How to prove that there is a continuous surjection from Cantor space onto a closed subspace?

Question

Let C be the Cantor space and let Y be a non-empty closed subspace of C. How do I go about showing that there exists a continuous surjection from C onto Y?

This is from a topology course, any hints/guidance is appreciated

Answer 1

In Kechris' Classical Descriptive Set Theory, p. 7-8 there is a very elegant argument based on pruned trees that represent closed sets in countable products. Look at that, it's good stuff.

Answer 2

You could use a big theorem, namely that every compact metric space is a continuous image of the Cantor set. It is proved in many topology (or analysis) books, for example Wikipedia gives Willard's General topology book, section 30.7. I also found it in Real Analysis notes by Saeed Zakeri presently available online, see Theorem 2.77 (p.59) there. The proof uses "dyadic filtrations" which kind of mimic the construction of the Cantor set. Applications of this "universal surjectivity" are discussed in a paper by Yoav Benyamini (it refers to the topology books by Engelking, and by Hocking and Young, and I think I first read this proof from Dugundji's Topology book). Another proof, via the Hilbert cube is outlined by Elena Gurevich here (apparently a seminar talk), it gives the book Topology without tears as a reference. The latter book is freely and officially made available online by the author, Sidney Morris, see Proposition 2.5.2 and Theorem 2.5.4 (p.247-249) there, the treatment is comprehensive.

Here is an attempt at a direct proof, without verifying whether it works. For each $y\in Y$ let $r(y)=y$ (so $r$ will end up being a retraction). If $x\in C\setminus Y$ then there is a sequence of closed intervals $I_0,I_1,...$ such that $I_0=[0,1]$, $x\in\cap_nI_n$, and for each $n$ the interval $I_{n+1}$ is that closed "third" interval of $I_n$, either the left or the right, that contains $x$ (after the open middle-third of $I_n$ has been removed).

Clearly $I_0\cap Y=Y\neq\varnothing$, and since $C\setminus Y$ is open there is an $n(x)\ge1$ such that $I_{n(x)-1}\cap Y\neq\varnothing$ but $I_{n(x)}\cap Y=\varnothing$ (use compactness, and that $\{x\}=\cap_nI_n\subseteq C\setminus Y$). If $I_{n(x)}$ is the "right closed third" of $I_{n(x)-1}$ then let $J_n(x)$ be "left closed third", and if, on the other hand, $I_{n(x)}$ is the "left closed third" of $I_{n(x)-1}$ then let $J_n(x)$ be "right closed third". Map all of $I_{n(x)}$ to the closest point in $Y\cap J_n(x)$.

Answer 3

In general, for every metrizable and separable space $X$ are equivalent, $X$ is zero dimensional and every nonempty closed set is a retract of $X$. I only know the proof with trees prunes and the universal Baire space by zero dimensional metrizable and separable.