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!


1 Answer 1


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.

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

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