```Date: Jun 14, 2013 4:32 AM
Author: Tucsondrew@me.com
Subject: Re: Matheology § 285

On Thursday, June 13, 2013 11:53:18 PM UTC-7, muec...@rz.fh-augsburg.de wrote:> On Thursday, 13 June 2013 18:24:14 UTC+2, Zeit Geist wrote:> > > > > I said for each particular q e Q, you must be able to produce a natural number m_(1/2) such that at step m(1/2) we have for at least one-half of all rationals p, p < q, p is in the natural order.> > > > Why?> > > > Why should that be required? Is there some axiom demanding it? > No axiom, but some is less than all.> >  > For a general well-ordering of the rationals, not in order of magnitude, we don't have to worry about, say, 1/2 of the rations less than q.> > > > With exactly the same right you could demand that there is an n such that half of all natural numbers have been attached to rationals. And that there is an m such that 2/3 of all natural numbers have been attached, and so on. In fact that would make sense and we could trust in countability - as your naive idea shows. But that is not possible. Quite the reverse! Enumerate the rationals as far as you like. Let n be the largest natural number that ever a human being will think about. Then with 1, 2, ..., n you will have enumerated less than> 10^-10000000000000000000000000000000000 of all rationals. Much less!Read what I wrote, dummy.For any particular rational, you must be able to find a step where 1/2 all the rationals lessthan that particular rational, not 1/2 of all rationals, must be placed in its final place.Now change "rational" to "natural".You should ( but probably lake the intellect to ) be able to see the difference.And, BTW, the largest natural anyone can think of is irrelevant.> > > > Your idea should show you very clearly what a nonsense notion countability is.> You are a fool.The cases are not analogous.For all n e N,  { m e N | m < n } is finite.For all q e Q, { p e Q | p < q } is infinite.Its about ordinality not cardinality.Please stop talking about things you don't or won't understand.> > Regards, WMZG
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