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Re: N^N
Posted:
May 12, 2013 9:34 AM


On Sat, 11 May 2013 20:41:12 0700, William Elliot <marsh@panix.com> wrote:
>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
Leon



