[proofplan] Fix a base point $x\in U$ and collect all points of $U$ that can be reached from $x$ by a polygonal path contained in $U$. Openness of $U$ lets us append a short line segment inside a small Euclidean ball, so this reachable set is open in $U$. The same segment-appending argument shows its complement is open in $U$. Since $U$ is connected and the reachable set is nonempty, the reachable set must be all of $U$. [/proofplan] [step:Define the set of points polygonally reachable from $x$] Fix $x\in U$. Define $A\subset U$ by declaring that $a\in A$ if and only if there exist $m\in\mathbb{N}$ and points $p_0,\dots,p_m\in U$ such that $p_0=x$, $p_m=a$, and for every $k\in\{1,\dots,m\}$, \begin{align*} \{(1-t)p_{k-1}+tp_k:t\in[0,1]\}\subset U. \end{align*} Thus $A$ is the set of points in $U$ reachable from $x$ by a polygonal path contained in $U$. The set $A$ is nonempty. Indeed, choose $m=1$ and $p_0=p_1=x$. For every $t\in[0,1]$, \begin{align*} (1-t)p_0+tp_1=x\in U, \end{align*} so $x\in A$. [/step] [step:Show that the reachable set is open in $U$] Let $a\in A$. Since $U$ is open in $\mathbb{R}^n$, there exists $r>0$ such that $B(a,r)\subset U$, where \begin{align*} B(a,r):=\{z\in\mathbb{R}^n:|z-a|<r\}. \end{align*} We prove that $B(a,r)\subset A$. Let $z\in B(a,r)$. Since $a\in A$, choose $m\in\mathbb{N}$ and points $p_0,\dots,p_m\in U$ witnessing a polygonal path from $x$ to $a$, so $p_0=x$ and $p_m=a$. Define new points $q_0,\dots,q_{m+1}\in U$ by $q_k=p_k$ for $k\in\{0,\dots,m\}$ and $q_{m+1}=z$. The only new segment is the segment from $a$ to $z$. For $t\in[0,1]$, define \begin{align*} \sigma_{a,z}(t):=(1-t)a+tz. \end{align*} By the triangle inequality, \begin{align*} |\sigma_{a,z}(t)-a|=|t(z-a)|=t|z-a|\le |z-a|<r. \end{align*} Hence $\sigma_{a,z}(t)\in B(a,r)\subset U$ for every $t\in[0,1]$. Therefore the old polygonal path from $x$ to $a$, followed by the line segment from $a$ to $z$, is a polygonal path in $U$ from $x$ to $z$. Thus $z\in A$, and so $B(a,r)\subset A$. Hence $A$ is open in the [subspace topology](/page/Subspace%20Topology) on $U$. [guided] We want to show that once a point $a$ is reachable, all sufficiently nearby points are also reachable. The reason is that openness of $U$ gives room around $a$ to draw a small straight segment without leaving $U$. Let $a\in A$. Since $U$ is open in $\mathbb{R}^n$, there exists $r>0$ such that \begin{align*} B(a,r):=\{z\in\mathbb{R}^n:|z-a|<r\}\subset U. \end{align*} We claim that every $z\in B(a,r)$ also belongs to $A$. Fix such a point $z$. Because $a\in A$, there are $m\in\mathbb{N}$ and points $p_0,\dots,p_m\in U$ with $p_0=x$ and $p_m=a$ such that each segment from $p_{k-1}$ to $p_k$ lies in $U$. To reach $z$, keep this polygonal path and append one additional segment. Define $q_0,\dots,q_{m+1}\in U$ by $q_k=p_k$ for $k\in\{0,\dots,m\}$ and $q_{m+1}=z$. We only need to verify that the segment from $q_m=a$ to $q_{m+1}=z$ remains inside $U$. Define \begin{align*} \sigma_{a,z}:[0,1]\to\mathbb{R}^n,\quad \sigma_{a,z}(t)=(1-t)a+tz. \end{align*} For every $t\in[0,1]$, the triangle inequality and homogeneity of the Euclidean norm give \begin{align*} |\sigma_{a,z}(t)-a|=|t(z-a)|=t|z-a|\le |z-a|<r. \end{align*} Thus $\sigma_{a,z}(t)\in B(a,r)\subset U$ for all $t\in[0,1]$. The appended segment is therefore allowed, so $z\in A$. Since every $z\in B(a,r)$ lies in $A$, we have found an open ball around $a$ contained in $A$. This proves that $A$ is open in $U$. [/guided] [/step] [step:Show that the complement of the reachable set is open in $U$] Let $b\in U\setminus A$. Since $U$ is open in $\mathbb{R}^n$, there exists $r>0$ such that $B(b,r)\subset U$. We prove that $B(b,r)\subset U\setminus A$. Suppose, for contradiction, that there exists $z\in B(b,r)\cap A$. Since $z\in A$, there is a polygonal path in $U$ from $x$ to $z$. Define \begin{align*} \sigma_{z,b}:[0,1]\to\mathbb{R}^n,\quad \sigma_{z,b}(t)=(1-t)z+tb. \end{align*} For every $t\in[0,1]$, \begin{align*} |\sigma_{z,b}(t)-b|=|(1-t)(z-b)|=(1-t)|z-b|\le |z-b|<r. \end{align*} Hence the segment from $z$ to $b$ is contained in $B(b,r)\subset U$. Appending this segment to the polygonal path from $x$ to $z$ gives a polygonal path in $U$ from $x$ to $b$, so $b\in A$. This contradicts $b\in U\setminus A$. Therefore $B(b,r)\cap A=\varnothing$, and hence $B(b,r)\subset U\setminus A$. Thus $U\setminus A$ is open in $U$. [/step] [step:Use connectedness to conclude every point is reachable] We have shown that $A$ is open in $U$ and that $U\setminus A$ is open in $U$. Hence $A$ is both open and closed in the subspace topology on $U$. By the definition of connectedness, a connected [topological space](/page/Topological%20Space) has no nonempty proper subset that is both open and closed. Since $U$ is connected and $A$ is nonempty, the only possibility is \begin{align*} A=U. \end{align*} Therefore every $y\in U$ belongs to $A$. By the definition of $A$, for every $y\in U$ there exists a polygonal path contained in $U$ from $x$ to $y$. Since $x\in U$ was arbitrary, the conclusion holds for every pair $x,y\in U$. [/step]