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  "body": "\n## Proof\n\nWe prove $1 \\Rightarrow 2$, $2 \\Rightarrow 3$, $3 \\Rightarrow 2$ and $2 \\Rightarrow 1$. All curves below are piecewise $C^1$ curves in $U$.\n\n1. Two facts about [[line integrals|0YS8RX87F4]] of $F$ come first. If $\\kappa$ is a constant curve, then every chord $\\kappa(t_i) - \\kappa(t_{i-1})$ is $0$, so every defining sum is $0$ and $\\int_\\kappa F \\cdot d\\mathbf{r} = 0$.\n2. Let $\\eta$ be a curve on $[c, d]$, let $s \\in \\R$, and let $\\eta_s(t) = \\eta(t - s)$ on $[c + s, d + s]$. Translating a partition by $s$ keeps its mesh and its chords, so the defining sums of $\\eta_s$ and $\\eta$ agree. Hence $\\int_{\\eta_s} F \\cdot d\\mathbf{r} = \\int_\\eta F \\cdot d\\mathbf{r}$.\n3. ($1 \\Rightarrow 2$.) Let $F = \\nabla \\phi$ with $\\phi$ of class $C^1$, as in the definition of a [[conservative field|05R9NF2J49]]. Let $\\gamma_1, \\gamma_2$ run from $A$ to $B$. By the [[gradient theorem|0SD237FEJD]], both integrals equal $\\phi(B) - \\phi(A)$.\n4. ($2 \\Rightarrow 3$.) Let $\\gamma \\colon [a, b] \\to U$ be closed, with $\\gamma(a) = \\gamma(b) = A$. The constant curve $\\kappa \\colon [a, b] \\to \\set{A}$ has the same initial and end point. By statement 2 and step 1, $\\int_\\gamma F \\cdot d\\mathbf{r} = \\int_\\kappa F \\cdot d\\mathbf{r} = 0$.\n5. ($3 \\Rightarrow 2$.) Let $\\gamma_1 \\colon [a, b] \\to U$ and $\\gamma_2 \\colon [c, d] \\to U$ both run from $A$ to $B$. Let $\\gamma_2^-$ be the [[reversed curve|0FSCXYG856]] on $[c, d]$. It runs from $B$ to $A$, and $\\int_{\\gamma_2^-} F \\cdot d\\mathbf{r} = -\\int_{\\gamma_2} F \\cdot d\\mathbf{r}$.\n6. Let $\\eta(t) = \\gamma_2^-(t - b + c)$ for $t \\in [b, b + d - c]$. Then $\\eta(b) = B = \\gamma_1(b)$, and $\\eta$ ends at $A$. By step 2, $\\int_\\eta F \\cdot d\\mathbf{r} = -\\int_{\\gamma_2} F \\cdot d\\mathbf{r}$.\n7. The concatenation of $\\gamma_1$ and $\\eta$ is a closed curve at $A$. By statement 3 and [[additivity|00ZEBSMNNV]], $0 = \\int_{\\gamma_1} F \\cdot d\\mathbf{r} + \\int_\\eta F \\cdot d\\mathbf{r} = \\int_{\\gamma_1} F \\cdot d\\mathbf{r} - \\int_{\\gamma_2} F \\cdot d\\mathbf{r}$.\n8. ($2 \\Rightarrow 1$.) If $U$ is empty there is nothing to prove. Otherwise fix $p \\in U$. A polygonal path is a concatenation of finitely many segments $t \\mapsto y + t (z - y)$, each shifted to the right parameter interval. It is piecewise $C^1$.\n9. Each [[open ball|0DC3MPR7C9]] $B(x, r)$ is convex. Indeed, for $y, z$ in it and $t \\in [0, 1]$, the [[triangle inequality|03HF5PMY5T]] gives $\\norm{(1 - t) y + t z - x} \\le (1 - t) \\norm{y - x} + t \\norm{z - x} \u003c r$.\n10. Let $A$ be the set of $x \\in U$ that are joined to $p$ by a polygonal path in $U$. Then $p \\in A$, using a constant segment.\n11. The set $A$ is open. Let $x \\in A$ and choose $r \u003e 0$ with $B(x, r) \\subseteq U$. For $y \\in B(x, r)$, the segment from $x$ to $y$ lies in the ball by step 9. Appending it to a path from $p$ to $x$ shows $y \\in A$.\n12. The set $U \\setminus A$ is open. Let $x \\in U \\setminus A$ and $B(x, r) \\subseteq U$. If some $y \\in B(x, r)$ were in $A$, appending the segment from $y$ to $x$ would give $x \\in A$. So $B(x, r) \\subseteq U \\setminus A$.\n13. Since $U$ is [[connected|0R37DFJKFW]], it is not the union of two disjoint nonempty open sets. With steps 10 to 12, $A = U$.\n14. For $x \\in U$, let $\\phi(x) = \\int_\\gamma F \\cdot d\\mathbf{r}$ for a curve $\\gamma$ in $U$ from $p$ to $x$. Such a curve exists by step 13, and by statement 2 the value does not depend on the choice of $\\gamma$.\n15. Fix $x \\in U$ and $i \\in \\set{1, \\dots, n}$. Choose $r \u003e 0$ with $B(x, r) \\subseteq U$, and let $0 \u003c \\abs{h} \u003c r$. Let $\\gamma \\colon [a, b] \\to U$ run from $p$ to $x$.\n16. Let $\\sigma(t) = x + (t - b) h e_i$ for $t \\in [b, b + 1]$, where $e_i$ is the $i$-th standard basis vector. It lies in $B(x, r)$, and the concatenation of $\\gamma$ and $\\sigma$ runs from $p$ to $x + h e_i$.\n17. By step 14 and additivity, $\\phi(x + h e_i) - \\phi(x) = \\int_\\sigma F \\cdot d\\mathbf{r}$.\n18. By the [[parametric formula|0W3W1VYWBP]] and the [[substitution|0D37QFPEZE]] $u = t - b$, $\\int_\\sigma F \\cdot d\\mathbf{r} = \\int_b^{b+1} F(\\sigma(t)) \\cdot h e_i \\, dt = h \\int_0^1 F_i(x + u h e_i) \\, du$.\n19. Hence $\\abs{\\frac{\\phi(x + h e_i) - \\phi(x)}{h} - F_i(x)} \\le \\int_0^1 \\abs{F_i(x + u h e_i) - F_i(x)} \\, du$. This is at most the supremum of $\\abs{F_i(w) - F_i(x)}$ over $\\norm{w - x} \\le \\abs{h}$.\n20. Since $F_i$ is continuous at $x$, this bound tends to $0$ as $h \\to 0$. So the [[partial derivative|05AXG3QRKR]] $\\partial \\phi / \\partial x_i$ exists at $x$ and equals $F_i(x)$.\n21. The partial derivatives $F_1, \\dots, F_n$ of $\\phi$ are continuous on $U$. So $\\phi$ is of [[class $C^1$|0TCBZMEYGN]], and it is differentiable by [[continuous partial derivatives imply differentiability|0KXSK8TEEB]].\n22. By the coordinate form of the [[gradient|0NS2QYVC4Y]], $\\nabla \\phi = (F_1, \\dots, F_n) = F$ on $U$. So $F$ is conservative, which is statement 1.\n\n## Idea\n\nA potential gives path independence by the gradient theorem. Conversely, path independence lets one define a potential as the integral from a fixed base point. A short straight last step in the direction $e_i$ then shows that the partial derivatives of this potential are the components of $F$.\n",
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