# Why in a totally disconnected seperable metric space, every open set is a disjoint union of clopen sets?

On page 14, Probability Measures on Metric Spaces,Parthasarath(1967):

We recall that a totally disconnected seperable metric space, every open set can be expressed as a countably disjoint union of closed and open sets.

I'm only able to show it's true in some examples, say, $\Bbb Q$ and Cantor set, but with no luck in general.

Added:Here's a screenshot of relevant part of the statement: • The validity of the result Parthasarathy sets out to prove is not affected by this slip. The only thing he actually uses is that the unit interval is a continuous one-to-one image of a completely metrizable zero-dimensional space (the complement of a countable subset of the Cantor set). – Martin Jul 4 '13 at 10:51
• @Martin: Thank you. If you don't tell me that, I would probably skip that proof. – Metta World Peace Jul 4 '13 at 11:09

If the stated result were true, every totally disconnected, separable metric space would have a base of clopen sets and would therefore be zero-dimensional. However, Example $6.2.19$ of Engelking, General Topology, is an example of a totally disconnected, separable metric space that is not zero-dimensional. I shan’t go through the details, which are a bit lengthy, but the space itself is the subspace of $\ell_2$ consisting of the square-summable sequences of rational numbers, with the usual $\ell_2$ metric.
• @MettaWorldPeace: You’re welcome. I doubt it, because if that were the intent, the result would be trivial: just express each open set as itself! (There are, I think, a couple of editions of Engelking; mine is the $1977$ English-language edition published in Poland.) – Brian M. Scott Jul 4 '13 at 10:41