
Re: Return to Hilbert's Hotel ...
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 restatement 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.

