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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 12, 2013 9:34 AM
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On Sat, 11 May 2013 20:41:12 -0700, William Elliot <>

>On Fri, 10 May 2013, Leon Aigret wrote:

>> 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.

It would. An initial segment of length n by itself would map to some
p/q with gcd(p, q) = 1 and certainly q >= n and for every irational x
in the interval one has |x - p/q| < 1 / q^2


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