In the Kechris book on (Classical) Descriptive Set Theory there is a claim that a separable metrisable space is zero dimensional if and only if every closed set is a cts retract of the whole space (Theorem 7.3).

Zero dimensional is equivalent to: there is a basis of clopen sets.

One direction ($\Rightarrow$) is easy: one can easily place the space into an homeomorphism with a subset of cantor space. I would like to know how to prove the other direction. The reference in Kechris is to Kuratowski: but I looked at this old topology handbook and it only contained the ($\Rightarrow$) direction (in fact it did not even claim the reverse direction).

Is Kechris’ claim correct? If so, and you know the proof or a sketch thereof, it would be great if you could post it.

I tried setting up a contradiction; let $x\in U$ with $U$ open so that there is no clopen $A$ with $x\in A\subseteq U$. Then letting $x\in V_{n}\subseteq U$ be a neighbourhood basis for $x$ with $(V_{n})_{n}$ monotone decreasing, I took the inverse limit of a sequence of (compatible) retracts:

$X\setminus V_{0}\twoheadleftarrow X\setminus V_{1}\twoheadleftarrow\ldots \breve{X}$

where for each $n$, there is no clopen set $A$, such that $X\setminus V_{n}\subseteq A\not\ni x$. But didn't get far with this$\ldots$

  • 3
    $\begingroup$ Hint: Effectively, every zero-dimensional seperable metrizable space is homeomorphic to the cantor middle thirds set on [0,1]. This should be a good starting point for intuition to take over. $\endgroup$
    – Not Mike
    Jan 20 '11 at 7:40
  • $\begingroup$ Problem 1. If $Y$ is a closed subset of a compact metric space $X$ and the nearest point mapping from $X$ to $Y$ is single valued, then it is continuous. Problem 2. Put an equivalent metric on the Cantor set $X=\{0,1\}^N$ so that the hypothesis of (1) is satisfied for every closed subset $Y$ of $X$. Problem 3. Show that Michael and I gave essentially the same hint. $\endgroup$ Jan 20 '11 at 9:15
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    $\begingroup$ @Michael: no, only if there are no isolated points and it is compact. Otherwise Q, R\Q, a convergent sequence etc. are counterexamples. $\endgroup$ Jan 21 '11 at 8:06
  • $\begingroup$ @Brandsma: wasn't meant to a perfect answer, just a fact that came off the top of my head and seemed helpful. That being said, you are correct, I should have added the assumption of compactness, and not having isolated points. $\endgroup$
    – Not Mike
    Feb 2 '11 at 23:48

The implication $\Leftarrow$ is nigh on trivial: given $x$ and $U$ take a retraction $r$ of $X$ onto $\lbrace x\rbrace\cup(X\setminus U)$ and then let $A=r^\gets\bigl[\lbrace x\rbrace\bigr]$.


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