# $C$ 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.

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Perhaps you have seen that the elements of the Cantor set can be thought of as branches in an infinite binary tree, where at each step we decide whether a given element is in the left or right half of some subinterval. Related to this idea is the fact that each element has a (not necessarily unique) ternary expansion where the coefficients are all either $2$ or $0$. Since $C$ is closed, you know that for each element $x$ of the Cantor set there is an element $y$ of $C$ which minimizes the distance $|x-y|$. The idea then is two make your function "locally constant" in some sense. –  rondo9 Dec 4 '12 at 5:19
Out of curiosity, did you ask the professor who mentioned it? –  Jonas Meyer Dec 4 '12 at 6:26
I did briefly after the class in which the professor stated it and he said something the the effect that rondo9 discussed above. –  jimmywho Dec 4 '12 at 6:32

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 $F$.

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Thank you for your response. I am a little confused on your construction of $\mathscr{I}$. Is it the union of the disjoint open intervals of [0,1]$\setminus\Delta$ or [0,1]\F? –  jimmywho Dec 4 '12 at 6:03
@jimmywho: Horrible typo! Thanks for catching it: that is indeed $[0,1]\setminus F$. –  Brian M. Scott Dec 4 '12 at 6:07
Okay the function makes sense now. To show continuity I was considering showing the preimage of an open set in F is open. What I see from the function f is that $f^{-1}(x)$ = $(x_{I-1},x_I)\bigcap \Delta$ for x $\in F$; $x=a_I.$ –  jimmywho Dec 4 '12 at 6:18
Another question, suppose 1 $\not\in$ F then the last open interval in $\mathscr{I}$ would be ($a_I$, 1] so f(x)=1 for x$\in (x_I,1]$? –  jimmywho Dec 4 '12 at 7:01
@jimmywho: No, in that case you want to set $x_I=1$, even though $1\in\Delta$. Similarly at the other end. –  Brian M. Scott Dec 4 '12 at 7:12