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:


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:

  1. $f$ is continuous
  2. $\lim_{z\to\infty} f(z)=f(\infty)$ in your sense.*
  3. 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 "$\lim_{z\to\infty} f(z)$".

  • $\begingroup$ Thanks! I debated whether to specify that $Y$ be Hausdorff so that the limit, as you pointed out, would be unique (if it exists). $\endgroup$ – Selrach Dunbar Aug 14 at 22:20
  • $\begingroup$ This is a very large topology on $Z$. After thinking some more about your anser, it occurrs to me that a much smaller topology on $Z$ (that would also allow the given definition of limit to encompass the concept of limit of a net) would be the topology that takes just the upsets, i.e. $[a,\infty]:=\{z \in Z | a \leq z \} \cup \{\infty\}$, as a basis. $\endgroup$ – Selrach Dunbar Aug 15 at 21:36
  • $\begingroup$ True, if that's all you care about. In practice, though, I find that the characterization in terms of condition (1) is actually more useful than than the characterization using the limit at $\infty$, and for condition (1) you need the topology on $Z$ to be discrete. $\endgroup$ – Eric Wofsey Aug 15 at 21:49
  • $\begingroup$ Another way to think about it is that there are two ways of measuring the "size" of a topology: by how many open sets there are, and by how many nets converge. My topology on $Z^*$ is very large in open sets but dually it is very small in nets (essentially no nontrivial nets converge except that the identity net on $Z$ converges to $\infty$). $\endgroup$ – Eric Wofsey Aug 15 at 21:50
  • $\begingroup$ With the original question answered, now I am wondering if there is another way to topologize an arbitrary directed set $Z$ so that the limit of a net is as described by the definition above and reduces to the usual (order) topology when applied to the specific directed set $[0,1)$ (in the standard simple order). $\endgroup$ – Selrach Dunbar Aug 15 at 21:50

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