Question about 0-dimensional Polish spaces 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.  
 A: 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 $3\cdot 2^{-n-1}$.
A: 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 $x \in B_{n(x)}$. 
