Ribenboim’s remark on posets. I am reading a book of Ribenboim “Rings and Modules” and I am trying to understand his remark on p.37:
Remark. In any partially ordered set $X$ (e.g., the set of all submodules of a module M, ordered by inclusion) the following assertions are equivalent: 
(i) Every strictly increasing (decreasing) sequence of elements of X is finite. 
(ii) Every nonempty subset of $X$ has a maximal (minimal) element. 
I tried (i) $\Rightarrow$ (ii) : Let $Y \subset X$ be a non-empty subset of X. Let C be a non-empty increasing chain in Y. We may suppose that C is strictly increasing (because, for instance, if there is a subset {a, b, c} with a = b = c, then we can write {a} instead). By hypothesis, C is finite, therefore C has a upper bound $y \in Y$, so $c \leq y$ for all $c \in C$ (in fact, $y \in C$ because C is finite).
By Zorn’s lemma, Y has a maximal element $m \in Y$.
But now I want to show (ii) $\Rightarrow$ (i).
I started with: let $S \subset X$ be a strictly increasing sequence of elements of $X$. Suppose $S$ is non-finite.
But I am stuck now. Can anybody please help me to finish this proof ?
 A: First of all, I think you ought to be more careful in your proof that $(i) \implies (ii)$.
Let $X$ be a nonempty subset of your poset. What's your (precise) justification that you can assume $X$ to be an increasing sequence?
Here is what I would do: Suppose $X$ does not have a maximal element. Consider $(X; \prec)$, where $\prec$ is the strict partial order associated to your poset. Since $X$ does not have a maximal element there is for every $x \in X$ some $y \in X$ such that $x \prec y$. Using dependent choice there is hence a function $f \colon \mathbb N \to X$ such that for all $n \in \mathbb N$
$$
x_n \prec x_{n+1}.
$$
But now $(x_n \mid n \in \mathbb N)$ is a strictly increasing sequence. Contradiction!
$(ii) \implies (i)$. This is easier: If $(x_n \mid n \in \mathbb N)$ is an infinite, strictly increasing sequence then the set $X = \{x_n \mid n \in \mathbb N \}$ doesn't have a maximum. (It may have an upper bound but it certainly hasn't a maximal element.) But that's a contradiction right away.
