The boundary of a bounded smooth domain is a compact embedded hypersurface

$\mathrm{\partial }D$ is closed. By [F6], $\mathrm{\partial }D=\bar{D}\setminus \mathrm{int}(D)=\bar{D}\cap ({\mathbb{R}}^3\setminus \mathrm{int}(D))$, an intersection of the closed set $\bar{D}$ [F8] with the complement of the open set $\mathrm{int}(D)$ [F6], hence closed [F7].

$\mathrm{\partial }D$ is bounded. Since $D$ is bounded, fix $x_0\in {\mathbb{R}}^3$ and $r>0$ with $D\subseteq B(x_0,r)$ [F10]. Let $x\in \bar{D}$. By [F6] the ball $B(x,1)$ meets $D$, so there is $y\in D$ with $d_2(x,y)<1$; then $d_2(y,x_0)<r$ and the triangle inequality gives $d_2(x,x_0)\le d_2(x,y)+d_2(y,x_0)<1+r$ [F10]. Hence $\bar{D}\subseteq B(x_0,1+r)$ is bounded, and $\mathrm{\partial }D\subseteq \bar{D}$ is bounded [F10, F6].

Chart patches. Let $x\in \mathrm{\partial }D$ and let $G$ be an open neighbourhood of $x$ in ${\mathbb{R}}^3$. Apply [F1] to get $W$, $R$, $B$, $h$ with the chart identity; replacing $W$ by the open set $W\cap G$ preserves it [F1], so assume $W\subseteq G$. Write $R(p)=(y,s)$, $\pi (y,s)=y$, put $y_x:=\pi (R(x))$, and set $B^{\mathrm{\prime }\mathrm{\prime }}:=\pi (R(W)\cap \{(y,s):s=h(y)\})$. This is an open subset of $B$ containing $y_x$: the set $R(W)\cap \{s=h(y)\}$ is relatively open in the graph $\mathrm{\Gamma }=\{(y,h(y)):y\in B\}$, and $\pi {\mathrm{\mid }}_{\mathrm{\Gamma }}:\mathrm{\Gamma }\to B$ is a homeomorphism: its inverse $y\mapsto (y,h(y))$ and the projection are continuous by [F17], so its image is open; and $y_x\in B^{\mathrm{\prime }\mathrm{\prime }}$ because $x\in \mathrm{\partial }D\cap W$ maps to a point of the graph [F1]. Define $X:B^{\mathrm{\prime }\mathrm{\prime }}\to {\mathbb{R}}^3$, $X(y):=R^{-1}(y,h(y))$. Then $X(B^{\mathrm{\prime }\mathrm{\prime }})=\mathrm{\partial }D\cap W$: every $X(y)$ lies in $\mathrm{\partial }D\cap W$ by [F1], and conversely for $p\in \mathrm{\partial }D\cap W$ the point $R(p)\in R(W)$ has $s=h(y)$ [F1], so $\pi (R(p))\in B^{\mathrm{\prime }\mathrm{\prime }}$ and $X(\pi (R(p)))=R^{-1}(y,h(y))=R^{-1}(R(p))=p$. Moreover $X$ is injective with inverse $\pi \circ R$ restricted to $X(B^{\mathrm{\prime }\mathrm{\prime }})$, because $\pi (R(X(y)))=y$ and $X(\pi (R(p)))=p$ for $p=X(y)$.

$\mathrm{\partial }D\mathrm{\ne }\mathrm{\varnothing }$. Suppose $\mathrm{\partial }D=\mathrm{\varnothing }$. Then $\bar{D}\subseteq \mathrm{int}(D)$ by [F6], and since $\mathrm{int}(D)\subseteq D\subseteq \bar{D}$ [F6] we get $D=\bar{D}$, so $D$ is closed [F8] as well as open; that is, $D$ is clopen, and $D\mathrm{\ne }\mathrm{\varnothing }$ by hypothesis. Moreover $D\mathrm{\ne }{\mathbb{R}}^3$: if $D={\mathbb{R}}^3$ then ${\mathbb{R}}^3=D\subseteq B(x_0,r)$ for the ball of step 1.2, but the point $w:=x_0+(r+1)e_1$ satisfies $d_2(w,x_0)=\parallel (r+1)e_1{\parallel }_2=(r+1)\parallel e_1{\parallel }_2=r+1>r$, so $w\in {\mathbb{R}}^3\setminus B(x_0,r)$ [F18, F19]. Thus $D$ is a proper nonempty clopen subset of the connected space ${\mathbb{R}}^3$ [F11], contradicting [F13]. Hence $\mathrm{\partial }D\mathrm{\ne }\mathrm{\varnothing }$.

