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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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    I did briefly after the class in which the professor stated it and he said something the the effect that rondo9 discussed above.2012-12-04

2 Answers 2

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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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    @BrianM.Scott ,Does this generalize? What if we replace $\Delta$ with any compact metrizable space which is strongly zero-dimensional, in which every neighborhood is uncountable?2015-11-28
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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 $A$.