There is an exercise that asks to show that:
If a countable union of closed sets has nonempty interior in a complete metric space, then at least one set of the union has nonempty interior.
Is this result valid? Doesn't the complete metric space that should be the countable union itself?
How could I prove it?
So far, I have the Baire theorem in hand:
Let $(E,d)$ be a complete metric space. Then every intersection of countably many dense open sets is dense.