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The definition in the title probably needs explaining. I should say that the question itself was an idea I had for someone else's undergraduate research project, but we decided early on it would be better for him to try adjacent and less technical questions. So it's not of importance for my own work per se, but I'd be interested to know if it easily reduces to a known conjecture/fact/counterexample in number theory.

Apologies if the question is too technical/localized/unappealing/bereft of schemes.

Given a subset $X$ of the natural numbers $N$, and given $n \in N$, we write $X-n$ for the backward translate of $X$, i.e. the set {$\{x-n : x\in X\}$}.

We say that $X$ is translation-finite if it has the following property: for every strictly increasing sequence n1 < n2 < in $N$, there exists k (possibly depending on the sequence) such that

$(X-n_1) \cap (X-n_2) \cap \dots\cap (X-n_k)$

is finite or empty.

Thus every finite set is trivially translation-finite: and if the elements of $X$ form a sequence in which the difference between successive terms tends to infinity, then $X$ is translation-finite and we can always take k=2. Moreover:

  • if $X$ contains an infinite arithmetic progression, or if it has positive (upper) Banach density, then it is NOT translation finite;

  • there exist translation-finite sets which, when enumerated as strictly increasing sequences, grow more slowly than any faster-than-liner function.

  • there exist translation-finite sets containing arbitrarily long arithmetic progressions.

These resultlets suggest the question in the title, but I don't know enough about number theory to know if it's a reasonable question. Note that if, in the definition, we were to fix k first (i.e. there exists k such that for any sequence (nj)...) then we would get something related to Hardy-Littlewood conjectures; but I was hoping that this might not be necessary to resolve the present question.

EDIT (2nd Nov) It's been pointed out below that the question reduces in some sense to a pair of known, hard, open problems. More precisely: if the answer to the question is yes, then we disprove the Hardy-Littlewood k-tuples conjecture; if the answer is no, then there are infinitely many prime gaps bounded by some absolute constant, and this is thought to be beyond current techniques unless one assumes the Eliott-Halberstam conjecture.

Added in 2013: Stefan Kohl points out that the latter is Yitang Zhang's famous recent result. However, as Will Sawin points out in comments, a negative answer to the main question would imply there are 3-tuple configurations occurring infinitely often in the primes, and (see thelink in Will's comment) this is thought to be out of reach even if we assume the EH conjecture holds.

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What do you mean by "the elements of X form a sequence tending to infinity"? Does that imply something other than X being an infinite set? –  Darsh Ranjan Oct 30 '09 at 0:07
    
aargh, typo/brain spasm - I missed out some important words. I meant that the gap size has to tend to infinity. I've edited the original post to correct this. –  Yemon Choi Oct 30 '09 at 2:00
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Yay for non-scheme questions! –  Theo Johnson-Freyd Oct 30 '09 at 2:12
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@Scott: makes sense to me. @Yemon: thanks for nicely summarizing th results in the question body! –  Ilya Nikokoshev Nov 27 '09 at 23:52
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Because the existence of any $3$-tuples which occurs infinitely often in primes is still out of reach, this is still open. See: mathoverflow.net/questions/132731/… –  Will Sawin Jul 4 '13 at 16:26
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1 Answer

up vote 19 down vote accepted

As you mention, this is related to the Hardy-Littlewood k-tuple conjecture. In particular, if their conjecture is true, then the primes are not translation-finite. Indeed, it is possible to find an increasing sequence n1 < n2 < n3 < ⋯ so that for every k, the first k nis form an admissible k-tuple. (For example, I think ni = (i+1)! works.) Then, by the k-tuple conjecture, infinitely many such prime constellations exist and so for all k, (X-n1) ∩ (X-n2) ∩ ⋯ ∩ (X-nk) is infinite. (Here and below, X is the set of primes.)

However, maybe we can prove that the primes are not translation finite by some other means. Unfortunately, the technology is not quite good enough to do that. Proving that the primes are not translation finite would, in particular, prove that there exist n1 < n2 such that (X-n1) ∩ (X-n2) is infinite. In particular, this implies that the gap n2-n1 occurs infinitely often in primes, and so pn+1-pn is constant infinitely often. (The standard notation pn indicates the nth prime.)

The best known upper bound for the size of small gaps in primes is that lim infn→∞ (pn+1-pn)/log pn = 0. This was established by Goldston and Yildirim around 2003 and the proof was later simplified. To the best of my knowledge, the best conditional result is by the same authors; they show that given the Elliott-Halberstam conjecture, the prime gap is infinitely often at most 20 or so.

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Thanks! That's very clear. So, if I understand you correctly: a proof of HL's k-tuple conjecture would imply the primes are not translation finite; while a proof that the primes are not translation finite would imply a bound on prime gaps that is beyond current knowledge. So my question essentially reduces to known open problems I guess. –  Yemon Choi Nov 1 '09 at 23:52
    
Which probably means it deserves a CW status + open-problems tag. –  Ilya Nikokoshev Nov 2 '09 at 0:04
    
As well as rewriting the question so the two open problems become clear. –  Ilya Nikokoshev Nov 2 '09 at 0:04
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@Yemon Choi: That's correct. Proving the primes are translation finite would disprove the k-tuples conjecture, while proving the primes aren't translation finite would prove bounds on small prime gaps. –  aorq Nov 2 '09 at 3:27
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Why does the infinitude of $(X-n_1)\cap(X-n_2)$ imply that the gap $n_2-n_1$ occurs infinitely often? Couldn't some smaller gaps occur infinitely often instead? –  François G. Dorais Dec 27 '12 at 4:59
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