Solution to the Chain of Sets puzzle

[This is a solution to the Chain of Sets puzzle posted earlier]

A brief description of the problem:

Is there an uncountable set $F$ of subsets of $\mathbb{N}$ such that for any distinct $A, B \in F$ either $A \subset B$ or $B \subset A$?

Solution

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A solution uses the fact that the dense set of rationals $\mathbb{Q}$ is countable i.e. there is some numbering of the rationals as $$\mathbb{Q} = \{q_1, q_2, q_3, \dots \}$$
For each real number $r$, let $Q_r$ be the set of rationals strictly less than $r$.

Each such $$Q_r = \{q_{i_1}, q_{i_2}, q_{i_3}, \dots \}$$ can be associated with a subset of naturals (by taking the subscripts) $$N_r = \{i_1, i_2, i_3, \dots\}$$
If $y \lt x$, then we have $N_y \subset N_x$. Since the rationals are dense, $N_x \neq N_y$.

So the answer is, yes, there is an uncountable set with the required property: $$F = \{N_r : r \in \mathbb{R}\}$$