TITLE: Why is it meaningless for a closed set to be polygonal path connected?
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: Besides to @Dominic's neat answer, I am noting a good Theorem below. However it is about $\mathbb R$, you can get the points in another dramatized answer better:

Theorem: Let $E\subset\mathbb R$ including at least two points. Then $E$ is connected if and only if it is an interval.

In fact, if $E$ is not an interval,so $$\exists~p\notin E, a,b\in E~~ a<p<b$$ Now if you set $$G=(-\infty,p),~~H=(p,+\infty)$$ then $a\in G, b\in H$ and then $E\cap G, H\cap E$ are disjoint different non empty sets that can make $E$. This shows that $E$ is not connected!