Nicest Way to Put a Topology on a Directed Set so that Definition of Limit Remains Unchanged? What is the nicest way ot put a topology on an arbitrary directed set, $Z$, so that the following definition of limit remains unchanged?
Suppose $f:X \rightarrow Y$ where $X$and $Y$ are topological spaces.
$\textbf{Definition:}$  We say ''the limit of $f$ as $x$ approaches $p$ is $L$'' and write $\lim\limits_{x \rightarrow p}f(x)=L$ provided:
For every neighborhood of $L$ (in $Y$), $N(L)$, there is some punctured neighborhood of $p$, $\overset{\cdot}{U}(p)$, such that $f[\overset{\cdot}{U}(p)] \subseteq N(L)$. (See here.)
This definition makes rigorous all the kinds of limits one encounters in a typical college algebra or calculus course (i.e. limits of functions and limits of sequences -- provided one extends the domain/range to include $\pm \infty$ and puts the order topology on them).
Is there a nice way to put a topology on an arbitrary directed set, $Z$, so that the above definition encompasses the limit of a net as well?

Note:
Special cases of the above definition that come up in college algebra and calculus courses include:


*

*limits of the form $x \rightarrow + \infty$ and $x \rightarrow - \infty$:  Here we consider functions from $\mathbb{R}$ to $Y$ and put the (total) order topology on $\overline{\mathbb{R}}:=\mathbb{R}\cup \{ - \infty, + \infty\}$.

*limits of the form $x \rightarrow \infty$:  Here we consider functions from a metric space $X$ to $Y$ and put the one-point compactification topology on $X^* := X \cup \{\infty\}$.

*limits of sequences:  Here we consider a sequence to be a function from $\mathbb{N}$ to $Y$ and put (equivalently):
(i) the (total) order topology on $\mathbb{N} \cup \{\infty\}$ or
(ii) one-point compactification topology on $\mathbb{N}^* = \mathbb{N} \cup \{\infty\}$ or
(iii) the subspace topology on $\mathbb{N} \cup \{\infty\}$ (considered as a subspace of $\overline{\mathbb{R}}$).
 A: Let $Z$ be a directed set and let $Z^*=Z\cup\{\infty\}$ where $\infty\not\in Z$.  Topologize $Z^*$ by saying that a set $U$ is open iff either $\infty\not\in U$ or there exists $i\in Z$ such that $j\in U$ for all $j\geq i$.  (Such sets are closed under finite intersections since $Z$ is directed.)  Then given a map $f:Z^*\to X$ for some other topological space $X$, the following are equivalent:


*

*$f$ is continuous

*$\lim_{z\to\infty} f(z)=f(\infty)$ in your sense.*

*The net $(f(z))_{z\in Z}$ converges to $f(\infty)$.


To prove this, note that $1\to 2$ is immediate since $\lim_{z\to\infty} f(z)=f(\infty)$ is just another way of saying that $f$ is continuous at $\infty$.  The equivalence of $2$ and $3$ is immediate from the definitions, since the deleted neighborhoods of $\infty$ in $Z^*$ are exactly sets which contain all sufficiently large elements of $Z$.  Finally, $2\to 1$ since $Z$ is an open discrete subset of $Z^*$ and so any map on $Z^*$ is automatically continuous at every point of $Z$, so to be continuous you just have to check continuity at $\infty$.
*Beware that you should be very careful with this notation in general, since there can be different values $L$ and $L'$ such that both $\lim_{z\to\infty} f(z)=L$ and $\lim_{z\to\infty} f(z)=L'$.  So unless you know the limit is unique, you should consider "$\lim_{z\to\infty} f(z)=L$" as an atomic predicate, rather than an actual statement of equality about some object "$\lim_{z\to\infty} f(z)$".
