in\mathscr{I}$ of the form $[0,b)$; in that case take $x_I=0$. Similarly, if there is an $I\in\mathscr{I}$ of the form $(a,1]$, set $x_I=1$. In these two cases you want $f$ to squash all of $I\cap\Delta$ to the endpoint of $I$ that is in $F$.
A: Let the Cantor set be $\{0,1\}^{\mathbb{N}}$. Consider the following metric defining the topology 
$$d((a_n), (b_n)) = \sum_{n\ge 0} \frac{|a_n - b_n|}{3^n}$$
(any number $\lambda >2$ would work instead of $3$). 
Notice that if for three points $b$, $a$, $a'$ we have $d(a,b) = d(a',b)$ then $a=a'$. 
Let $A$ be a closed non-void subset of the Cantor set. Define $p_A(b)$ to be the closest point in $A$ to the point $b$ (this is unique by the above observation).
Then $p_A$ is a retract of the Cantor set onto $A$. 
