# Baire's theorem: category and density for complements of first category sets

I'm working with the following version of Baire's category theorem:

If a non-empty complete metric space $$(M,d)$$ is the countable union of closed sets, then one of these closed sets has non-empty interior.

I want to show that if $$A\subset M$$ is a set of first category then $$A^c := M\setminus A$$ is a set of second category and dense in $$M$$.

The equivalent versions of Baire's theorem have me confused as I am very new to the concept of Baire categories. I tried working with the following statement:

$$A$$ is a set of first category (i.e. $$A = \bigcup_{n \in \mathbb{N}} A_n$$ and for all $$n$$ holds $$A_n$$ is nowhere dense) iff for all $$n$$ the set $$(\overline{A_n})^c$$ is dense in $$M$$.

The obvious proof by taking $$A$$ to the complement needs to assume that in a complete metric space the intersection of countably many dense open sets is dense. I read that this is the implication of Baire's lemma, so I guess I cannot just assume this holds true. The necessary step should relate to the statement of the theorem, however, even after reading the referenced post, I do not see how this is in accordance with this version of it.

• In your link the result is called Baire's lemma, not Baire's property. The Baire property is a property of sets: a set $U$ has the BP if it differs from an open set by a meagre set. – MacRance Jun 6 '20 at 22:39

## 1 Answer

Assume $$A$$ is of first category, and assume, toward a contradiction, that its complement $$A^c$$ is also of first category. So $$A$$ is the union of countably many nowhere dense sets $$A_n$$ and $$A^c$$ is the union of countably many nowhere dense sets $$B_n$$. By definition of "nowhere dense", the closures $$\overline{A_n}$$ and $$\overline{B_n}$$ have empty interiors. But the union of all these closures inlcudes both $$A$$ and $$A^c$$, so it is the whole space, and that contradicts Baire's theorem.

• Why is $A^c$ dense in $M$? – Friedrich Jun 7 '20 at 11:33
• @Friedrich If $A^c$ were not dense in $M$, there would be an open ball included in $A$. Consider a slightly smaller closed ball with the same center. On the one hand this closed ball is, as a closed subspace of $M$, a complete metric space. On the other hand, it's included in the first-category set $A$. That contradicts the Baire category theorem. – Andreas Blass Jun 7 '20 at 12:17
• From Baire's category theorem we know that the closed subspace of $M$ has an interior point. Why do I know that $A$ does not have an interior point? From my understanding, we only know that the subsets which make up $A$ have empty interiors, not necessarily $A$ itself. – Friedrich Jun 7 '20 at 13:17
• To repeat much of my previous comment: If $A$ had an interior point, it would include an open ball and would therefore also include a slightly smaller closed ball (with the same center). That closed ball, being a closed subset of $M$ wold be a complete metric space. But it's included in the first-category set $A$. That contradicts the Baire category theorem. – Andreas Blass Jun 7 '20 at 16:21
• My question is: why is the fact that such a closed ball is contained in a first-category set a contradiction of Baire's theorem. – Friedrich Jun 7 '20 at 16:23