I want to show any point $z_0$ of a domain is a limit point of that domain working with the following elementary definitions:
domain – a nonempty open (not containing any of its boundary points) connected set
limit point – a point $z_0$ is a limit point of a set $D$ if each deleted nbhd of $z_0$ contains at least one point of $D$
I proceeded as follows:
Let $D$ be a domain with $z_0\in D$. Since $D$ is connected, there exists a sequence of joined end-to-end line segments entirely contained in $D$ connecting $z_0$ to another point $z_1$ in $D$. Since these line segments connecting $z_0$ and $z_1$ are entirely contained in $D$, there exist infinitely many points along these line segments as arbitrarily close to $z_0$ (and $z_1$ if desired) that are all contained in $D$. Thus $z_0$ is a limit point of $D$.
Does this argument make sense?
Best Answer
The definition of connectedness is confusing IMO. It's a restricted form of connectedness, connectedness by straight line segments (or polygonally connected), a special form of path-connectedness which is again more specialised than connectedness in the topological sense. These 3 notions happen to coincide for complex domains (open and connected subsets of $\Bbb C$; technically because $\Bbb C$ is locally convex etc.), so in complex analysis one tends to be more sloppy with the distinctions.
All we need of connectedness (in any of its forms) is that no set of the $\{z_0\}$ in a complex domain $\Omega$ is open in that domain (it's already closed, so would be a non-trivial clopen set otherwise), and so for any $n \in \Bbb N$ the ball $B(z_0, \frac{1}{n})=\{z \in \Bbb C: |z-z_0| < \frac1n\}$ contains points from $\Omega$ that are unequal to $z_0$. If we pick one such $z_n$ for each $n$, we have the desired sequence in $\Omega$ that converges to $z_0$. If we like we can pick all terms to be distinct as well.