Does $\gcd(I)=1$ imply the monoid generated by $I$ is $\mathbb{N}$ minus finitely many numbers?

Question

This is true if $I=\{a_1,\dots,a_n\}$ is a finite set of positive integers. Namely, if $\gcd(a_1,\dots,a_n)=1$, then for all sufficiently large $N$ there is a non-negative integer solution $(k_1,\dots,k_n)$ to $$k_1a_1+\cdots+k_na_n = N.$$ In other words, the monoid generated by $I$ consists of every natural number except for possibly finitely many exceptions.

I want to consider an infinite set $I=\{a_1,a_2,\dots\}$ an infinite set of positive integers with $\gcd(a_1,a_2,\dots)=1$. Then is it true that for all sufficiently large $N$ there is a non-negative integer solution $(k_1,k_2,\dots)$ to $$k_1a_1+k_2a_2+\cdots = N$$ where $k_i=0$ for all but finitely many $i$?

My attempt: It's enough to find a finite subset of $I$ with gcd 1, and then we can apply the result of the finite case. To do this, set $b_1=a_1$. Then $b_1$ has finitely many prime factors, and we can let $p$ be the smallest. Since $\gcd(a_1,a_2,\cdots)=1$, there exists $a_i$ such that $p \nmid a_i$. Set $b_2=a_i$. Now $\gcd(b_1,b_2)$ has strictly fewer prime factors than $b_1$ (since $p$ is not one of them), and we can let $p'$ be the smallest. Again, there must be $a_j$ such that $p' \nmid a_j$, so set $b_3=a_j$. Then $\gcd(b_1,b_2,b_3)$ has strictly fewer prime factors than $\gcd(b_1,b_2)$. Continue in this fashion, and since the number of prime factors of $\gcd(b_1,\dots,b_t)$ is strictly decreasing with $t$, there must be $T$ such that $\gcd(b_1,\dots,b_T)=1$. Is this correct? Is there a simpler way to arrive at this result?

Answer 1

Your idea is good. It can be formalized in a clearer way.

For a finite subset $F$ of $I$, define $d(F)$ to be the gcd of the members of $F$. It is easy to show that if $F_1\subseteq F_2$, then $d(F_2)$ is a divisor of $d(F_1)$.

Then there exists $G$ such that $d(G)$ is minimal. I contend that $d(G)$ is $1$. Indeed, if $p$ is a prime divisor of $d(G)>1$, we can find $b\in I$ such that $p\nmid b$, otherwise every element of $I$ would be divisible by $p$.

Then $p\nmid d(G\cup\{b\})$, so $d(G\cup\{b\})$ is a proper divisor of $d(G)$, hence smaller. Contradiction.

Finally, the submonoid generated by $G$ is contained in the submonoid generated by $I$. Since the former contains all positive integers from some point on, the same is true for the latter.