
Re: N^N
Posted:
May 3, 2013 7:00 AM


On May 3, 3:51 am, William Elliot <ma...@panix.com> wrote: > On Thu, 2 May 2013, Butch Malahide wrote: > > On May 2, 9:59 pm, William Elliot <ma...@panix.com> wrote: > > > > > > > Why in N^N homeomorphic to R\Q? > > > > > > R\Q is homeomorphic to the space of *positive* irrational numbers. The > > > > > simple continued fraction expansion is a natural bijection between the > > > > > positive irrationals and the space N^N. > > > > > Where by "positive irrationals" I mean irrationals between 0 and 1. > > > > How so? The continued fraction for positive integers a1, a2,.. > > > [a1, a2,.. ] = a1 + 1/(a2 + 1/(a3 + ..)) > > > > Would not those continued fractions not be > > > limited to (0,1) but to (0,oo)? > > > By (0,oo) do you mean (1,oo)? > > Whoops. Indeed yes, (1,oo).
Well, (n1, n2, n3, ...) > a1 + 1/(a2 + 1/(a3 + ... is a bijection from N^N to (1,oo); on the other hand, (n1, n2, n3, ...) > 1/(a1 + 1/(a2 + 1/(a3 + ... is a bijection from N^N to (0,oo). I guess you could do it either way and get a homeomorphism.

