The normal component of the curl is the limiting circulation per unit area of shrinking discs

Since ${\sum }_k{n}_k^2=\parallel n{\parallel }_2^2=1$ by [F1], not all three coordinates can have $n_k^2>1\mathrm{/}3$; fix $k$ with $n_k^2\le 1\mathrm{/}3$. Then $n$ and $e_k$ are linearly independent: a relation $e_k=\lambda n$ would force $\mid \lambda \mid =1$ by comparing norms, hence $n=\pm e_k$ and $n_k^2=1$, contradicting $n_k^2\le 1\mathrm{/}3$; and $n\mathrm{\ne }0$.

For all $u,v,w\in {\mathbb{R}}^3$, expanding by [F1] and [F2] gives $⟨u\times v,w⟩=(u_y{v}_z-u_z{v}_y)w_x+(u_z{v}_x-u_x{v}_z)w_y+(u_x{v}_y-u_y{v}_x)w_z,$ $⟨w\times u,v⟩=(w_y{u}_z-w_z{u}_y)v_x+(w_z{u}_x-w_x{u}_z)v_y+(w_x{u}_y-w_y{u}_x)v_z,$ and the six signed monomials of the first expression are those of the second, matched as $u_y{v}_z{w}_x$ with $w_x{u}_y{v}_z$, $-u_z{v}_y{w}_x$ with $-w_x{u}_z{v}_y$, $u_z{v}_x{w}_y$ with $w_y{u}_z{v}_x$, $-u_x{v}_z{w}_y$ with $-w_y{u}_x{v}_z$, $u_x{v}_y{w}_z$ with $w_z{u}_x{v}_y$ and $-u_y{v}_x{w}_z$ with $-w_z{u}_y{v}_x$. Hence $⟨u\times v,w⟩=⟨w\times u,v⟩$.

The set $O$ is open and $p\in O$, so by [F7] there is a real $r_0>0$ with every $q$ satisfying $\parallel q-p{\parallel }_2\le r_0$ lying in $O$; take such an $r_0$.

Fix $r$ with $0<r\le r_0$. The function $(\rho ,\theta )\mapsto \rho$ is continuous on the compact Jordan rectangle $D_r$, hence integrable by [L9] and [F3]. Its sections in $\theta$ are the continuous functions $\rho \mapsto \rho$ on $[0,r]$, so [L6] gives ${\int }_{D_r}\rho ={\int }_0^{2\pi }\left({\int }_0^r\rho d\rho \right)d\theta$; by [L7] with $G(\rho )={\rho }^2\mathrm{/}2$ the inner integral is $r^2\mathrm{/}2$, and again by [L7] with $G(\theta )=r^2\theta \mathrm{/}2$ the outer integral is $\pi r^2$. So ${\int }_{D_r}\rho =\pi r^2$.

By step 1.1 and [L2] the number $\parallel n\times e_k{\parallel }_2^2$ is positive, so $n\times e_k\mathrm{\ne }0$; put $a:=\frac{n\times e_k}{\parallel n\times e_k{\parallel }_2}\mathrm{.}$ Then $\parallel a{\parallel }_2=1$ by [F1], and $⟨a,n⟩=0$ because $n\times e_k$ is orthogonal to $n$ by [L1].

Put $b:=n\times a$. By [L1] it is orthogonal to $n$ and to $a$, so $⟨b,n⟩=⟨b,a⟩=0$; and by [L2] with step 2.1, $\parallel b{\parallel }_2^2=\parallel n{\parallel }_2^2\parallel a{\parallel }_2^2-⟨n,a{⟩}^2=1$, so $\parallel b{\parallel }_2=1$.

