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I've tried a bunch of different approaches to this question (and would particularly appreciate some hints that don't give the whole thing away immediately).

I've noticed that if $S$ contains some finite subsets, then for each $n \in \mathbb{N}$ there is at most one subset of size $n$ in the collection $S$ as $S$ can't contain duplicates (in order to satisfy the given condition). Certainly if $A_i$ is a finite subset contained in $S$ of size $i$, then $A_i \subset A_j$ if $i > j$.

I've tried considering the possibilities where there are some finite subsets in the collection $S$, and then we can also add in infinite-size subsets that are each a superset of all of the finite subsets. However it seems rather difficult to visualise what happens when you consider infinite subsets of $\mathbb{N}$ that can't be obtained by removing finitely many elements from $\mathbb{N}$. What are good approaches to take to figure out whether such a set can be uncountable - is there an element of construction here?

perfect4th
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Giving just a hint is hard because the idea is the whole thing. The point is that $(\Bbb R,\le)$ can be constructed by identifying $\Bbb R$ with the family of intervals of $\Bbb Q$ which are bounded above, unbounded below and without maximum, with the ordering being $\subseteq$. And $\mathcal P(\Bbb Q)$, as a poset, is essentially the same as $\mathcal P(\Bbb N)$.

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    Nice. A slightly less informative (somewhat cryptic) version of this hint would be to look for a solution with rationals rather than integers, then observe that the rationals are countable so ... – Ethan Bolker Jan 03 '21 at 15:34
  • What does the notation $(\mathbb{R}, \leq)$ represent? – perfect4th Jan 03 '21 at 15:58
  • The (partially) ordered set $\Bbb R$, with the usual order $\le$. @Integral12 –  Jan 03 '21 at 16:02