primes as far as the eye can see, discrete continua

[Justin]

On 2004-12-08T10:30:22-0500, Tyler Durden wrote:
> >From: "Major Variola (ret)" <mv at cdc.gov>
> >
> >Saw in a recent _Science_ that Ben Green of Cambridge proved
> >that for any N, there are an infinite number of evenly spaced
> >progressions
> >of primes that are N numbers long.   He got a prize for that.
> 
> What about where N=1?
> 
> I don't understand. You can only have an infinite number (or number of 
> progressions) where the number of numbers in a number is inifinite.

True for N=1 trivially, because it's easily proven that there are
infinitely many primes.  (For a set of primes S, find the product of
them all and add 1.  The result is obviously not divisible by any prime
in S, so it's either a prime or a composite that factors into at least
two smaller primes not in S.  Either way, add the new prime(s) to S, and
repeat.)

I looked at B. Green's paper, but got lost around page 10 (of 50).

He apparently proves that there are arbitrarily long progressions of
primes.  From that, you can cut some such arbitrarily long progression
of primes into k-length progressions, and as N->infinity, you end up
approaching an infinite number of k-length progressions.

It's even easier (conceptually) if you accept two different progressions
that have different spacing.  for instance, when N=3,

5,11,17
17,23,29
31,37,43
would be a set of equal-spacing progressions.

5,11,17
17,53,89
would be a set of unequal-spacing progressions.  Different progressions
have different spacings.

The paper was giving me a headache so I don't want to try to figure out
which he meant.  Clearly, the former is stronger.