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  "body": "\n## Proof\n\nWe show that the set of points reached from a fixed point by polygonal chains is open, closed in $U$ and nonempty.\n\n1. Write $\\|v\\|$ for the Euclidean norm on $\\mathbb{R}^n$, the [[2-norm|0KNJWEVSMG]], so that the [[Euclidean metric|0E1MMCAY5B]] is $\\|x - y\\|$. For $p, q \\in \\mathbb{R}^n$, the segment $[p, q]$ is the set of points $(1 - t)p + tq$ with $0 \\le t \\le 1$.\n2. If $U$ is empty, there is nothing to prove. Let $U$ be nonempty and fix $x \\in U$.\n3. A chain in $U$ from $x$ to $y$ is a finite sequence $p_0 = x, p_1, \\ldots, p_m = y$ of points, with $m \\ge 0$ and $[p_{k-1}, p_k] \\subseteq U$ for $1 \\le k \\le m$. Let $P$ be the set of $y \\in U$ with such a chain.\n4. Each [[open ball|0DC3MPR7C9]] $B(p, r)$ is [[convex|0K06X9QXDX]]. For $q \\in B(p, r)$ and $0 \\le t \\le 1$, we have $\\|(1 - t)p + tq - p\\| = t \\|q - p\\| \u003c r$.\n5. $x \\in P$, by the chain with $m = 0$.\n6. $P$ is open. Let $p \\in P$. As $U$ is open, [[the ball criterion|0A26X79RTB]] gives $r \u003e 0$ with $B(p, r) \\subseteq U$. For $q \\in B(p, r)$, step 4 gives $[p, q] \\subseteq U$. So appending $q$ to a chain for $p$ gives a chain for $q$, and $B(p, r) \\subseteq P$.\n7. $U \\setminus P$ is open. Let $p \\in U \\setminus P$ and $B(p, r) \\subseteq U$. If some $q \\in B(p, r)$ were in $P$, then $[q, p] \\subseteq U$ by step 4. Appending $p$ to a chain for $q$ would put $p$ in $P$.\n8. So $B(p, r) \\subseteq U \\setminus P$, and $U \\setminus P$ is open.\n9. The sets $P$ and $U \\setminus P$ are open in $\\mathbb{R}^n$ and lie in $U$, so they are open in the [[subspace topology|0D5YHQGDJY]] of $U$. They are disjoint, and their union is $U$.\n10. $U$ is [[connected|0R37DFJKFW]] and $P \\neq \\varnothing$. By the definition of a [[connected space|0D939CMME7]], $U \\setminus P = \\varnothing$. So $P = U$.\n11. Let $y \\in U$ and take a chain $p_0, \\ldots, p_m$ from $x$ to $y$. If $m = 0$, then $y = x$ and the constant map is a path. Let $m \\ge 1$.\n12. Define $\\gamma \\colon [0, 1] \\to \\mathbb{R}^n$ as follows. For $1 \\le k \\le m$ and $t \\in [\\tfrac{k-1}{m}, \\tfrac{k}{m}]$, let $\\gamma(t) = p_{k-1} + (mt - k + 1)(p_k - p_{k-1})$.\n13. At $t = k/m$, the formulas for the pieces $k$ and $k + 1$ both give $p_k$, so $\\gamma$ is well defined. Also $\\gamma(0) = x$, $\\gamma(1) = y$, and $\\gamma(t) \\in [p_{k-1}, p_k] \\subseteq U$.\n14. Let $L = m \\max_k \\|p_k - p_{k-1}\\|$. If $s, t$ lie in the same piece $k$, then $\\|\\gamma(t) - \\gamma(s)\\| = m |t - s| \\, \\|p_k - p_{k-1}\\| \\le L |t - s|$.\n15. If $s \u003c t$ lie in different pieces, let $s \u003c b_1 \u003c \\cdots \u003c b_j \u003c t$ be the points $k/m$ between them. The triangle inequality and step 14 on each subinterval give $\\|\\gamma(t) - \\gamma(s)\\| \\le L (t - s)$.\n16. So $\\gamma$ is [[Lipschitz|0TMZ21WKCN]] with constant $L$. Given $\\varepsilon \u003e 0$, the choice $\\delta = \\varepsilon / (L + 1)$ works in the epsilon-delta condition. By [[continuity via open sets|0YF82GBTND]], $\\gamma$ is continuous into $\\mathbb{R}^n$.\n17. The open sets of $U$ are the sets $U \\cap O$ with $O$ open in $\\mathbb{R}^n$. Since $\\gamma$ takes values in $U$, $\\gamma^{-1}(U \\cap O) = \\gamma^{-1}(O)$ is open. So $\\gamma$ is a path in $U$ from $x$ to $y$.\n18. The point $x \\in U$ was arbitrary. So any two points of the nonempty set $U$ are joined by a path in $U$, and $U$ is [[path-connected|0K0GE6BCYW]].\n\n## Idea\n\nBalls are convex, so inside a ball any point can be reached from any other by one straight segment. Hence the points reachable from $x$ by broken lines form an open set, and so do the other points. Connectedness of $U$ forces every point to be reachable.\n",
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