Consider an infinite set $S$, of positive integers. If all the finite subsets of $S$ have GCD $>$ $1$, is it necessary that the GCD of $S$ is greater than $1$ as well?
closed as off topic by Dan Petersen, GH from MO, Andres Caicedo, quid, Alain Valette Nov 9 '12 at 17:59Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question. 


Consider the set of finite linear combinations with integral coefficients from $S$. It is an ideal in $\mathbb{Z}$, hence it equals the set of multiples of some integer $d>0$. The integer $d$ divides every element of $S$, hence it cannot exceed the GCD of $S$. On the other hand, $d$ is a finite linear combination with integral coefficients from $S$, hence it is divisible by the GCD of some finite subset of $S$, which of course is divisibe by the GCD of $S$ as well. This shows two things at one stroke: $d>1$ by assumption, and $d$ equals the GCD of $S$. P.S. I voted to close, but then could not resist to write down what I consider the most natural proof. 


Yes. Fix some $x\in S$. It has finitely many factors, and one of these needs to be a common factor with every other number in $S$. Otherwise, let $y_i$ be such that $d_i\not\vert y_i$ where $\lbrace d_i: i\le m\rbrace$ is a list of the divisors of $x$ which are $>1$; then $\lbrace y_i: i\le m\rbrace\cup\lbrace x\rbrace$ is a finite subset of $S$ with common denominator 1. 

