Originally Posted by

**tarheelborn** To clarify that I am on the right track in the formal proof, may I run the final version by you?

Proof. (by contradiction). Suppose that $\displaystyle A$ is disconnected. Then, by definition, there exist two disjoint sets, $\displaystyle U$ and $\displaystyle V$, such that $\displaystyle A\subseteq U \cup V$, $\displaystyle A \cap U \neq \emptyset $, and $\displaystyle A \cap V \neq \emptyset $. Now suppose that $\displaystyle z_1 \in A \cap V$ and $\displaystyle z_2 \in A \cap V$. Thus there is a line segment between $\displaystyle z_1$ and $\displaystyle z_2$ that lies partially in $\displaystyle V$ and partially in $\displaystyle U$ such that $\displaystyle [z_1,z_2]$ is connected by a finite union of segments of the form $\displaystyle [z_0,z_1]\cup[z_1,z_2]\cup \cdots \cup [z_{n-1},z_n]$, where the terminal point of one corresponds to the initial point of the next (when applicable). Thus, $\displaystyle [z_1,z_2]$ forms a polygonal path from $\displaystyle V$ to $\displaystyle U$. Hence $\displaystyle A$ is polygonally connected and, therefore, connected.$\displaystyle \square$