I don't know if it's true, I would like to prove (or disprove) if $S\subset \mathbb N$ is an infinite set, then there exists an finite set $S'\subset S$, such that $\text{gcd}\ S'=\text{gcd}\ S$.
I need this result to use in a theorem I'm proving.
I've already tried to prove it without success. Maybe someone could help me with a counterexample?
Any help is welcome
Write out the prime factorization of $\gcd(S)=p_{1}^{k_1}...p_{n}^{k_{n}}$. For each $p_{i}$ there exists an $s_{i}\in S$ so that $p_{i}^{k_{i}+1} \nmid s_{i}$ otherwise $p_{i}^{k_{i}+1}|s_{i}$ for all $s_{i}\in S$ and then $p_{i}^{k_{i}+1}|\gcd{S}$ (which it does not). Then take $S^{\prime}=\{s_1, s_2, ..., s_{n}\}$. (Note that it may be possible that $s_{i}=s_{j}$ for some pairs $i$ and $j$.)
Fix $n\in S$; clearly $\gcd S\mid n$. For each divisor $d$ of $n$ that does not divide $\gcd S$ there is an $n_d\in S$ such that $d\nmid n_d$. Let $D$ be the set of such divisors of $n$, and let $S'=\{n\}\cup\{n_d:d\in D\}$; then $\gcd S'=\gcd S$.
Hint $ $ A simple 2-line Bezout proof shows that $\,d = \gcd S\,$ is the least positive element of the set $\,\{ a_1 s_1 + \cdots+ a_k s_k : a_i\in\Bbb Z,\, s_i \in S\},\,$ so $\,d = a_1 s_1 + \cdots+ a_j s_j\,$ $\Rightarrow$ $\,d = \gcd(s_1,\ldots s_j)$
Write $S = \{s_1, s_2, s_3, \ldots, \}$ and set $d_i = \gcd\{ s_1, s_2, \ldots, s_i \}$ for $i \ge 1$. Then $d_1 \ge d_2 \ge d_3 \ldots$ and so there exists $j \ge 1$ such that $d_j = d_k$ for all $k \ge j$. It is easy to see that $d_j$ is a divisor of all numbers in $S$ and any other divisor is divided by $d_j$, i.e., $d_j = \gcd(S)$.
