Let $\mathcal{A}$ be an uncountable almost disjoint family (not necessarily maximal) of infinite subsets of $\mathbb{N}$. Denote by $\mathcal{A}_{\subseteq}=\{ B\subseteq\mathbb{N}:|B|=\omega \wedge \exists A\in\mathcal{A}(B\subseteq A) \}$.

Question 1: Must there exist a $B\in\mathcal{A}_{\subseteq}$ such that for any $n$, $\{C\in\mathcal{A}:|C\cap B|\geq n\}$ is uncountable?

If this is true, then:

Question 1': Must there exist a $B\in\mathcal{A}_{\subseteq}$ such that for any $n$, $\{C\in\mathcal{A}:B\upharpoonright n\subset C\}$ is uncountable? (By $B\upharpoonright n$, I mean the first $n$ elements of $B$ listed in increasing order.)

I don't have an intuition as to whether this should be true or false, but the question stems from the observation that for any uncountable family $\mathcal{A}$ of subsets of $\mathbb{N}$, there is an $n\in\mathbb{N}$ such that $\{A\in\mathcal{A}:n\in A\}$ is uncountable.

EDIT: As seen in the solutions below, the passage to $\mathcal{A}_{\subseteq}$ is unnecessary.

EDIT: In light of the positive answer to the questions above, I'll add an additional question that I am interested in.

Question 2: Must there be a sequence $(A_n)_{n\in\omega}$ of distinct elements of $\mathcal{A}$ (or elements of $\mathcal{A}_{\subseteq}$ which are below pairwise distinct elements of $\mathcal{A}$) such that for any $n$, $\{B\in\mathcal{A}:\forall i\leq n(A_i\upharpoonright n\subset B)\}$ (or $\{B\in\mathcal{A}:\forall i\leq n(A_i\upharpoonright n-i\subset B)\}$) is uncountable. (Again, $A \upharpoonright n$ means the first $n$ elements of $A$ listed in increasing order, it does not mean the elements of $A$ below $n$ here.)

  • $\begingroup$ In question 1, why do we consider subsets instead of sets theirselves? $\endgroup$ – Fedor Petrov Nov 23 '15 at 17:32
  • $\begingroup$ @FedorPetrov I had a related application in mind, and this is all that was necessary. Certainly a witness in $\mathcal{A}$ would be fine. $\endgroup$ – Iian Smythe Nov 23 '15 at 17:38

Yes to both: Just take a condensation point $A$ of $\mathcal{A}$. This means every clopen neighborhood of $A$ contains uncountably many members of $\mathcal{A}$.


For any finite $M\subset {\mathbb N}$ which is contained in at most countably many sets from $\mathcal{A}$ we just remove them all out from $\mathcal{A}$ . Now any finite subset of any of our sets is contained in uncountably many sets.


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