Metric space and Borel sets

Question

Excuse me can you see this question

prove that

In a metric space Borel sets satisfying :

a) $G$ contains the open sets

b) countable intersection of elements of $G$ belong to $G$

c) countable union of elements of $G$ belong to $G$

" Open can be replaced by closed "

Answer 1

An important observation is that in a metric space every closed set can be written as a countable intersection of open sets, and dually, every open set as a countable union of closed sets. This will allow you to show ( I think) that any family that fulfills $a,b,c$ minimally will have to be closed under complementation too, and will thus be a $\sigma$-algebra containing the open sets (hence, in my definition, if minimally so, equal to the Borel sets).