# Translation of a non-standard analysis formulation

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.

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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.

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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.

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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".

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