Is the product of second category spaces second category?

132 Views Asked by At

The following is an exercise of Gemignani's Elementary topology. (I'm studying for an exam.) Suppose $Y$ and $Z$ are subspaces of some space $X,\tau$ and $Y$ and $Z$ are both of the second category in $X$. Is $Y\times Z$ of the second category in $X\times X$?

My intuition is that it is: I assumed $Y\times Z$ is not (suspecting that $\mathrm{Int}_{X\times\{z\}}\mathrm{Cl}_{X\times\{z\}}(Y\times\{z\}\cap A_n)\subseteq\mathrm{IntCl}A_n$, where $Y\times Z=\cup A_n$), for getting to a contradiction, but I'm not getting to it. I'm not able to find a counterexample neither.

1

There are 1 best solutions below

0
On BEST ANSWER

There are counterexamples, but all of the ones that occur to me so far require some pretty sophisticated set-theoretic or measure-theoretic background; if the question is in Gemignani, there ought to be a relatively elementary example, but so far I’ve not found one. The first of the three that I’ll mention is a consistency result, requiring the Continuum Hypothesis; the others require no such extra hypothesis.

Theorem $5$ of J.C. Oxtoby, Cartesian products of Baire spaces [PDF], shows that under the assumption of the Continuum Hypothesis there is a Tikhonov Baire space $X$ such that $X\times X$ is meagre (i.e., of first category).

In Barely Baire spaces Bill Fleissner and Ken Kunen construct a Baire space $X$ such that $X^2$ is nowhere Baire, i.e., such that $X^2$ has a family $\mathscr{D}=\{D_i:i\in\omega\}$ of dense open subsets such that $\bigcap\mathscr{D}=\varnothing$.

In Note on category in Cartesian products of metrizable spaces Roman Pol shows that if $X$ and $Y$ are second category metric spaces in which no point has a separable nbhd, then there are second category sets $A\subseteq X$ and $B\subseteq Y$ such that $A\times B$ is first category in $X\times Y$.