2

Let $(X, \tau)$ be Cantor space, the collection of binary sequences equipped with the product of discrete topologies.

Does there exist a collection $(U_i)_{i \in I}$ of subsets of $X$ such that $I$ is an uncountable set, $U_i$ is second category for all $i \in I$, and $U_i \cap U_j = \emptyset$ for all $i \neq j \in I$?

aduh
  • 8,662
  • Maybe $U_i$ is the subset of the Cantor space where the even-place digits of $x$ in base 3 match $i$? – Daniel Schepler Jan 26 '18 at 20:52
  • @DanielSchepler: It's closed and nowhere dense. – Nate Eldredge Jan 26 '18 at 20:53
  • 1
    For a start, the sets $U_i$ are going to have to be pretty weird. Suppose they all have the property of Baire, so that $U_i = V_i \triangle M_i$ with $M_i$ meager and $V_i$ open (and not empty, else $U_i$ is meager). Then $V_i \cap V_j$ has to be open and meager, hence empty by Baire category. So now you have uncountably many pairwise disjoint nonempty open sets, and that contradicts separability. So you will need sets that don't have the BP, and that is going to need the Axiom of Choice in an essential way. – Nate Eldredge Jan 26 '18 at 21:05
  • @NateEldredge Very interesting, thanks. – aduh Jan 26 '18 at 21:12
  • 1
    You can get continuum many such pairwise disjoint sets (not just uncountably many) by using Bernstein sets. See tomasz's answer to What's application of Bernstein Set? – Dave L. Renfro Jan 26 '18 at 21:16

1 Answers1

2

Yes, such a family can easily be constructed by transfinite induction. There are only $\mathfrak{c}$ countable unions of closed nowhere dense subsets of $X$; let $(C_\beta)_{\beta<\mathfrak{c}}$ be an enumeration of them. Also, let $f:\mathfrak{c}\to\mathfrak{c}\times\mathfrak{c}$ be a bijection. Now construct a family $(U_\gamma)_{\gamma<\mathfrak{c}}$ by an induction of length $\mathfrak{c}$, where in the $\alpha$th step you choose a new element of $X\setminus C_\beta$ to be in $U_\gamma$, where $f(\alpha)=(\beta,\gamma)$. This is always possible since $X\setminus C_\beta$ has cardinality $\mathfrak{c}$, and you have only chosen $|\alpha|<\mathfrak{c}$ elements so far. In the end, you get a family $(U_\gamma)$ such that each $U_\gamma$ is not contained in any $C_\beta$, and thus has second category.

Eric Wofsey
  • 330,363
  • This is very nice, thank you. One question: Can you please explain why there are $\mathfrak{c}$ countable unions of closed nowhere dense sets? – aduh Jan 26 '18 at 21:44
  • There are only $\mathfrak{c}$ open sets (since $X$ has a countable basis and every open set is a union of some subcollection of the basis) so there are only $\mathfrak{c}$ closed sets, so there are only $\mathfrak{c}^{\aleph_0}=\mathfrak{c}$ countable collections of closed nowhere dense sets. – Eric Wofsey Jan 26 '18 at 21:58
  • Thanks very much. Say, does there happen to be a publicly available version of this old talk of yours? Seems like something I could learn a lot from. – aduh Jan 28 '18 at 17:15
  • I'm afraid not. I do have notes from some similar lectures I gave which I could email you if you want. – Eric Wofsey Jan 28 '18 at 17:34