# Is this Baire theorem a special case of Baire category theorem?

I'm reading a chinese text book "Real Analysis" (by Zhou Minqiang), one of it's conclusion is "Baire theorem"

For any $E\subset R^n$ is a $F_\sigma$ set: $$E=\bigcup_{k=1}^{\infty}F_k,$$ where $F_k$ is close set. If each $F_k$ has no interior point, $E$ also has no interior point.

I couldn't find such "Baire theorem" on wikipedia?

There is a "Baire category theorem":

Statement of the theorem[edit source | editbeta]

A Baire space is a topological space with the following property: for each countable collection of open dense sets $U_n$, their intersection $\bigcap_n U_n$ is dense.

• (BCT1) Every complete metric space is a Baire space. More generally, every topological space which is homeomorphic to an open subset of a complete pseudometric space is a Baire space. Thus every completely metrizable topological space is a Baire space.

• (BCT2) Every locally compact Hausdorff space is a Baire space. The proof is similar to the preceding statement; the finite intersection property takes the role played by completeness.

Note that neither of these statements implies the other, since there are complete metric spaces which are not locally compact (the irrational numbers with the metric defined below; also, any Banach space of infinite dimension), and there are locally compact Hausdorff space which are not metrizable (for instance, any uncountable product of non-trivial compact Hausdorff spaces is such; also, several function spaces used in Functional Analysis; the uncountable Fort space). See Steen and Seebach in the references below.

• (BCT3) A non-empty complete metric space is NOT the countable union of nowhere-dense closed sets.

This formulation is equivalent to BCT1 and is sometimes more useful in applications. Also: if a non-empty complete metric space is the countable union of closed sets, then one of these closed sets has non-empty interior.

But I don't understand what this BCT is about. Is the Baire Category Theorem in wiki a more general conclusion, the "Baire theorem" I quoted from the book is a more specific one? of which one? BCT1, BCT2 or BCT3?

• It's a particular instance of the Baire category theorem. This one specifically says that $\mathbb{R}^n$ is of the second category in itself. – Daniel Fischer Sep 8 '13 at 12:23

Let $G_k$ be the complement of $F_k$. Since $F_k$ has no interior point (and is closed), $G_k$ is open and dense (otherwise, the complement of its closure would be an open set contained in $F_k$). Either version of the Baire Category Theorem implies that the intersection of the $G_k$'s also open and dense, but they are contained in the complement of $E$, so $E$ cannot have an interior point.
• This answer is correct, except for the statement that the intersection of the $G_k$s is open, it's only a $G_\delta$ in general. But we do not need it to be open, only dense, to conclude that $E$ has empty interior. So just remove "also open" and receive your upvote. – Robert Furber Jul 6 at 3:14