Proving Ascending Ideals Stabilize in PID I am trying to understand the proof to the following theorem:

Let $R$ be a principal ideal domain and let $I_{1}, I_{2}, ...$ be ideals in $R$ such that $I_{1} \subset I_{2} \subset \cdots$. Then there exists an integer $N$ such that $I_{n} = I_{N}$ for all $n \geq N$.

Here is the proof I typed a while ago:
 
In the stablizing part of the proof, I don't understand why I let $x\in I_{N}$ then suddenly let $(x)=I_{N}$. I tried writing it again and got:

If $\bigcup I_{i}=(x)$, then $x\in I_{N}$ for some integer $N$ since $R$ is a principal ideal domain. Then for all $n\geq N$, $I_{N}\subset I_{n}\subset (x)$. Since $x\in I_{N}$, then it must also be that $I_{n}\subset I_{N}$. Then $I_{n}=I_{N}$. 

However, I could also be wrong with this approach.
 A: It strikes me that most if not all of the individual assertions our OP numericalorange puts forth in his efforts to construct a proof that principal ideal domains are Noetherian are in fact correct, and enter into a coherent proof at some point, but that in one or two places the exposition of the pattern of logical inference could perhaps be a little more clear.  For example, I found it a bit of a challenge to follow the statement,
"Since $x \in I_{N}$, then it must also be that $I_{n}\subset I_{N}$. Then $I_{n}=I_{N}.$"
Of course this is true, for the reason that $x \in I_N$ implies $(x) \subset I_N$ and then we have
$(x) \subset I_N \subset I_n \subset J = (x), \tag 0$
which then leads directly to  $I_n = I_N$ for $n \ge N$; but I think explicitly inserting the intermediate steps lends clarity to the argument.
I think it is also worth pointing out that the result binds under the somewhat weaker hypothesis that $R$ is merely a principal ideal ring; that $R$ is a domain nowhere need be invoked in what follows.
Having said these things, here is the way I present my demonstration of this fact.  In the following, I take as given that the union of a nested sequence of ideals is itself and ideal; this is a very well-known and oft-used result.  Now we have
$i < j \Longrightarrow I_i \subset I_j, \tag 1$
since the $I_i$ are given to be a nested sequence of ideals.  Then per assumption,
$J = \displaystyle \bigcup_i I_i \; \text{is an ideal}; \tag 2$
now since every ideal in $R$ is principal,
$\exists j \in R, \; J = (j); \tag 3$
which in the light of (2) implies
$\exists N \in \Bbb N, \; j \in I_N; \tag 4$
thus,
$(j) \subset I_N \subset J = (j) \Longrightarrow I_N = (j); \tag 5$
also,
$n \ge N \Longrightarrow (j) \subset I_N \subset I_n \subset J = (j); \tag 6$
therefore,
$n \ge N \Longrightarrow I_n = I_N = (j). \tag 7$
$OE\Delta$, what we sought to prove.  
In closing, I would say that the two key ideas here are ensconced in (2) and (6), which as it were "traps" $I_N$ 'twixt $(j)$ and itself.
