Is this set of integer sequences countable?

Question

I'm faced with a set of strictly increasing functions $\Bbb N\to \Bbb N$, i.e. positive integer valued sequences. The only thing I know about them is that they are pairwise eventually disjoint, by which I mean that given $U$, $V$ in this set, $U(\Bbb N)\cap V(\Bbb N)$ is finite.

Is this family countable? Uncountable? Does it depend?

Obviously the set of all positive integer sequences is uncountable, so this feels like a bit of a long shot, but I'm hoping the pairwise eventually disjoint condition might put a cap on the cardinality.

One thing I've been able to do, using a sort of diagonal argument, is to show that given such a set, if it's countable, I can find a new sequence that's eventually pairwise disjoint to every element in the set, and yet has no number assigned to it. But this new sequence won't actually be in the set, so this doesn't produce a contradiction like a diagonal argument is supposed to.

If the set is countable, then I can "decapitate" each sequence by a finite amount to obtain a set of pairwise disjoint sequences, which I'd really like to be able to do.

Answer 1

You can have as many as $2^\omega=\mathfrak{c}$ such sequences. First, we can identify strictly increasing sequences with their ranges. Two subsets of $\Bbb N$ are almost disjoint if their intersection is finite. There are families of $2^\omega$ pairwise almost disjoint infinite subsets of $\Bbb N$: see my answer to this question and the answers to the question that it duplicated for a variety of constructions.

Answer 2

There is no concrete answer. If the sets are pairwise disjoint, it is easy to show that the family is countable, but it will also satisfy your requirements.

However, it is possible to find an uncountable family of sets which are pairwise eventually disjoint, or almost disjoint. The key point is to note that there is no uniform bound on when two sets become disjoint.

Answer 3

For each real number $r>1$ we can define $f_r(n)=\lfloor rn\rfloor$. If $x\neq y$ then there exists natural number $N$ such that $|x-y|>1/N$, it forces $f_x(N)\neq f_y(N)$. Also, $\lfloor r(n+1)\rfloor=\lfloor rn+r \rfloor\ge \lfloor rn+1\rfloor>\lfloor rn\rfloor $ so $f_r$ is strictly increasing. Therefore $r\mapsto \lfloor rn\rfloor$ gives $\mathfrak{c}$ many strictly increasing sequences of natural numbers.