## Sylvy’s weekly puzzle #4

This is an old one, but good fun.

**Sylvy’s weekly puzzle #4**

Let denote the powerset of the natural numbers. For clarity, we *will* consider 0 to be a natural number. Then the pair

is a *partial order*: the (weak) ordering is reflexive, antisymmetric, and transitive. But it is very definitely **not** a *total order*: given we may have neither nor . A *chain* is a subset of on which **is** a total order. For example:

is a chain, but note that is countable.

This week’s problem is to find an *uncountable* chain, that is a subset of which is totally ordered by 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!

## Sylvy’s weekly puzzle #4a | sylvy's mathsy blog 9:16 pm

onNovember 4, 2015 Permalink |[…] are unfamiliar with the `countable’ and `uncountable’ terminology that I used in puzzle #4. Instead of introducing it, I’ll give a completely different puzzle which was originally […]

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## Puzzle corner: Solution to Sylvy’s puzzle #4 | sylvy's mathsy blog 11:48 am

onDecember 18, 2015 Permalink |[…] this post I want to discuss the solution to Puzzle #4. Let’s recall the puzzle: it was to find an uncountable chain in […]

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