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!

2 thoughts on “Sylvy’s weekly puzzle #4”

  1. Pingback: Sylvy’s weekly puzzle #4a | sylvy's mathsy blog
  2. Pingback: Puzzle corner: Solution to Sylvy’s puzzle #4 | sylvy's mathsy blog

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