Given, say a perfect polish space $P$ that contains a Cantor set $C$.

Let $\mathcal A$ be $\mathcal P(\mathcal P(C))$. Given a sequence $\langle U_n : n\in\omega\rangle\in{}^\omega\mathcal{A}$, $U_i \ne U_j$, is there a way to pick a finite family of sets $F_n$ from $U_n$, given some conditions on each $U_i$ such that $$ \bigcup_{i=1}^{\infty}F_i = C $$ That is to say for each sequence $\langle U_n : n\in\omega\rangle$ of elements of $^\omega\mathcal{A}$, we pluck a finite family of sets $F_n$ that are contained in $\bigcup U_n \subset \mathcal A$. Doing this, is there a strategy that gets us the a copy of the cantor set?

I am genuinely just curious and I may end up doing a project associated with questions such as these. However, at the present moment, this question was just something that popped into my head.

If we take $U_i$ to be a sequence of singletons, there should be countably many distinct $U_i$, we'd pick finitely many singletons each time, but the countable union of finite sets is at most countable, so we'd never end up with a copy of the cantor set.

So now we impose the condition that each $U_i$ be a sequence such that at least one thing in the sequence is a countably infinite subset of $C$. Can we do it now?

The answer is STILL no, because we can just keep the same countably infinite subset for each $U_i$ but to satisfy that the $U_i$ are distinct, we can make it so that each $U_i$ contains a different singleton but other than that, $U_i = U_j$ $\forall i\ne j$.

What is the minimum restriction we place on $U_i$ so that no matter what sequence, under such restrictions, we can always pick finitely many from each $U_i$ and geta copy of the cantor set?
Certainly it cannot be that each $U_i$ is a sequence of countably infinite subsets of $\mathcal A$, because the countable union of countable sets is at most countable. What sort of things must each $U_i$ contain ? Is there a way to characterize how little I need to ask from each $U_i$?

  • $\begingroup$ I have no idea why we need that new tag. $\endgroup$ – Asaf Karagila Sep 22 '13 at 21:04
  • $\begingroup$ I think you want to say that $\langle U_n : n\in\omega\rangle\in{}^\omega\mathcal{A}$ in the beginning. $\endgroup$ – Shehzad Ahmed Sep 22 '13 at 21:21
  • $\begingroup$ @ShehzadAhmed Agreed $\endgroup$ – Rustyn Sep 22 '13 at 21:25
  • $\begingroup$ @AsafKaragila Sorry--Lol $\endgroup$ – Rustyn Sep 22 '13 at 21:29
  • $\begingroup$ I texted you this, but I figured I should add in the comments: Additionally, if I'm reading this correctly, you seem to be asking if you can pick finitely many elements from a countable sequence of subsets of $C$. But $C$ has cardinality continuum, and a countable union of finite sets would be countable. I may be misreading this. $\endgroup$ – Shehzad Ahmed Sep 22 '13 at 21:31

My conjecture is that the least conditions for each $U_i$ is that each $U_i$ admits something from a partition $\mathbb{P}$ of $C$, and that this partition is at most countably infinite, but I am unsure if there's something less explicit. Even if each $U_i$ admitted something from $\mathbb{P}$, wouldn't there be no way of finding this particular element from $\mathbb{P}$? Would there be a way to construct a partition of $C$, (in the end, by choosing carefully each member of $F_n$), given that each $U_i$ ONLY contained members from countably infinite partitions $\mathbb{P}$ of $C$ ?


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