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Hello everybody,

I'm stuck with proving (or disproving) the following statement.

Statement: For every $0$-dimensional Polish space $(X,\mathcal{T}\ )$, and a countable basis of clopen sets $\mathcal{B}$ for $\mathcal{T}$, every open set is the disjoint union of clopen sets in $\mathcal{B}$.

Every open set is the union $O=\cup_{n}B^{0}_{n}$ of the basic clopen sets contained in it (say ordered with a given numbering of $\mathcal{B}$). The idea is to make it a disjoint union by considering, iteratively,

$O= B^{0}_{1} \cup O^{1}$

where $O^{1} = O\setminus B^{0}_{1}$, which is open. Then again we have

$O^{1}=\cup_{n}B^{1}_{n}$.

So consider $O^{2}= O^{1}\setminus B^{1}_{1}$

etcetera. The resulting union

$\cup_{m} B^{m}_{1}$

is open. However, I'm stuck in proving that, in general, a point $x\in O$ ends up necessarily in some $B^{k}_{1}$, for $k\in \mathbb{N}$, i.e., I can't prove that

$O= \cup_{m}B^{m}_{1}$.

Googling around I found this interesting paper [1]. The author says that it is a known fact (unfortunately he doesn't give a reference) that for every $0$-dimensional Polish space, the Borel sets are generated from the clopens by closing under countable disjoint unions and complements. This does not solve my problem, but still, I would be interested in reading a proof. Could you point me to some relevant literature?

Thanks in advance,

[1] Abhijit Dasgupta. Constructing $\Delta^{0}_{3}$ using topologically restrictive countable disjoint unions.

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2 Answers 2

up vote 6 down vote accepted

Disjointify from the bottom up instead of from the top down, as you do in Real Analysis 1.

The question is at the level of homework, but I give a hint because you identify yourself and explain what you have tried, including looking at the literature, and I see why you are stuck on the problem.

EDIT 11.1.11: First an apology. Like Henno, I misread your question.

Let's examine what happens when $X$ is compact. Your question has an affirmative answer when $X$ is countable. Observe that it is enough to show that if $X$ is a countable compact metric space and $\mathcal{B}$ is a basis of clopen sets, then $X$ itself is the disjoint union of sets from $\mathcal{B}$. Indeed, from this special case you get that every clopen set is the disjoint union of basic clopen sets, and every open set is the disjoint union of clopen sets.

Now every countable compact metric space is homeomorphic to a ray $[1,\alpha]$ of ordinals, so you can use transfinite induction. Take a basic clopen set $A$ that contains $\alpha$ and apply the inductive hypothesis to $X\sim A$.

However, with some help from Gideon Schechtman I checked that your question has an negative answer when $X$ is the Cantor set $\{-1,1\}^\Bbb{N}$. Consider $X$ as a compact group with Haar measure and let $\mathcal{B}$ be the collection of clopen sets that have, for some $n$, measure $2^{-n} + 2^{-n-1} = 3\cdot 2^{-n-1}$. It is easy to check that this is a base for the topology. If $X$ were a disjoint union of sets from $\mathcal{B}$ then by compactness it would be just a finite disjoint union, which would imply that $1$ is a finite sum of numbers of the form $3\cdot 2^{-n-1}$.

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Nice one; I was also looking for some base in the Cantor space as well for a counterexample. –  Henno Brandsma Nov 1 '11 at 22:19
    
I kept getting lost in the complicated set theoretic combinatorics of bases but eventually realized that I could avoid the combinatorics entirely. Compactness is important in the argument. At first I thought the irrationals would be easier, but there also the combinatorics gave me problems. In fact, the simple bases for the irrationals I looked at did not give a counterexample. –  Bill Johnson Nov 1 '11 at 23:32

If you have such a union of $O = \cup_n B_n$, consider the sets $B'_n = B_n \setminus \cup_{i=0}^{n-1} B_i$, where $B'_0 = B_0$. Show these sets are disjoint, clopen, and have the same union as the original sets, as for every $x \in O$ there is a first index $n(x)$ such that $x \in B_{n(x)}$.

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I agree: an answer should be an answer. Hints should be left as comments. –  Todd Trimble Oct 29 '11 at 12:41
    
Hello Henno. Thanks for your ansewer. Your argument seems conclusive only if every clopen set of the form B'_{n} is itself expressible as a disjoint union of basic clopens. Note that in the statement of my question I didn't say that the basis is a boolean algebra. Of course I see it is possible to iterate the same procedure to "disjointify" B'_{n} (for every n>0), but again I don't see why a point should end up in a basic clopen at any finite stage. –  Matteo Mio Oct 29 '11 at 18:07
    
@Matteo: do you really need exactly that statement? In other words, is there a problem with modifying the statement, taking into account that every basis of clopen sets can be refined to a basis of clopen sets that is a Boolean algebra, and working with a such a refinement? –  Todd Trimble Oct 29 '11 at 20:00
    
@Todd: yes. Sometimes it is convenient to work with a basis which is not closed under, say, unions or intersections etcetera, even if the basis can always be refined to a boolean algebra. I'm not sure if the assertion, as it is formulated in the question, holds or not. –  Matteo Mio Oct 29 '11 at 22:19

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