[Math] Why is it meaningless for a closed set to be polygonal path connected

Question

My textbook (Complex Analysis by Saff & Snider) defines connectedness for open sets; the given definition of a connected open set is: a set in which every pair of points can be joined by a polygonal path that lies entirely in the set.

Using the given definition of connected set, I don't understand why it isn't similarly defined for closed sets too?

Answer 1

Because there is a difference between path connected and conected. And there is even a difference between polygonal connected and path connected. For example taking the set \[ \{ z \in \mathbb{C}: |z|=1\}\] has no polygonal paths connecting two points. But you intuitively says that this set should be connected. Connected in the pure topologic version sound this way:

A Topologic Space $(T,\tau)$ is called connected when there aren't any nonempty sets $O_1 \in \tau, O_2\in \tau$ with $O_1\cap O_2=\varnothing$ and $O_1 \cup O_2=T$. A subset of a Topological space is connected when it is connected as a topological space with the trace topology.

Every path connected space is connected but not every connected space is path connected. In $\mathbb{R}^n$ path connection, polygonal path connection and connection fall together for open sets, they are all equivalent when you know your set is already open.

This defintion is for sure much more complicated than your special case, and as you are interested in complex analysis about meromorphic functions most time you usually only use open sets anyway. so he wanted to avoid a complicated definition by using the special case of a more general one.