A very interesting riddle for those of you with some basic background in Set Theory.

Prove or Disprove the Following Claim:

\[\begin{align*} & \exists B \subset \mathcal{P}(\mathbb{N}) \text{ s.t. }\\ & |B| = \aleph \text{ and }\\ & B \text{ is completely ordered by the} \subset \text{relation}.\\ \end{align*}\]

Notes:

\(\mathbb{N}\) denotes the set of natural numbers.

\(\aleph\) denotes the cardinality of the continuum (i.e. \(\aleph = 2^{\aleph_0}\)).

“B is completely ordered by the \(\subset\) relation” means that for every two elements a, b of B, either a is a subset of b or b is a subset of a.

Spoiler Alert - Solution Ahead

While it may seem a little counterintuitive the answer is yes - this set exists. It is really hard to construct one directly over the naturals numbers, but the solution is actually really simple:

Take a bijection \(f: \mathbb{Q} \rightarrow \mathbb{N}\). For each \(x \in \mathbb{R}\) let \(A_x = \{q \in \mathbb{Q} \mid q < x\}\). Then let \(B = \{f(A_x) \mid x \in \mathbb{R}\}\). \(\blacksquare\)