Could a complete metric space be a union of uncountably many nowhere dense subsets of it?

Question

According to Baire's theorem, Any complete metric space can't be written as a union of a sequence of nowhere dense subsets of it.

So, this assumes that the union is a union of countably many subsets.

Now, Is that true in the case of a union of uncoutably many nowhere dense subsets of the metric space?

I mean, given a complete metric space which is uncountable, could it happen that this metric space is a union of uncountably many nowhere dense subsets of it?

If yes, How to show that? If no, could you provide a counterexample?

Answer 1

It is indeed possible.

Let $X$ be a complete metric space, in which every point is not an open subset ($\mathbb{R}$ can be a good example). It follows that every point is a nowhere dense subset. Take $$X=\bigcup_{x\in X}\{x\}.$$