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Topic: Cantor's first proof in DETAILS
Replies: 3   Last Post: Nov 17, 2012 10:49 PM

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Ben Bacarisse

Posts: 1,149
Registered: 7/4/07
Re: Cantor's first proof in DETAILS
Posted: Nov 17, 2012 7:45 AM
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"Ross A. Finlayson" <ross.finlayson@gmail.com> writes:

> On Nov 16, 7:50 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> "Ross A. Finlayson" <ross.finlay...@gmail.com> writes:
>> > On Nov 16, 8:12 am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> >> "LudovicoVan" <ju...@diegidio.name> writes:
>> >> > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote in message
>> >> >news:c2ac7c88-5567-408d-a520-df4f143f840f@googlegroups.com...

>>
>> >> >> <http://www.tiki-lounge.com/~raf/finlayson_injectrationals.pdf>
>>
>> >> > I'll see if I can understand it: for now, thanks for sharing.
>>
>> >> A first step is to remove the indexes from the irrationals.  In a
>> >> argument about the supposed countability of the irrationals, to refer to
>> >> them with indexes (e.g. p_i) looks like begging the question.

>>
>> >> As it happens, I don't think the indexes do anything but add a layer of
>> >> confusion.  I think you can rename the various quantities without
>> >> altering the meaning, i.e. rather than talk about irrationals p_i and
>> >> p_h just use p and r (q is taken).

>>
>> > It's constructive, that.
>>
>> The key set, Q_<i (which would then be called Q_<p) is empty though.

>
> Note the emphasized bit.


I did. That the set in question is empty simply implies that for every
rational q < p, there is an irrational r < p with r > q. That the set
in question is *not* empty would give any member of it the property of
being a rational least upper bound of all the irrationals less than p.

--
Ben.



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