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.