# Intersection of infinitely many intervals with increasing number of elements

I have a system of intervals: $[A_1, +\infty)$; $[A_2, +\infty)$; $[A_3, +\infty)$; ... $A_1 < A_2 < A_3 < ...$ etc.

Let $NP(i)$ denote the number of primes of the form "s^2+d" , "d" is constant integer, (+infinity is also acceptable) occurring in interval $[A_i, +\infty)$, $i = 1, 2, 3,$ ... Let $NP(1) + NP(2) + NP(3) + ... = + \infty$.

Is deduction that there has to exist such number $k$ that $NP(k) = +\infty$ OK ?.

Consider that intersection of all intervals $[A_i, +\infty)$, $i = 1, 2, 3,$ ... is null set.

-
What does NP(k) mean here, and what does it have to do with your system of intervals? – Henning Makholm Apr 4 '13 at 16:50
What does $<$ means for the intervals? Is it $[$ ? – 1015 Apr 4 '13 at 16:54
NP(i) denote the number of primes (infinite is also acceptable) occurring in i-th interval <Ai,+∞), i=1,2,3, ... – pgalik Apr 4 '13 at 16:59
mark "<" means that interval is closed – pgalik Apr 4 '13 at 17:01
I have to note that primes in interval <Ai,+∞) have the form s^2+d, where "d" is some constant integer and "s" is integer variable – pgalik Apr 4 '13 at 17:06

If $NP(1)<+\infty$, then only a finite number of primes of the form $s^2+d$ occur in $[A_1,+\infty)$. Suppose that there are $r$ such primes and that the biggest such prime is $P$. Then since $A_1<A_2<\cdots$, there must be some $k$ with $A_k>P$. Now $$r=NP(1)\ge NP(2)\ge NP(3)\ge\cdots \ \ \text{and}\ \ NP(k)=NP(k+1)=\cdots=0,$$ so $$NP(1)+NP(2)+NP(3)+\cdots\le r(k-1) <+\infty,$$ a contradiction. Therefore, under the assumptions given, $NP(1)+NP(2)+\cdots=+\infty$ implies that $NP(1)=+\infty$.