$\mathrm{\partial }D$ is compact by [F9], being closed [step 1.1] and bounded [step 1.2].

$X$ is of class $C^k$, regular, and its inverse is $C^k$. The map $g(y):=(y,h(y))$ has components the coordinate functions and $h$, all $C^k$, so $g$ is $C^k$ with all partial derivatives continuous [F17]; $R^{-1}(z)=Q^{\mathsf{T}}(z-b)$ is affine with orthogonal linear part, hence $C^{\mathrm{\infty }}$, continuous, with continuous inverse [F18]. By the chain rule $X=R^{-1}\circ g$ is $C^k$, hence continuous, with $DX(y)=Q^{\mathsf{T}}\circ Dg(y)$ [F17]. The differential $Dg(y)$ has first two rows the identity, hence is injective, and $Q^{\mathsf{T}}$ is invertible, so $DX(y)$ is injective at every $y$; thus $X$ is a regular injective parametrisation, regularity being exactly the injectivity of $DX(y)$ just proved. The inverse $\pi \circ R{\mathrm{\mid }}_{X(B^{\mathrm{\prime }\mathrm{\prime }})}$ is the restriction of the continuous map $\pi \circ R$ [F18] and is continuous at every point of $X(B^{\mathrm{\prime }\mathrm{\prime }})$; so $X$ is a homeomorphism onto $X(B^{\mathrm{\prime }\mathrm{\prime }})$.

Connected sub-patches. Let $B_1$ be an open ball with $y_x\in B_1$ and $\overline{B_1}\subseteq B^{\mathrm{\prime }\mathrm{\prime }}$, and put $O_1:=W\cap R^{-1}(B_1\times \mathbb{R})$, an open neighbourhood of $x$. Then $X(B_1)=\mathrm{\partial }D\cap O_1$: the inclusion $\subseteq$ is [step 1.3] with $y\in B_1\subseteq B^{\mathrm{\prime }\mathrm{\prime }}$; conversely $p\in \mathrm{\partial }D\cap O_1$ has $\pi (R(p))\in B^{\mathrm{\prime }\mathrm{\prime }}\cap B_1=B_1$ [step 1.3]; this set is relatively open in $\mathrm{\partial }D$ because $O_1$ is open in ${\mathbb{R}}^3$ [F7]. The ball $B_1$ is path-connected: for $u,v\in B_1$ and $t\in [0,1]$ the triangle inequality and homogeneity of the norm give $\parallel (1-t)u+tv-y_x{\parallel }_2\le (1-t)\parallel u-y_x{\parallel }_2+t\parallel v-y_x{\parallel }_2<(1-t)\rho +t\rho =\rho$ when $B_1=B(y_x,\rho )$, so the segment $t\mapsto (1-t)u+tv$ stays in $B_1$ [F18]; thus $B_1$ is path-connected, hence connected [F11]. As the continuous image of $B_1$ under $X$ [step 2.3] the set $X(B_1)$ is connected [F12]. Since $B_1$ can be chosen with arbitrarily small radius, $\mathrm{\partial }D$ is locally connected [F16].

