Jim told me this cute problem about two years ago. I was reminded of it while reading the recent posts over at Ars Mathematica. For your enjoyment:

Prove that in the poset of subsets of , there exists an uncountable chain(!)

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Jim told me this cute problem about two years ago. I was reminded of it while reading the recent posts over at Ars Mathematica. For your enjoyment:

Prove that in the poset of subsets of , there exists an uncountable chain(!)

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August 29, 2007 at 4:49 am |

Heh. Cute indeed. Here’s a partial spoiler: Replace the natural numbers by some other countable set. And a total spoiler (rot13 to spoil): Hfr gur engvbany ahzoref. Qrqrxvaq phgf.

August 29, 2007 at 11:05 am |

It is a nice problem. One curious thing is that the first uncountable ordinal, which is the staple ingredient of most counterexamples in this spirit, is a total red herring here. Indeed, the chain involved cannot be well-ordered.

August 30, 2007 at 1:26 pm |

Wow! This (1) seems impossible, yet (2) has a trivial proof. Very thought-provoking.

July 15, 2010 at 11:35 am |

Let the countble set be the set of rational numbers.

with each real number x, let A(x) be the set of rationals smaller than x.

The chain of sets {A(x): x in R} is as required (its cardinaliy is two to aleph 0, which is uncountable).

No use of the axiom of choice is needed.

June 23, 2014 at 11:45 pm |

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March 25, 2017 at 7:08 pm |

A really good answer, full of raltnoatiiy!