Usefulness of upper bounds for directed sets A directed system $\langle A_i, f_{ij} \rangle$ is defined in terms of a directed set $\langle I, \leq \rangle$. I was thinking, where in the definition of directed systems and the relevant directed limit do we use the fact that there is an upper bound for every pair of elements. That is, for any $x, y \in I$, $\exists z$ such that $x \leq z$ and $y \leq z$. This upper bound condition doesn't strike me as all that useful so I was wondering:
Why were directed sets defined with this condition. Mathematicians must have thought this had some use. What was it? Where does it come into play?
 A: As you have used the tag ring-theory, my answer is based on that. You may assume that the rings below are commutative with $1$ if you want.

Suppose you have a nonempty collection $\{I_{\lambda}\}_{\lambda \in \Lambda}$ of (left) ideals in a ring $R$.
In general, the union $U := \bigcup_{\lambda \in \Lambda} I_{\lambda}$ need not be an ideal. The issue is that the sum of two elements of $U$ need not be in $U$. (And this is the only issue. It is easy to see that $U$ contains $0$, is closed under inverses, and closed under multiplication by elements of $R$.)
Indeed, if $x$ belongs to some $I_{\alpha}$ and $y$ to some $I_{\beta}$, then there is no way to conclude that $x + y$ belongs to some $I_{\gamma}$ since $x$ and $y$ need not be contained in the some common ideal $I_{\lambda}$.
However, now assume that $\Lambda$ is a directed poset and we have $I_{\alpha} \subset I_{\beta}$ whenever $\alpha < \beta$ in $\Lambda$.
In this case, we see that $U$ is indeed an ideal. The proof is simple:

If $x, y$ belong to the union. Then, $x$ belongs to some $I_{\alpha}$ and $y$ to some $I_{\beta}$. By the directedness, one can choose a common $I_{\lambda}$ containing both. Now, since $I_{\lambda}$ is an ideal, see that $x + y \in I_{\lambda} \subset U$. $\Box$

In general, this is a common theme for having directedness. We may have a collection of substructures (ideals, submodules, subfields, etc.) that is ordered by inclusion. If this poset is directed (upwards), then given any finite number of elements in the union, one can actually find a single structure that contains all of them. Now, the required closure/existence property can be verified in this structure and we are done. (Note that algebraic properties/axioms typically only involve finitely many elements.)

At times, one may even start off with a collection of objects which are not necessarily substructures of something. As a concrete example, suppose that you have a collection of fields $\{F_{i}\}_{i \in I}$ that is ordered by inclusion (a priori, there is no superfield containing them all). I am going to assume that the inclusions are literal inclusions as sets, and as subfields. Then, if this collection is directed, then one can form the field $$F = \bigcup_{i \in I} F_i.$$
To define the operations, one does the following: given any $x, y \in F$, one can find a common $F_i$ containing both. Then $x + y$ and $xy$ is defined to be the corresponding element from $F_i$. The fact that this is well-defined (i.e., does not depend on the choice of $i$) also follows from the directed-ness. Indeed, if $x, y$ also belong to $F_j$ as well, then one finds an $F_k$ such that both $F_i$ and $F_j$ are subfields of $F_k$. Thus, the values of $x + y, xy$ coincide in both $F_i$ and $F_j$.
The field axioms are checked similarly.

Another useful fact is that in a directed set, a maximal element is actually a maximum. (Why?)
Now, define a module $M$ to be Noetherian, if every nonempty collection of submodules has a maximal element. If this collection is also directed, then one can conclude that the collection has a maximum element.

Corollary. Let $M$ be a Noetherian module. Let $\{M_i : i \in I\}$ be a collection of submodules of $M$, where $I$ is an arbitrary indexing set. Then, there exists a finite subset $J \subset I$ such that
$$\sum_{i \in J} M_i = \sum_{i \in I} M_i.$$
In particular, every submodule is finitely generated.

Proof sketch. Consider the collection of submodules which are finite sums of the $M_i$s. This is a directed collection of submodules (consider the sum of two submodules in this collection). Thus, by the Noetherian hypothesis, there is a maximum element, which by definition is of the form $\sum_{J} M_i$ for some finite $J \subset I$. This maximum element contains every $M_i$ (why?) and we are done. $\Box$
