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:

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> > 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.

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> Why?

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> Why should that be required? Is there some axiom demanding it?

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No axiom, but some is less than all.

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> > 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.

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> 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 less

than 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.

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> Your idea should show you very clearly what a nonsense notion countability is.

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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, WM

ZG