Products of Baire spaces

Question

I could not find any references about this fact. I apologize if this is completely trivial, but is the product of two Baire spaces, or for that matter of finitely many of them a Baire space? Now is a countable product of Baire spaces a Baire space?

What about an uncountable product of Baire space? This fact seems to be treated in an article I can't access.

It seems to work for the Sorgenfrey line: $S$ is a Baire space and $SxS$ is a Baire space since if you consider the diagonal $A$={($-x$,$x$): $x\in S$ and $x\in$ℚ} then this is a closed discrete subspaces which is the union of countably many closed nowhere dense sets but its interior is empty.

Is that true?

Thx

Answer 1

See:

Cohen, Paul E.
Products of Baire spaces.
Proc. Amer. Math. Soc. 55 (1976), no. 1, 119--124.

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MathSciNet review by Douglas Censer: A topological space is said to be Baire if any countable intersection of its dense open sets is dense. Assuming the continuum hypothesis (CH), J. C. Oxtoby [Fund. Math. 49 (1960/61), 157--166; MR0140638 (25 #4055); errata, MR 26, p. 1543] constructed a Baire space whose square is not Baire. The author shows that the assumption of CH is unnecessary here. The spaces considered in this paper are partially ordered $(P,\leq)$ with topology generated by the initial segments. The construction of the desired Baire space with non-Baire product is based on the study of $P$ as a set of forcing conditions, as outlined by R. M. Solovay [Ann. of Math. 92 (1970), 1--56; MR0265151 MR0265151 (42 #64)].

Note that googling "products of Baire spaces" returns Cohen's article as the second hit. (The first hit is this question!)

For a positive result, see

Zsilinszky, László
Products of Baire spaces revisited.
Fund. Math. 183 (2004), no. 2, 115--121.

Excerpted from the MathSciNet review by Paul Bankston: Without extra assumptions, the product of two Baire spaces need not be Baire [see, e.g., J. C. Oxtoby, Fund. Math. 49 (1960/1961), 157--166; MR0140638 (25 #4055); P. E. Cohen, Proc. Amer. Math. Soc. 55 (1976), no. 1, 119--124; MR0401480 (53 #5307)]. This brings us to the notion of a $\pi$-base; i.e., a collection of open sets such that every nonempty open set in the space contains a member of the collection. A $\pi$-base each of whose members contains only countably many members of the $\pi$-base is called countable-in-itself.

Oxtoby's theorem, from his 1961 paper, states that any Tikhonov (resp., finite) product of Baire spaces with countable (resp., countable-in-itself) $\pi$-bases is a Baire space. The main result of the present paper is a significant strengthening of this; in particular it implies that arbitrary Tikhonov products of Baire spaces with countable-in-itself $\pi$-bases are Baire spaces.

A space $X$ is called universally Kuratowski-Ulam (uK-U for short, first considered in [C. Kuratowski and S. Ulam, Fund. Math. 19 (1932), 247--251; Zbl 0005.18301]) if whenever $Y$ is a space and $E$ is a meager subset of $X\times Y$, the set $ Y \setminus \{y\in Y\:\ \{x\in X\:\ (x,y)\in E\}\ \text{is meager in}\ X\} $ is meager in $Y$. The author now defines a space $X$ to be almost locally uK-U if the set $\{x\in X\:\ x \text{has an open uK-U neighborhood} \}$ is dense in $X$.

After showing that the property of being almost locally uK-U is a proper generalization of having a countable-in-itself $\pi$-base, the author proves his main theorem: only a Tikhonov (or countable box) product of Baire spaces that are almost locally uK-U is a Baire space.

The proof for the box product case is a variant of that for the Tikhonov case, and partially answers a question raised in [W. G. Fleissner, in General topology and its relations to modern analysis and algebra, IV (Proc. Fourth Prague Topological Sympos., Prague, 1976), Part B, 125--126, Soc. Czechoslovak Mathematicians and Physicists, Prague, 1977; MR0464181 (57 #4116)].

Answer 2

There are even two normed Baire spaces whose product is not Baire. See J. van Mill and R Pol, The Baire category theorem in products of linear spaces and topological groupsn, Topology Appl. 22 (1986) 267--282.

Answer 3

In addition to previous answers:

The easiest way of proving that space is Baire is using one of following theorems:

1)Any locally compact space is Baire 2)Any complete metric space is Baire

In fact, there is a notion of Cech completeness which generalises both theorems. (A space is called Cech-complete if remainder of its Stone-Cech compactification $\beta X\setminus X$ is a $F_{\sigma}$ in Stone-Cech compactification, every locally compact is Cech-complete and every complete metric space is Cech-complete).

Then, while product of Baire spaces need not to be Baire, the product of ANY(even uncountable!) collection of Cech-complete spaces is Baire.

Answer 4

There are two metric Baire spaces, whose product is not Baire; a counterexample is given in the slides An absolute barely Baire space, L.F. Aurichi and G.A.A. Medina, 2020. We present Fleissner and Kunen's proof with applications of Clubs and stationary sets in $\omega_{1}$.

Fleissner, W. G.; Kunen, K., Barely Baire spaces, Fundam. Math. 101, 229-240 (1978). ZBL0413.54036.