# Polygonally connected open sets

I cannot understand the following theorem: An open set $S$ in $\Re^n$ is connected if and only if it is polygonally connected. I would be thankful if some one could present an intuitive proof of this theorem. Thanks for reading!

Any two points in an open ball can be connected by a straight line segment.

Let $$U$$ be your connected open set and let $$p \in U$$ be any point.

Consider the set $$V = \{ u \in U \mid u \text{ can be connected to } p \text{ by a polygonal path in } U\}.$$ We want to prove $$V = U$$. We prove:

1. $$V$$ is non-empty since $$p \in V$$.

2. $$V$$ is open: if $$q \in V$$ then there is a polygonal path $$\gamma$$ in $$U$$ connecting $$q$$ to $$p$$. Let $$B \subseteq U$$ be a small open ball having $$q$$ as a center. Since every $$b \in B$$ can be connected to $$q$$ via a straight line segment, we can go from $$b$$ to $$p$$ by first going to $$q$$ along that straight line segment and then to $$p$$ along $$\gamma$$.

3. $$V$$ is closed in $$U$$: Suppose $$q_n \to q \in U$$ is a convergent sequence with $$q_n \in V$$ for all $$n$$. We need to prove that $$q\in U$$.

Let $$B \subseteq U$$ be a small ball with center $$q$$. Let $$N$$ be so large that $$q_N \in B$$: such an $$N$$ exists because $$q_n\to q$$. Fix a polygonal path $$\gamma$$ inside $$U$$ from $$q_N$$ to $$p$$, it exists because $$q_N\in V$$. Then we can go from $$q$$ to $$q_N$$ along a straight line segment inside $$B$$ and then to $$p$$ along $$\gamma$$. It follows that $$q\in V$$.

Since $$V$$ is non-empty, open and closed in $$U$$, we deduce that $$V=U$$ from $$U$$'s connectedness.

• Thanks for this awesome answer. In second last para you were showing $q\in V$ and not $q\in U$. Feb 12, 2019 at 13:27