Question about proof of chain conditions Here is a proof from Atiyah-Macdonald:

For i) $\implies$ ii) could one not write "If $(x_n)$ is such that $x_m = x_{m+1} = \dots$ then obviously $x_m$ is a maximal element"?
I am asking because the book has been getting terser with proofs and has reached super-terse by now but "i) $\implies$ ii)" seems much longer than necessary. So I must be missing something.
Note: for the equivalence of these two statements we need choice (which is not mentioned in the book).
 A: It seems to me that the countability issue is something of a red herring.
$\Sigma$ is a partially ordered set, and $T$ is a nonempty subset of $\Sigma$.  Even if $T$ is countable, it need not be an increasing chain, i.e., of the form $\{x_n\}_{n=1}^{\infty}$ with $x_1 \leq x_2 \leq \ldots \leq x_n \leq \ldots$.  So if an increasing chain $\{x_n\}_{n=1}^{\infty}$ in $T$ stabilizes then the eventual value of the sequence is a maximal element of the subset $\{ x_n\}_{n=1}^{\infty}$ of $T$, which need not be a maximal element of $T$ itself.  Thus the "longer" proof given is necessary.
Added: For my take on the result in question, see Lemma 10 of these notes on factorization.  (This seems to be currently missing from my commutative algebra notes, which only mostly include the material in this shorter set of notes.)  To my eye what I say is the same as what Atiyah-Macdonald say, but someone learning the material for the first time might (possibly) think otherwise.
A: If $\Sigma$ is not countable, then not every subset $T$ is the set of terms of some sequence $(x_n)$.
A: I think the confusion is coming from the fact that $\Sigma$ a priori is any set that we don't know is countable or not - how on earth does not get a "sequence" out of this to contradict the ACC say in a proof? Perhaps the proof of (1) $\implies$ (2) from Isaacs' Algebra may be more illuminating:
Suppose $\Sigma$ satisfies the ACC and that there exists $S \subseteq \Sigma$ not empty such that $S$ has no maximal element. Then for every $a \in S$, the set 
$$\{b\in S : b > a\}$$ 
is not empty. Now we invoke the axiom of choice: given any $a \in S$ I can always choose $b$ such that $b > a$. Viz, there exists a "choice function" $f : S \rightarrow S$ such that $f(a) > a$ for all $a$. Now choose just some $a \in S$ and set $a_1 = a$,  $a_2 = f(a_1)$, in general we have $a_{n+1} = f(a_n)$. Then
$$a_1 < a_2 < a_3 < \ldots $$
is an increasing sequence that does not stabilise, contradicting the ACC.
