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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – MacRance
    Jun 6, 2020 at 22:39

1 Answer 1


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.

  • $\begingroup$ Why is $A^c$ dense in $M$? $\endgroup$
    – Friedrich
    Jun 7, 2020 at 11:33
  • $\begingroup$ @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. $\endgroup$ Jun 7, 2020 at 12:17
  • $\begingroup$ 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. $\endgroup$
    – Friedrich
    Jun 7, 2020 at 13:17
  • $\begingroup$ 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. $\endgroup$ Jun 7, 2020 at 16:21
  • $\begingroup$ 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. $\endgroup$
    – Friedrich
    Jun 7, 2020 at 16:23

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