# Is it necessary that gcd > 1 of an infinite set? [closed]

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, Andrés E. Caicedo, user9072, Alain ValetteNov 9 '12 at 17:59

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• I believe a compactness argument applies. The answer should be yes. Gerhard "Ask Me About System Design" Paseman, 2012.11.09 – Gerhard Paseman Nov 9 '12 at 17:36
• This is a nice question, but not one suitable for this site. You could reask it on math.stackexchange.com a similar site with a broader scope than this one, which is dedicated to question with a connection to current research. – user9072 Nov 9 '12 at 17:38
• You can consider a filtration $A_1\subset A_2\subset A_3 ...$ of your set of natural numbers $S$ such that $\cup A_i = S$ and every $A_i$ is finite. Then take the $\mathrm{gcd}$. – Jason Polak Nov 9 '12 at 17:42

## 2 Answers

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.