[proofplan] We prove first that connectedness forces polygonal reachability. Fix a basepoint $a \in U$ and let $P$ be the set of points reachable from $a$ by finitely many line segments lying in $U$; openness of $U$ makes $P$ open, and the same local segment argument makes its complement open. Connectedness then forces $P=U$, and arbitrary endpoints are joined by reversing one polygonal chain through the basepoint and appending the other. Conversely, a polygonal chain is a continuous path, so polygonal reachability implies path-connectedness; we include the standard separation argument showing that this implies connectedness. [/proofplan] [step:Define the polygonally reachable set from a basepoint] Choose a point $a \in U$, which exists because $U$ is nonempty. Define $P \subset U$ to be the set of all points $x \in U$ for which there exist an integer $m \geq 1$ and points $p_0,p_1,\dots,p_m \in U$ such that $p_0=a$, $p_m=x$, and, for every $j \in \{1,\dots,m\}$, \begin{align*} [p_{j-1},p_j] := \{(1-t)p_{j-1}+tp_j : t \in [0,1]\} \subset U. \end{align*} The point $a$ belongs to $P$: take $m=1$ and $p_0=p_1=a$, so the only segment is the singleton $\{a\}$, which is contained in $U$. [/step] [step:Show that the reachable set is open in $U$] Let $x \in P$. Since $U$ is open in $\mathbb{R}^n$, there exists $r>0$ such that $B(x,r) \subset U$. If $y \in B(x,r)$, then the segment $[x,y]$ is contained in $B(x,r)$, because for every $t \in [0,1]$, \begin{align*} |(1-t)x+ty-x| = t|y-x| < r. \end{align*} By the definition of $P$, there is a polygonal chain in $U$ from $a$ to $x$. Appending the segment $[x,y] \subset U$ gives a polygonal chain in $U$ from $a$ to $y$. Thus $B(x,r) \cap U \subset P$. Since $x \in P$ was arbitrary, $P$ is open in the [subspace topology](/page/Subspace%20Topology) on $U$. [guided] We want to prove that reachability by polygonal chains is a local property inside an open subset of Euclidean space. Fix $x \in P$. By definition, $x$ is already reachable from $a$ by a finite polygonal chain lying in $U$. Because $U$ is open in $\mathbb{R}^n$, there is a radius $r>0$ such that the Euclidean open ball $B(x,r)$ is contained in $U$. Now take any point $y \in B(x,r)$. The direct line segment from $x$ to $y$ stays inside the same ball: for each $t \in [0,1]$, \begin{align*} |(1-t)x+ty-x| = |t(y-x)| = t|y-x| < r. \end{align*} Therefore $[x,y] \subset B(x,r) \subset U$. Since a polygonal chain from $a$ to $x$ already lies in $U$, adding this final segment from $x$ to $y$ produces a polygonal chain from $a$ to $y$ lying entirely in $U$. Hence every $y \in B(x,r)$ belongs to $P$, so $B(x,r)\cap U \subset P$. This proves that $P$ is open in $U$. [/guided] [/step] [step:Show that the complement of the reachable set is open in $U$] Let $x \in U \setminus P$. Since $U$ is open in $\mathbb{R}^n$, choose $r>0$ such that $B(x,r) \subset U$. We claim that $B(x,r) \cap P = \varnothing$. If there were a point $y \in B(x,r) \cap P$, then the segment $[y,x]$ would be contained in $B(x,r) \subset U$ by the same convexity calculation as above. Appending $[y,x]$ to a polygonal chain from $a$ to $y$ would give a polygonal chain from $a$ to $x$ in $U$, contradicting $x \notin P$. Hence $B(x,r)\cap U \subset U\setminus P$, and $U\setminus P$ is open in $U$. [/step] [step:Use connectedness to make every point reachable from the basepoint] Assume $U$ is connected. The set $P$ is nonempty, open in $U$, and has open complement in $U$. Thus $P$ is a nonempty clopen subset of the [connected space](/page/Connected%20Space) $U$. By the definition of connectedness, the only clopen subsets of $U$ are $\varnothing$ and $U$, so $P=U$. Therefore every point of $U$ is reachable from $a$ by a polygonal chain contained in $U$. [/step] [step:Join arbitrary endpoints by passing through the basepoint] Let $x,y \in U$. Since $P=U$, there is a polygonal chain in $U$ from $a$ to $x$, and there is a polygonal chain in $U$ from $a$ to $y$. Reversing the first chain gives a polygonal chain from $x$ to $a$, because each segment $[p_{j-1},p_j]$ is equal to $[p_j,p_{j-1}]$. Appending the second chain gives a polygonal chain from $x$ to $y$ contained in $U$. This proves the forward implication. [/step] [step:Convert polygonal reachability into connectedness] Assume now that every pair of points in $U$ can be joined by a polygonal chain contained in $U$. Suppose, for contradiction, that $U$ is disconnected. Then there exist nonempty disjoint subsets $A,B \subset U$ that are open in the subspace topology on $U$ and satisfy $U=A\cup B$. Choose $x \in A$ and $y \in B$. By hypothesis, there are an integer $m \geq 1$ and points $p_0,p_1,\dots,p_m \in U$ with $p_0=x$, $p_m=y$, and $[p_{j-1},p_j]\subset U$ for every $j$. Define \begin{align*} k := \min\{j \in \{1,\dots,m\} : p_j \in B\}. \end{align*} This set is nonempty because $p_m=y\in B$, so $k$ is well-defined. Since $p_0=x\in A$, we have $k\geq 1$, and by minimality $p_{k-1}\in A$. Define the segment path $\sigma:[0,1]\to U$ by \begin{align*} \sigma(t) := (1-t)p_{k-1}+tp_k. \end{align*} Its image lies in $U$ by the polygonal-chain hypothesis. Let \begin{align*} S := \{t \in [0,1] : \sigma(t)\in A\}. \end{align*} Then $0\in S$ and $1\notin S$. Because $A$ and $B$ are open in $U$ and $\sigma$ is continuous, the sets $S=\sigma^{-1}(A)$ and $[0,1]\setminus S=\sigma^{-1}(B)$ are both open in the subspace topology on $[0,1]$. This separates the interval $[0,1]$ into two nonempty disjoint open subsets, contradicting the connectedness of the interval $[0,1]$. Therefore no such separation of $U$ exists, and $U$ is connected. [/step]