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Topic: Return to Hilbert's Hotel ...
Replies: 15   Last Post: Dec 24, 2013 6:34 AM

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Mike Terry

Posts: 664
Registered: 12/6/04
Re: Return to Hilbert's Hotel ...
Posted: Dec 22, 2013 10:29 AM
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"quasi" <quasi@null.set> wrote in message
news:sdldb9t34jo43l4q1ue6882ofhat6q8vbi@4ax.com...
> Timothy Murphy wrote:
> >quasi wrote:
> >>David Bernier wrote:
> >>>
> >>>At Chez Hilbert, there are countably infinitely many rooms,
> >>>and countably infinitely many guests.
> >>>
> >>>Is it true that we can always find infinitely many guests
> >>>which are mutual acquaintances, or infinitely many guests
> >>>that are mutual strangers?.

> >>
> >> Yes, I think so.
> >>
> >> Here's the argument I came up with.
> >>

> >I'm not sure if your argument
> >

> >> ...
> >
> >is basically different from that in

>
><http://en.wikipedia.org/wiki/Ramsey%27s_theorem#Infinite_Ramsey_theorem>.
>
> I didn't look.
>
> At first I couldn't prove the claim, so instead I tried to
> construct a counterexample. But the way in which my attempt at
> a construction failed guided me towards a proof.
>

> >In any case the following re-statement of the Wikipedia argument
> >in the case n=2 (which it was pointed out is all that is needed)
> >seems to me to be as simple (or at least as short) as possible.
> >
> >Choose any guest x_0.
> >Either he knows an infinite number of guests,
> >or doesn't know an infinite number of guests (or both).
> >Call x_0 happy in the first case, and unhappy in the second,
> >and call the infinite set V_0 in either case.
> >
> >Now choose any guest x_1 in V_0,
> >and carry out the same argument in V_0.
> >
> >Continue in this way, choosing guests x_0,x_1,x_2,...
> >Either an infinite number of these guests are happy,
> >or an infinite number are unhappy.
> >In the first case the guests know each other,
> >in the second they don't know each other.

>
> Yes, definitely simpler.
>
> quasi


plus, I don't see any requirement for Axiom of Choice, given that the guests
are given as countable. (Where Timothy says "choose any guest ..." we can
specify the lowest numbered guest.)

Mike.





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