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

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Leon Aigret

Posts: 31
Registered: 12/2/12
Re: N^N
Posted: May 10, 2013 12:53 PM
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On Fri, 3 May 2013 04:00:40 -0700 (PDT), Butch Malahide
<fred.galvin@gmail.com> wrote:

>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.


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?

Leon



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