Background
Definition 1: Let $R$ be a commutative ring with identity, $c\in R$ and let $I$ be the set of all multiples of $c$ in $R$, that is, $I=\{rc\mid r\in R\}$. Set $I$ is an ideal and is called the principal ideal generated by c and hereafter will be denoted by $(c)$.
Theorem 1: Let $R$ be a PID, and let $a_1,a_2,\ldots,a_n$ be nonzero eleements of $R$. Then $d\sim \text{gcd}(a_1,a_2,\ldots,a_n)$ exists, and there exist $r_1,\ldots,r_n\in R$ such that
$$\text{gcd}(a_1,a_2,\ldots,a_n)=r_1a_1+r_2a_2+\cdots+r_na_n.\quad\textbf{(a)}$$
...In a PID we can say more. If $A_1,A_2,\ldots,A_n$ are ideals of any ring $R$, their sum is defined by $$A_1+A_2+\cdots+A_n=\{x_1+x_2+\cdots+x_n\mid x_i\in A_i \text{ for all }i\}\quad\textbf{(b)}$$.
This is easily verified to be an ideal of $R$ containing each $A_i$. In particular, if $a_1,a_2,\ldots,a_n$ are nonzero elements of $R$, then
$$\langle a_1\rangle + \langle a_2\rangle + \ldots +\langle a_n\rangle=\{r_1a_1+r_2a_2+\cdots+r_na_n\mid r_i\in R\}\quad\textbf{(c)}$$
is the ideal considered in the proof of Theorem 1. Hence, that proof shows that $(a)$ below is true. The dual holds too:
$$d\sim\text{gcd}(a_1,a_2,\ldots,a_n)\quad\text{ if and only if }\quad\langle a_1\rangle + \langle a_2\rangle + \ldots +\langle a_n\rangle=\langle d\rangle\quad\textbf{(d)}$$
$$m\sim\text{lcm}(a_1,a_2,\ldots,a_n)\quad\text{ if and only if }\quad\langle a_1\rangle \cap \langle a_2\rangle \cap \ldots \cap\langle a_n\rangle=\langle m\rangle\quad\textbf{(e)}$$
Exercsie 26: An ideal $A$ of a commutative ring $R$ is said to be finitely generated if we have $A=\{r_1a_1+r_2a_2+\cdots+r_na_n\mid r_i\in R\}$ for some $a_1,a_2,\cdots,a_n\in A$. We write $A=\langle a_1,\ldots,a_n\rangle$ in this case and say that $a_1,a_2,\ldots,a_n$ generate $A$.
$(a')$ Show that the following conditions are equivalent for an integral domain $R$ (then called a Bezout domain).
$(1)$ Every 2-generated ideal $A=\langle a,b \rangle$ is principal.
$(2)$ If $a\neq 0$ and $b\neq 0$, then $d=\mathrm{gcd}(a,b)$ exists and $d=ra+sb$ for some $r,s\in R$.
$(b')$ If $R$ isa Bezout domain, show that every finitely generated ideal is principal; in fact, for all $a_1,\dots,a_n$ in $R$, show that $d\sim \mathrm{gcd}(a_1,\ldots,a_n)$ exists and that $\langle a_1,\ldots,a_n\rangle=\langle d\rangle$$.
Questions
I have some purely notations related questions.
Let $A=\langle a_1,\ldots,a_n\rangle$ and $A=\{r_1a_1+r_2a_2+\cdots+r_na_n\mid r_i\in R\}$ for some $a_1,a_2,\cdots,a_n\in A$. From $\textbf{(c)}$ and the definition of finitely generated ideal in Exercsie 26 abovve, Then in set builder notation, we have $A=\langle a_1,\ldots,a_n\rangle=\langle a_1\rangle + \langle a_2\rangle + \ldots +\langle a_n\rangle=\{r_1a_1+r_2a_2+\cdots+r_na_n\mid r_i\in R\}$
For simplicity, say $n=2$, then
$(1)$ for $(1)$ of Exercise 26 for $A=\langle a_,b\rangle$ consider as a principal ideal, that means there exists some $c$ so that $A=\langle a_,b\rangle=\langle c\rangle$. But from Definition 1 above, principal ideal generated by c is denoted as $\langle c\rangle=I=\{rc\mid r\in R\}$, then $A=\langle a_,b\rangle=\{r_1a_1+r_2a_2\mid r_1,r_2\in R\}=\langle c\rangle=\{rc\mid r\in R\}$?
$(2)$ For the case of gcd of (d) for $n=2$, $ \langle a_1\rangle + \langle a_2\rangle=\langle d\rangle$, then in set builder notation, it would be $ \langle a_1\rangle + \langle a_2\rangle=\{r_1a_1+r_2a_2\mid r_1,r_2\in R\}=\langle d\rangle=\{sd\mid s\in R\}$?
$(3)$ if $\langle d\rangle$ is the gcd of $\langle a,b\rangle$ with $c$ being the principal. ideal of $\langle a,b\rangle$ and $d\mid c$, then necessarily $\langle c\rangle \subset\langle d\rangle$?
Thank you in advance
