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Let $C$ be the Cantor set, and let $\{D_\alpha:\alpha<\kappa\}$ be a collection of pairwise disjoint subsets of $C$ such that each $D_\alpha$ is countable and dense in $C$, and $$\bigcup \{D_\alpha:\alpha<\kappa\}=C.$$

Is there a closed set $A\subseteq C$ such that $A\cap D_0=\varnothing$ and $A\cap D_\alpha\neq \varnothing$ for all $\alpha>0$?

I thought about doing something with measure theory here, using a "fat" cantor set of positive measure, and covering $D_0$ with open sets of small measure.

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  • $\begingroup$ $A=C\setminus D_0$? Bigger is not possible. Rest may or may not depend on $D_\alpha$ $\endgroup$ – SK19 Apr 8 '18 at 6:55
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    $\begingroup$ But I want $A$ to be closed. If it depends on the particular $D_\alpha$, I'd be interested in any positive example. $\endgroup$ – Forever Mozart Apr 8 '18 at 6:58
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You can construct a counterexample by transfinite induction. Start by letting $D_0$ be your favorite countable dense subset of $C$. Fix an enumeration $(U_\beta)_{\beta<\mathfrak{c}}$ of all the open subsets of $C$ which contain $D_0$. Now just choose disjoint countable dense subsets $D_\alpha$ for $1\leq\alpha<\mathfrak{c}$ one by one so that for each $\beta<\mathfrak{c}$, there is some $\alpha>0$ such that $D_\alpha\subseteq U_\beta$. We can do this since at any stage, we have defined fewer than $\mathfrak{c}$ of our sets $D_\alpha$, so we can choose a new countable dense subset which is disjoint from all the ones we've chosen so far and is contained in any particular $U_\beta$. We can also construct some extra countable dense subsets along the way to make sure that the union of all of them is all of $C$.

Now, suppose the set $A$ which you ask for existed. Then $C\setminus A$ would be an open set containing $D_0$, and so would be equal to $U_\beta$ for some $\beta$. But then there is some $\alpha>0$ such that $D_\alpha$ is contained in $U_\beta$, and $A$ cannot intersect that $D_\alpha$.

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