MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Usually, it is quite easy (if cumbersome) to translate a formula like "if $\varepsilon$ is infinitesimal, and if $f$ is differentiable at $a$, $f'(a)$ is the shadow of $\frac{f(a+\varepsilon)-f(a)}{\varepsilon}$". But I cannot find a standard translation of the conjecture :"There exists a galaxy of non-standard integers containing an infinite number of primes", where galaxies are the equivalence classes of $\mathbb N$ by the relation $m\simeq n \iff (m-n)$ is standard. The closest I can find is the k-tuple conjecture (the first of the Hardy-Littlewood conjectures), but it seems obvious this is quite stronger than mine.

share|cite|improve this question
up vote 2 down vote accepted

How about this ...

Let's say a finite set $V \subseteq \mathbb N$ is a Feldmann set iff there are infinitely many $n$ such that for all $u \in V$, $n+u$ is prime.

For example, $\{0,2\}$ is a Feldmann set iff there are infinitely many twin primes.

Your equivalent statment: there are arbitrarily large Feldmann sets.

share|cite|improve this answer
what if their are arbitrarily large Feldmann sets, but the size is bounded above by some function of the size of the smallest element? – Will Sawin Jan 23 '13 at 22:10
Any translate of a Feldmann set is a Feldmann set... – Gerald Edgar Jan 24 '13 at 2:26
What about the second-smallest element? – Will Sawin Jan 24 '13 at 5:56

There exists a sequence of numbers $a_i$, such that, for all $k$, there are infinitely many $n$ such that, for all $i$ between $1$ and $k$, $a_i+n$ is prime.

Proof this implies a galaxy: For each number $k$, choose an $n$ such that for all $i$ between $1$ and $k$, $a_i+n$ is prime, and has not been chosen before, then choose a non-principal ultrafilter on this sequence.

Proof of converse: Choose an element of your galaxy, and let $a_i$ be the differences between primes in the galaxy and that fixed element. Then the set of $n$ such that, for all $i$ between $1$ and $k$, $a_i+n$ is prime has a nonstandard element, so is infinite.

share|cite|improve this answer

This is more of a comment on the answers of Will Sawin and Gerald Edgar.

Let $P$ be the set of prime numbers, and given some $s\in {\mathbb N}$, let $P-s =\{ n\in {\mathbb N} \colon s+n\in P\}$.

If I have parsed their answers correctly, the negation of the standard translation should be: whenever $a_i$ is an (increasing?) sequence of numbers, there exists some $k$ such that $(P-a_1)\cap \dots \cap (P-a_k)$ is finite. This is precisely the assertion that the set $P$ is translation finite in the sense of my old MO question.

Hence the standard translation would seem to be phraseable as "the set $P$ is not translation-finite".

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.