ATAR Notes: Forum
Uni Stuff => Science => Faculties => Mathematics => Topic started by: humph on May 17, 2008, 10:01:21 pm
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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 ???
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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".
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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.
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yeah it is not, he didn't say it has to be total though.