Is there a sparse almost disjoint family over $\omega$ of cardinality $2^{\aleph_0}$?

Question

Is there an almost disjoint family $\mathcal{F}$ of subsets of $\omega$ of cardinality $2^{\aleph_0}$ satisfying the following property?

For all $A,B\in\mathcal{F}$ with $A\neq B$ and every $k\in\omega$, there is an $n\in\omega$ such that $|a-b|\geqslant k$ for all $a\in A\setminus n$ and $b\in B\setminus n$.

Answer 1

Yes. Let $\mathcal F$ be an almost disjoint family of subsets of $\{n^2:n\in\omega\}$ of cardinality $2^{\aleph_0}$.

Answer 2

Yes. Consider the function $f:2^{<\omega}\rightarrow 2^{<\omega}$ defined as follows:

• $f(\langle\rangle)=\langle\rangle$.

• Having defined $f$ on all strings of length $\le n$, we define $f$ on the strings of length $n+1$ as follows. Let $\{\sigma_i: 0\le i< 2^{n+1}\}$ enumerate the strings of length $n+1$, let $\tau^-$ be the immediate predecessor of $\tau$ (for $\tau$ nonempty), and let $$f(\sigma_i)=f(\sigma_i^-)0^{n\cdot (i+1)}(0^n10^n)0^{(n-i+1)n}$$ (where juxtaposition denotes concatenation). I've written that deliberately strangely to make the idea clear: each $\sigma_i$ gets one new $1$, but we add long blocks of $0$s to keep the new $1$s far apart from each other.

For $X\in 2^\omega$ let $\hat{X}=\{n: \exists\sigma\prec X(f(\sigma)(n)=1)\}$. The set $\{\hat{X}:X\in 2^\omega\}$ is then an almost disjoint family of size continuum satisfying the additional "spread-out" requirement. More generally, this sort of strategy lets us build large almost-disjoint families satisfying any "local" condition that doesn't get in the way of further "splitting."