Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Replies: 15   Last Post: Dec 24, 2013 6:34 AM

 Messages: [ Previous | Next ]
 Mike Terry Posts: 767 Registered: 12/6/04
Posted: Dec 22, 2013 10:29 AM

"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.

Date Subject Author
12/19/13 David Bernier
12/19/13 Josh Jaggard
12/19/13 William Elliot
12/19/13 William Elliot
12/20/13 Robin Chapman
12/20/13 m.michael.musatov@gmail.com
12/20/13 David C. Ullrich
12/20/13 Timothy Murphy
12/20/13 David Bernier
12/21/13 Timothy Murphy
12/24/13 David Bernier
12/24/13 Peter Percival
12/22/13 quasi
12/22/13 Timothy Murphy
12/22/13 quasi
12/22/13 Mike Terry