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Let $(X, \tau)$ be Cantor space. That is, $X = \{0,1\}^\omega$ and $\tau$ is the collection of open sets in the product of discrete topologies on $\{0,1\}$. Let $\mathcal{A}(\tau)$ be the algebra generated by $\tau$. So $\mathcal{A}$ contains the open and closed sets, as well as finite unions and intersections of these.

Does there exist an uncountable collection $\{D_i: i \in [0,1]\}$ of pairwise disjoint, dense, Borel subsets of $X$ such that $D_i \notin \mathcal{A}(\tau)$ for all $i$?

If we do not require that $D_i \notin \mathcal{A}(\tau)$, then the answer is affirmative, by a result that is apparently due to Ceder ("On maximally Borel resolvable spaces"). Unfortunately, I cannot access Ceder's paper, and so I cannot see about modifying his proof. So, in addition to the main question above, references to a proof of Ceder's result would be appreciated.

I have tried to answer this by mimicking the inductive technique used in the answer to this question. But I cannot see how to ensure by that method that my collection is Borel and not contained in $\mathcal{A}(\tau)$.

aduh
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1 Answers1

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The stipulation $D_i\notin\mathcal A(\tau)$ is redundant: if $D$ is a dense set with dense complement, then $D\notin\mathcal A(\tau).$ This is because, if a set is dense and has dense complement, its boundary is the whole space. On the other hand, every set in $\mathcal A(\tau)$ has a nowhere dense boundary. To see this, observe that the boundary of an open set is nowhere dense, and the class of sets with nowhere dense boundary is closed under complementation and finite union, since $\partial(X\setminus A)=\partial(A)$ and $\partial(A\cup B)\subseteq\partial(A)\cup\partial(B).$

Therefore, it will suffice to partition the Cantor space $X$ into countable dense subsets. Consider $X$ as a topological group in the usual way, as a product of groups of order $2.$ The set of all elements with finite support is a countable dense subgroup, and its cosets are countable dense sets, Q.E.D.

bof
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