## RE: [PrimeNumbers] Two infinite sets

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• Another thought. If we redue P by a prime, and Q by all numerators that have that prime as a factor, then P and Q are always 1-1, however P(inf)=inf, and
Message 1 of 4 , Oct 24, 2002
Another thought.

If we redue P by a prime, and Q by all numerators that have that prime as a
factor, then P and Q are always 1-1, however P(inf)=inf, and Q(inf)=0.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
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-----Original Message-----
From: Phil Carmody [mailto:thefatphil@...]
Sent: 23 October 2002 22:45
Subject: RE: [PrimeNumbers] Two infinite sets

--- Jon Perry <perry@...> wrote:
> Slightly off track, if we look at P=set of integers, and Q = set of
> rationals, then there is a 1-1 relation between them.
>
> If we knock out the first prime from P, and all rationals from Q with this
> prime as a numerator, is there a 1-1 relation?

Assuming you define 'numerator' correctly, then you don't need to change the
canonical diagonal traversal. It gets ugly pretty quickly but it's primitive
recursive.

> As P will end up at the set of composites, which is infinite, then I would
> say it makes sense to assume that Q at the end is infinite.

You do not need P in order to describe or deduce the behaviour of Q.
It seems trivial to create an injection from |N to the limit of Q, and
that's
all you need to prove the cardinality of that limit. Care is required with
defining the limit of Q, but it appears that taking the infinite
intersection of all the Qs would work. Perhaps someone fresher in
transfinite set theory can put me straight if I err.

Phil

=====
First rule of Factor Club - you do not talk about Factor Club.
Second rule of Factor Club - you DO NOT talk about Factor Club.
Third rule of Factor Club - when the cofactor is prime, or you've trial-
divided up to the square root of the number, the factoring is over.

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