Re: Aleph One Sets
From: Mike Oliver (mike_lists_at_verizon.net)
Date: 10/22/04
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Date: Fri, 22 Oct 2004 13:09:29 -0500
Mike Oliver wrote:
> Keith Ramsay wrote:
>> There are 2^c sets of reals that are dense in the reals. The
>> order-preserving bijections between dense sets of reals are
>> all restrictions of continuous functions from the reals to
>> the reals. Since continuous functions can be determined by their
>> values on the rationals, there are c^{aleph-0} = c continuous
>> functions. If x<=2^c and x*c>=2^c, it must be that x=2^c. I
>> don't know whether I needed the axiom of choice here.
>
>
> For this argument you do, yes. Essentially you're arguing
> that each Wadge class has cardinality at most c and there
> are 2^c representatives, so there must be at least 2^c
> Wadge classes.
>
> But in a model of AD, the Wadge classes can be wellordered
> whereas P(R) certainly cannot, so there can't possibly
> be an injection from P(R) into the collection of Wadge classes.
So on second thought maybe there's a little work left to do
here -- the Wadge stuff doesn't work quite as smoothly
on the "real reals". I can't believe it matters much;
my argument *would* work if you were looking at dense
subsets of the irrationals, and there's only a countable
set distinguishing them from the reals--but I don't
instantly see how to finish the argument off.
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