I refer to the proof of the statement “Every complete metric space is a Baire space.”
The proof of this statement, as given in “Introduction to Banach spaces and their Geometry”, by Bernard Beauzamy, is
Let be a countable set of _open_ dense subsets of complete metric space . Take any open set . We will prove that is non-empty.. For point , construct, . Here is the closure of . Not take the intersection of and , and construct . Similarly, for every , construct . Also, ensure that ; in other words ensure that converges to .
It is clear that the points form a cauchy sequence. As is complete, the limit of this point exists in . This point is present in all , and also in all (as a result, it is present in ). Hence, it is present in . Hence is not empty. This proves the theorem.
We have taken one open set , and proved, is not empty. Should we take another open set , we will again be able to prove is not empty. Note that this does not in any way imply that . All that it says is that any open set has a non-empty intersection with , which is sufficient for to be a dense set in .
Why do we not take instead of ? This is because if the balls are not closed, then it would be difficult to prove the limit point of the sequence is also present in . Try this for yourself. Also, it is ok to take closed balls because if has a non-empty intersection with , then it definitely has a non-empty intersection with , and we are only concerned with getting an intersection so that we can construct another closed ball inside it. It is elementary to prove that the limit point of has to lie inside the intersection of all the closed balls (hint: contains all of the cauchy sequence , where ).