Let $X$ be a locally compact Hausdorff topological space.
In 'Isomorphism with Small Bound' by Cambern, page $1063,$ first sentence in the proof of Proposition $1,$ he quoted the followings:
Let $x \in X$ be any point and let $\{ U_i:i\in I \}$ be the family of neighbourhoods of $x,$ where the set of indices $I$ is directed by the usual manner of set inclusion ($i_1 \leq i_2$ if $U_{i_2} \subseteq U_{i_1}$.)
Question: How to ensure that for every $x \in X,$ there exists the set $\{ U_i:i\in I \}$?
If $X$ is first countable, then the set $\{U_i:i\in I\}$ exists by first countability. But, Cambern removed the first countability in the paper. So, how to guarantee the existence of $\{U_i:i\in I\}$?
$(I, \le)$ is a directed set, as in the definition of a net. It is sort of a tautology here as neighbourhoods in reverse inclusion order, are the "canonical" example of a directed set defined from a topological space. Note that $I$ does not have a linear order, just a partial order, with the property that for any $i_1, i_2 \in I$ we have some $i_3 \in I$ with $i_1 \le i_3, i_2 \le i_3$ (which for neighbourhoods is just their intersection).
I think the $I$ is just introduced as a notational "sugar". In fact, the index set "is" the set of neighbourhoods of $x$ itself (or bijectively mapped with it), and the order is (or corresponds to) reverse inclusion. The $i$'s are easier to use in formulae, e.g. Also, in the proof of that proposition the author defines another net with that same index set.