2
$\begingroup$

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:

$\endgroup$
5
$\begingroup$

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 "$\lim_{z\to\infty} f(z)$".

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.