Is there a domaine discontinuity in the complex plane?

Question

Consider a domain deined by a rectangle in the complex plane. When the width (or length) goes to zero, the rectangle is reduced to a segment which is not a domain. Can this discontinuiy be removed and how? I found no refernces that discuss this issue. Any thoughts about it would be enlightening and useful

Answer 1

In order to discuss this, we need solid definitions, not vague concepts. So how can we talk about limits of sets?

Well, there is a concept of $\limsup$ and $\liminf$ for sequences of sets: Limits of sequences of sets. Read through Brian M. Scott's excellent answer. Note in particular that $\liminf A_n$ is the set of all $x$ such that $x\in A_n$ for all but finitely many $n$, while $\limsup A_n$ is the set of all $x$ such that $x\in A_n$ for infinitely many $n$. We can say that $\{A_n\}$ converges if $\liminf A_n = \limsup A_n$ and define $\lim A_n$ to be the common set. So $\{A_n\}$ converges if and only if every $x$ that is in infinitely many $A_n$ is in fact in all but a finite number of them. Note that if $\{A_n\}_n$ converges, any subsequence $\{A_{n_k}\}_k$ also converges and to the same limit.

Now for some set $X$, we define a subset $\scr C$ of ${\scr P}(X)$ (i.e. the set of all subsets of $X$) to be closed if for every convergent sequence $\{A_n\}\subset \scr C$, the limit is also in $\scr C$. Then we find that $\emptyset$ and ${\scr P}(X)$ are obviously both closed. Let $U$ be any collection of closed sets. If $\{A_n\}\subset \bigcap U$, then $\{A_n\}\subset \scr C$ for all ${\scr C} \in U$. Since all the $\scr C$ are closed, if $\{A_n\}$ is convergent, its limit is also in every $\scr C$, and so in $\bigcap U$ as well. Thus $\bigcap U$ must be closed. Conversely, if $\scr C_1, C_2$ are two closed sets and $\{A_n\} \subseteq \scr C_1 \cup C_2$ converges, then at least one of $\scr C_1, C_2$ must contain an infinite subsequence of $\{A_n\}$, which also converges and to the same limit. Therefore $\lim A_n \in \scr C_1 \cup C_2$. So $\scr C_1 \cup C_2$ is also closed.

Define a subset of ${\scr P}(X)$ to be open if its complement is closed. Then then by De Morgan's laws, the open sets of ${\scr P}(X)$ form a topology. So we can now talk about continuity and more general limits of sets, and all the other properties that topology brings us.

So let's apply it to your example. In this case $X = \Bbb C$ and we are concerned with the set of domains $\scr D \subset P(\Bbb C)$. What can we say about $\scr D$, or more generally, about the collection of all open sets $\scr T$?

• Are they closed? No. Consider $A_n = \{ z : |z| < 1/n\}$. Each $A_n \in \scr D \subset T$, but $\lim A_n = \{0\} \notin \scr T$.
• Are they open? No. Consider $B_n = \{ z : |z| \le 1 - 1/n\}$. Each $B_n \notin \scr T$, but $\lim B_n = \{ z : |z| < 1\} \in \scr D \subset T$.
• Are their closures useful? That depends on what you want to use them for. For measure theory, yes. For complex analysis, no. As noted above, $\{0\}$ is in $\scr \overline D$, and it should be clear every other single point set is as well (in fact, every finite set is, but that takes a little more creativity). But the main point of introducing domains for complex analysis is that a function needs to be differentiable in a neighborhood of each point, not just at the point itself, for Cauchy's theorem to hold and thus to have access to this incredibly rich theory. So extending $\scr D$ to its closure defeats the purpose.
• What about the interiors? Variations of the second example demonstrate that $\scr T$ has no interior. Every open set is the limit of a sequence of non-open sets. As a subset of $\scr T, D$ cannot have an interior either.

There is of course much more that can be investigated, but there you have the gist of it. You don't hear much about limits and topologies of sets because they are seldom useful. Like all of mathematics, they are interesting in their own right. But if you are studying something else, their usefulness is so limited that only special cases are ever needed (such as $\displaystyle \bigcap_{n=0}^\infty, \bigcup_{n=0}^\infty$) which can be used without bothering with a full theory.