## Sylvy’s weekly puzzle #4

Cross-reference to puzzle page on my website

This is an old one, but good fun.

Sylvy’s weekly puzzle #4

Let $\mathcal{P}(\mathbb{N})$ denote the powerset of the natural numbers. For clarity, we will consider 0 to be a natural number. Then the pair $(\mathcal{P}(\mathbb{N}),\subseteq)$

is a partial order: the (weak) ordering is reflexive, antisymmetric, and transitive. But it is very definitely not a total order: given $X,Y\in\mathcal{P}(\mathbb{N})$ we may have neither $X\subseteq Y$ nor $Y\subseteq X$. A chain is a subset $\mathcal{C}$ of $\mathcal{P}(\mathbb{N})$ on which $\subseteq$ is a total order. For example: $\mathcal{C}_{0}=\big\{\emptyset,\{0\},\{0,1,\},\{0,1,2\},...\big\}$

is a chain, but note that $\mathcal{C}_{0}$ is countable.

This week’s problem is to find an uncountable chain, that is a subset $\mathcal{C}$ of $\mathcal{P}(\mathbb{N})$ which is totally ordered by $\subseteq$ and uncountable.

Deadline: 10am Thursday 5th of November

(you can either send your solution by email or put a hand-written solution into the folder on my office door)

Prize: I’m very please to anounce that the winner will get a book of mathematical puzzles!!

Solution class: there will be a class at 10am on Thursday 5th of November in my office to go over the solution to this puzzle.

The webpage for these puzzles is http://anscombe.sdf.org/puzzle.html

Have fun!

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