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Topic: N^N
Replies: 10   Last Post: May 12, 2013 9:34 AM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: N^N
Posted: May 3, 2013 7:00 AM
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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.




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