By step 1.2 with $u=a$, $v=b$ and $w=n$, and then step 3.1, $⟨a\times b,n⟩=⟨n\times a,b⟩=⟨b,b⟩=1$. By [L2] and steps 2.1 and 3.1, $\parallel a\times b{\parallel }_2^2=\parallel a{\parallel }_2^2\parallel b{\parallel }_2^2-⟨a,b{⟩}^2=1$. Hence $\parallel a\times b-n{\parallel }_2^2=\parallel a\times b{\parallel }_2^2-2⟨a\times b,n⟩+\parallel n{\parallel }_2^2=1-2+1=0$ by [F1], and positive definiteness in [F1] gives $a\times b=n$.

By [L3] and [L7] the map ${\phi }_r$ is differentiable in each parameter with ${\phi }_{r,\rho }=\cos \theta a+\sin \theta b$ and ${\phi }_{r,\theta }=\rho (-\sin \theta a+\cos \theta b)$, and all its iterated parameter derivatives of order at most $2$ exist and are continuous, so ${\phi }_r$ is $C^2$ on the whole plane by [F7]. Expanding by bilinearity and the alternating law in [L1], ${\phi }_{r,\rho }\times {\phi }_{r,\theta }=\rho ({\cos }^2\theta (a\times b)-{\sin }^2\theta (b\times a))=\rho ({\cos }^2\theta +{\sin }^2\theta )(a\times b),$ which is $\rho (a\times b)$ by [L3].

Combining steps 4.1 and 4.2, ${\phi }_{r,\rho }\times {\phi }_{r,\theta }=\rho n$, which is nonzero exactly when $\rho >0$.

The rectangle $D_r$ is a Type I and a Type II region with $0<r$ and constant graphs $0<2\pi$, hence an elementary Green region and a nonempty finite elementary Green region with the one-piece decomposition, compact and Jordan measurable, and it is the closure of its nonempty convex, hence connected, interior $(0,r)\times (0,2\pi )$ ([F3], [F5]). The cross product of step 5.1 is nonzero on that interior. For injectivity, let $(\rho ,\theta )$ be interior and $({\rho }^{\mathrm{\prime }},{\theta }^{\mathrm{\prime }})\in D_r$ have the same image; pairing $\rho \cos \theta a+\rho \sin \theta b={\rho }^{\mathrm{\prime }}\cos {\theta }^{\mathrm{\prime }}a+{\rho }^{\mathrm{\prime }}\sin {\theta }^{\mathrm{\prime }}b$ with $a$ and with $b$ and using steps 2.1 and 3.1 gives $\rho \cos \theta ={\rho }^{\mathrm{\prime }}\cos {\theta }^{\mathrm{\prime }}$ and $\rho \sin \theta ={\rho }^{\mathrm{\prime }}\sin {\theta }^{\mathrm{\prime }}$; squaring and adding with [L3] gives ${\rho }^2={\rho }^{\mathrm{\prime }2}$, so ${\rho }^{\mathrm{\prime }}=\rho >0$ and $\cos \theta =\cos {\theta }^{\mathrm{\prime }}$, $\sin \theta =\sin {\theta }^{\mathrm{\prime }}$. Then [L4] and [L3] give $\cos (\theta -{\theta }^{\mathrm{\prime }})=\cos \theta \cos {\theta }^{\mathrm{\prime }}+\sin \theta \sin {\theta }^{\mathrm{\prime }}={\cos }^2{\theta }^{\mathrm{\prime }}+{\sin }^2{\theta }^{\mathrm{\prime }}=1$, so ${\sin }^2(\theta -{\theta }^{\mathrm{\prime }})=0$ and $\theta -{\theta }^{\mathrm{\prime }}=m\pi$ for an integer $m$ by [L3] and [L5]; since $\theta \in (0,2\pi )$ and ${\theta }^{\mathrm{\prime }}\in [0,2\pi ]$ we have $\mid \theta -{\theta }^{\mathrm{\prime }}\mid <2\pi$, so $m\in \{-1,0,1\}$, and $\cos (\pm \pi )=-1\mathrm{\ne }1$ by [L5], leaving $\theta ={\theta }^{\mathrm{\prime }}$. Finally $\parallel {\phi }_r(\rho ,\theta )-p{\parallel }_2=\rho \le r\le r_0$ by [F1], steps 2.1 and 3.1 and [L3], so the image lies in $O$ by step 1.3. Hence $(D_r,{\phi }_r)$ is a $C^2$ patch over a finite elementary Green region with image in $O$.

