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 on November 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 on December 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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