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Lethal Questions

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humph:
Post your überhard maths questions here for all to compare!


I may be a bit bitter about that kinda thing right now. A question on my Analysis 2 assignment:

Prove there is an ordering of with the property that for each the set is at most countable.


... which you need to assume the continuum hypothesis to prove. Crazy stuff  ???

kamil9876:
Doesn't the following ordering suffice?

iff is a non-negative rational number. Easy to check for transitivity and that other property whose name i keep forgetting. The countability follows from the countability of .

edit: just found out today it is called "anti-symmetry".

Ahmad:
I don't think that's a total order though

One idea is that there exists a minimal uncountable well-ordered set (by AC). Every section (the subsets in the initial post) is countable by minimality. We can find a bijection of this set with R assuming CH and the result follows.

kamil9876:
yeah it is not, he didn't say it has to be total though.

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