If $(M,d)$ is complete, is every non-empty open set a second category set?
I know Baire's Category Theorem for Metric Spaces, which says: "A complete metric space is of the second category in itself, i.e. if we write $M = \bigcup_{n\ge 1} E_n$, then the closure of some $E_n$ contains an open ball. Equivalently, if $(G_n)$ is a sequence of dense open sets in $M$, then $\bigcup_{n\ge 1} G_n \ne \varnothing$. In fact $\bigcap_{n\ge 1} G_n$ is dense in $M$."
Somehow, I need to use this (or more) to answer the question above. I feel that the answer is Yes, but I'm not sure how to prove/disprove it. Here's what I've tried so far:
Suppose $S \ne \varnothing$, and $S$ is open. So for every $x\in S$, there exists $r_x > 0$ such that $B(x,r_x) \subset S$. Now there are two possibilities, $S$ is either first category or second.
Attempt 1:
If $S$ is first category in $M$, then $S$ can be written as $$S = \bigcup_{n\ge 1} S_n$$ where $S_n$ is nowhere dense in $M$ for every $n\ge 1$ (so, $\text{int}(\overline{S_n}) = \emptyset$).
To find a contradiction, I must get hold of some $n$ such that $\text{int}(\overline{S_n}) \ne \emptyset$. What do I do?
Attempt 2:
Consider $M_n$ such that $M = \bigcup_{n\ge 1} M_n$ and $A \subset M$ is open. Since $M$ is complete, it is second category in itself, i.e. there is some $n$ such that $M_n$ is not nowhere dense. $M \cap A = A$, so we get $$A = A \cap \bigcup_{n\ge 1} M_n = \bigcup_{n\ge 1} (M_n \cap A)$$ Now if we can prove that $M_n \cap A$ is not nowhere dense, then we are done (maybe it isn't true, maybe this isn't the right way). We know that $\text{int}(\overline{M_n}) \ne \emptyset$, what can we say about $\text{int}(\overline{ M_n \cap A})$?
A link in the comments proves that every open subset of a Baire space is also a Baire space. How does that help?
- $(M,d)$ is a complete metric space, and so a Baire space.
- $A\subset M$ is open, and by the statement above, $A$ is also a Baire space.
- How do I know that $(A,d)$ is complete? Are complete metric spaces the only metric spaces which are Baire spaces? If yes, then knowing that $A$ is complete - I can use Baire's theorem to conclude that $A$ is second category.
