# Region minus a closed discrete set is still a region

Let $G \subseteq \mathbb{C}$ be an open connected set. $D$ be a discrete and closed set in $G$. (A set of isolated points). Then $G \setminus D$ is still open connected.

What I thought:

(i) $D$ most be countably infinite, as intersection of any compact set $K$ with $D$ is has at most finitely many element (limit point compactness). Also, $D$ is closed, so $G\setminus D$ is open.

(ii) Open connected is equivalent to path connected. Let $a,b \in G \setminus D$ Now cover the path (a compact subset of $G$) and there exists a $\varepsilon$ such that $B(z,\varepsilon) \subseteq G$ for all $z$ on the path.

(iii) As there only countably many points in $D$, there exists a polygonal path from $a$ to $b$ ( by ranging through all uncountably many angles. ) So $G \setminus D$ is path connected, hence connected.

• I like the idea of your proof, but I cannot see how the "countability" argument works. As you have worded it, I am convinced that every line having one of an uncountable set of angles contains at most one point of $D$. I would even be convinced that this uncountable set of angles is dense in the circle of angles. However, there's still the problem that a line with one of those angles can contain one point of $D$. – Lee Mosher Jul 4 '17 at 13:50
• @LeeMosher, you are right, my proof at (iii) doesn't actually work; I like the way skyking does it. – CL. Jul 4 '17 at 15:27
• If G has isolated points, how could it be connected? – William Elliot Jul 4 '17 at 20:18
• In your opening statement you did not state that $D$ is closed, but you assumed it later. $D$ must be closed in the space $G$ for $G$ \ $D$ to be open. For examplif if $G=\mathbb C$ and $D=\{1/n: n\in \mathbb N\}$ then $D$ is discrete but not closed in $G$, and $G$ \ $D$ is not open. – DanielWainfleet Jul 5 '17 at 20:57
• William Elliot, sorry, may you explain what you mean? @DanielWainfleet, yes, you are right, thanks. – CL. Jul 5 '17 at 21:00

Another approach would to also use the general definition for connectedness. If $G\setminus D$ were not connected it could be partitioned into open sets $U_1$, $U_2$.

Now consider any $c\in D$ that is not in $U_1\cup U_2$. Now we can construct a disc $B_c$ around $c$ such that it doesn't intersect $D$ except at $c$ and also is within $G$. Now we would have that $B\subset U_1\cup U_2$ which means that it's either subset of one of them or intersects both. Now the union of $U_1$ those balls that are within $U_1$ is open and the same applies to $U_2$. Assume that we have a ball that intersect both $U_1$ and $U_2$, then we would have that they partition $B_c\setminus\{c\}$, but we know that $B_c\setminus\{c\}$ is connected so this is an contradiction. So we see that

$$G = U_1 \cup U_2 \cup D = U_1 \cup U_2 \cup \bigcup_{c\in D} B_c = \bigcup_i\left( B_i \cup\bigcup_{B_c\subset U_1}B_c\right)$$

Which means that $G$ is not connected.

Otherwise your approach is workable too, but I don't really get the construct in (iii). Instead assume that you have a path in $G$ from $a$ to $b$. Now cover it with open sets $B(z)\subset G$ such that if $z\notin D$ $B(z)\cap D=\emptyset$ and otherwise $B(z)\cap D = \{z\}$. Now you just construct a path by simply avoiding the centerpoint of the balls and you have a path within $G\setminus D$

Both proofs rely on special topological structures of $\mathbb C$. It's obvious that the statement for example isn't true for $\mathbb R$. Both relies on a stronger variant of local connectedness, more precisely that for each point $c$ and each open set $U\ni c$ there's a punctured connected neighborhood of $c$ within $U$. The second also requires the equivalence of connectedness and path-connectedness. In the general case you need to replace the balls with neighborhoods.

• actually, I got your second proof, but I don't get the first one: won't we just have $B \subseteq U_i$ where $i=1$ or $2$. What do they necessarily partition $B$? – CL. Jul 4 '17 at 15:39
• @CWL I skipped to the problematic part immediately. We have possibly two kind of balls, those that are within only one of the sets $U_i$ and those that intersects both. My earlier proof (I've updated the proof) showed that the second sort can't be. The first kind on the other hand would together with it's superset create a open set that covers those parts of $D$. – skyking Jul 5 '17 at 6:14