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N^N
Posted:
May 11, 2013 11:41 PM


On Fri, 10 May 2013, Leon Aigret wrote: > <fred.galvin@gmail.com> wrote: > >On May 3, 3:51 am, 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. > >> > >> > > The continued fraction for positive integers a1, a2,.. > >> > > [a1, a2,.. ] = a1 + 1/(a2 + 1/(a3 + ..)) > > > >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. > > Just curious. Would it be possible to prove the homeomorphism part > without observing that a set of all natural number sequences with an > identical initial segment is open in N^N, that all such sets form a > base for its topology, that continued fraction theory shows that these > sets are mapped to open intervals of irrationals and, finally, that > there are sufficiently many of these open intervals to provide a base > for the neighborhood systems of all irrationals involved?
The mapping of open base sets of N^N to open intervals of R\Q shows the bijection is open. What's left to show is the bijection is continuous.
Wouldn't that be possible by showing the longer the initial segment, the smaller the mapped interval.



