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

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

$(X-n_1) \cap (x-n2) X-n_2) \cap ...\cap \dots\cap (x-nk) X-n_k)$

is finite or empty.

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

• if X $X$ 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.

4 edited tags
3 updated in light of info in comments

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. { 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-n1) \cap (x-n2) \cap ...\cap (x-nk) 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.

2 corrected a nonsensical slip
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