By [F4] the positive boundary chain of $D_r$ in its Type I description, with $\rho$ horizontal, is the four arcs ${\sigma }_1(t)=(t,0)$ on $[0,r]$, ${\sigma }_2(t)=(r,t)$ on $[0,2\pi ]$, ${\sigma }_3(t)=(r-t,2\pi )$ on $[0,r]$ and ${\sigma }_4(t)=(0,2\pi -t)$ on $[0,2\pi ]$. Composing with ${\phi }_r$ and using $\cos 0=\cos 2\pi =1$, $\sin 0=\sin 2\pi =0$ from [L3] and [L5]: ${\phi }_r\circ {\sigma }_1(t)=p+ta$, ${\phi }_r\circ {\sigma }_2(t)=C_r(t)$, ${\phi }_r\circ {\sigma }_3(t)=p+(r-t)a$ and ${\phi }_r\circ {\sigma }_4(t)=p$. The third is the reversal of the first in the sense of [F6], so [L10] makes their integrals cancel; the fourth is constant, so its derivative extension is $0$ and its integral is $0$ by [F6]. Hence the circulation of $F$ around the induced boundary chain of $(D_r,{\phi }_r)$ is ${\int }_{C_r}F\cdot d\mathbf{r}$.

By step 6.1 the pair $(D_r,{\phi }_r)$ satisfies the hypotheses of [L11], and $F$ is $C^1$ on the open $O$ containing ${\phi }_r[D_r]$. So [L11] and step 7.1 give ${\int }_{C_r}F\cdot d\mathbf{r}={\int }_{D_r}⟨(\mathrm{curl}F)\circ {\phi }_r,{\phi }_{r,\rho }\times {\phi }_{r,\theta }⟩={\int }_{D_r}\rho g_r,g_r(\rho ,\theta ):=⟨(\mathrm{curl}F)({\phi }_r(\rho ,\theta )),n⟩,$ using step 5.1 and the bilinearity of the inner product in [F1]; the integrand is continuous on the compact Jordan $D_r$, hence integrable by [L9].

Let $\epsilon >0$ be real. The field $\mathrm{curl}F$ is continuous on $O$ and $⟨\cdot ,n⟩$ is continuous, so [F7] gives ${\delta }_0>0$ such that $\mid ⟨\mathrm{curl}F(q),n⟩-⟨\mathrm{curl}F(p),n⟩\mid \le \epsilon$ for every $q\in O$ with $\parallel q-p{\parallel }_2<{\delta }_0$; put $\delta :={\delta }_0$. Let $0<r\le r_0$ with $r<\delta$. Every point of ${\phi }_r[D_r]$ is within $r<{\delta }_0$ of $p$ by step 6.1, so $\mid g_r-⟨\mathrm{curl}F(p),n⟩\mid \le \epsilon$ on $D_r$; hence by [L8] and step 1.4 $\mid {\int }_{D_r}\rho g_r-⟨\mathrm{curl}F(p),n⟩\pi r^2\mid =\mid {\int }_{D_r}\rho (g_r-⟨\mathrm{curl}F(p),n⟩)\mid \le \epsilon {\int }_{D_r}\rho =\epsilon \pi r^2\mathrm{.}$ Dividing by $\pi r^2>0$ and substituting step 8.1 gives $\mid \frac{1}{\pi r^2}{\int }_{C_r}F\cdot d\mathbf{r}-⟨\mathrm{curl}F(p),n⟩\mid \le \epsilon$, which by [F7] is the asserted limit; with steps 4.1, 6.1 and 7.1 every clause of the Statement is established.