1
$\begingroup$

Let $X$ be a compact metric space.

Let $\{X_\alpha:\alpha\lt \mathfrak c\}$ be a partition of $X$ into $\mathfrak c=|\mathbb R|$ dense first category $F_\sigma$-subsets of $X$.

Let $A$ be a non-empty closed subset of $X$ such that $A\cap X_\alpha$ is first category in $A$ for each $\alpha<\mathfrak c$.

It is clear that uncountably many of the $X_\alpha$ must intersect $A$. But what can be said about $$\bigcup \{X_\alpha:X_\alpha\cap A\neq\varnothing\}?$$

Is this set equal to $X$? Does it at least contain a dense $G_\delta$-subset of $X$?

$\endgroup$
1
$\begingroup$

Counterexample.

Let $\{\alpha:\alpha\lt\mathfrak c\}=I\cup J$ where $I\cap J=\emptyset,\ |I|=|J|=\mathfrak c.$

Let $A=\{t_\alpha:\alpha\in J\}$ be the Cantor ternary set; $t_\alpha\ne t_\beta$ for $\alpha\ne\beta$.

Let $S$ be a dense $G_\delta$-subset of $[0,1]$ which has Lebesgue measure zero and is disjoint from $A;$ then $T=[0,1]\setminus S$ is a first category subset of $[0,1].$

For every interval $(a,b)\subseteq[0,1]$ we have $|S\cap(a,b)|=|(T\setminus A)\cap(a,b)|=\mathfrak c.$

Let $\{S_\alpha:\alpha\in I\}$ be a partition of $S$ into countable dense sets.

Let $\{T_\alpha:\alpha\in J\}$ be a partition of $T$ into countable dense sets such that $T_\alpha\cap A=\{t_\alpha\}.$

For $\alpha\lt\mathfrak c$ define

$$X_\alpha=\begin{cases} S_\alpha\ \text{ if }\ \alpha\in I,\\ T_\alpha\ \text{ if }\ \alpha\in J. \end{cases}$$

$X=[0,1]$ is a compact metric space, and $\{X_\alpha:\alpha\lt\mathfrak c\}$ is a partition of $X$ into $\mathfrak c$ countable dense subsets.

$A$ is a nonempty closed subset of $X$ such that $X_\alpha\cap A=\emptyset$ for $\alpha\in I$ and $X_\alpha\cap A=\{t_\alpha\}$ for $\alpha\in J,$ so that $X_\alpha\cap A$ is first category in $A$ for each $\alpha\lt\mathfrak c.$

Finally, $\bigcup\{X_\alpha:X_\alpha\cap A\ne\emptyset\}=\bigcup\{X_\alpha:\alpha\in J\}=T$ is first category in $X.$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.