The normal field. In the coordinates of [step 1.3] the vector $(-D{h}_j(y),1)$ is orthogonal to every tangent column $(e_i,{\mathrm{\partial }}_i{h}_j(y))$ of [step 2.3] because $-({\mathrm{\partial }}_i{h}_j)(y)+({\mathrm{\partial }}_i{h}_j)(y)=0$, and it is nonzero, so $(-D{h}_j,1)\mathrm{/}\sqrt{1+\mid D{h}_j{\mid }^2}$ is a unit normal field of the graph; by [F2] it is the prescribed outward normal $\nu$, transported back by $Q_j^{\mathsf{T}}$. Since $h_j$ is $C^k$ with $k\ge 1$, $D{h}_j$ is continuous, and the norm of $(-D{h}_j(y),1)$ is $\sqrt{1+\mid D{h}_j(y){\mid }^2}\ge 1$, so $y\mapsto \nu (X_j(y))$ is continuous as a composition of continuous maps on each patch [F17, F19]; by [F5] the field $\nu$ is defined consistently on overlaps and is continuous on all of $\mathrm{\partial }D$. This proves (c).

Components are open. Let $C$ be a connected component of $\mathrm{\partial }D$ [F14] and let $x\in C$. By [step 3.1] and [F16] applied to the open set $\mathrm{\partial }D$ itself, there is a connected open-in-$\mathrm{\partial }D$ set $V$ with $x\in V$. Then $V\cup C$ is connected [F15], contains $x$, and contains the maximal connected set $C$ [F14], so $V\cup C=C$, i.e. $V\subseteq C$. Hence every point of $C$ has a neighbourhood contained in $C$, so $C$ is open in $\mathrm{\partial }D$ [F7].

Finite embedded presentation. The family of all open balls $B_1$ with $\overline{B_1}\subseteq B^{\mathrm{\prime }\mathrm{\prime }}$ for charts of [step 1.3] over all $x\in \mathrm{\partial }D$ produces open sets $O_1=W\cap R^{-1}(B_1\times \mathbb{R})$ whose patches $X(B_1)=\mathrm{\partial }D\cap O_1$ [step 3.1] cover $\mathrm{\partial }D$; compactness [step 2.2] gives a finite subcover $O_1,\dots ,O_N$ with parametrisations $X_j:B_{1,j}\to \mathrm{\partial }D\cap O_j$, each regular, injective and $C^k$ [step 2.3] with connected open parameter balls. Let $T:=X_{\mathrm{\ell }}^{-1}\circ X_j$ be a transition, defined on the open set $X_j^{-1}(\mathrm{\partial }D\cap O_j\cap O_{\mathrm{\ell }})\subseteq B_{1,j}$: it is a composition of the homeomorphisms of [step 1.3] and [step 2.3], hence a bijection onto its image, and it is given by $T(y)=\pi (R_{\mathrm{\ell }}(R_j^{-1}(y,h_j(y))))$, a composition of affine $C^{\mathrm{\infty }}$ maps with the $C^k$ map $g_j(y)=(y,h_j(y))$ and the linear projection $\pi$; so $T$ is $C^k$ with $C^k$ inverse $X_j^{-1}\circ X_{\mathrm{\ell }}$ [F17]. With $K:=\mathrm{\partial }D$ and the finitely many open sets $O_j$ covering it, [F4] supplies smooth ${\chi }_j$ with compact support in $O_j$ and ${\sum }_j{\chi }_j=1$ on a neighbourhood of $\mathrm{\partial }D$. This is exactly the finite presentation used by the surface integral [F3]; by [F5] the chart integral then defines the finite Borel surface measure of $\mathrm{\partial }D$, independent of the charts and partition, with graph density $\sqrt{1+\mid D{h}_j{\mid }^2}$, and $\mathrm{\partial }D$ is a compact embedded $C^k$ hypersurface. This proves (b).

Finitely many components. The components of $\mathrm{\partial }D$ are nonempty, pairwise disjoint, cover $\mathrm{\partial }D$ [F14], and are open [step 4.1]; being pairwise disjoint, every subfamily with the same union has the same members. Hence they form an open cover of the compact space $\mathrm{\partial }D$ [step 2.2], which therefore has a finite subcover [F9]; any component meets that finite union and is therefore one of its finitely many members. So $\mathrm{\partial }D$ has finitely many components; each is closed [F14] and bounded (being a subset of bounded $\mathrm{\partial }D$), hence compact [F9]. This proves (a).

The complement need not be connected. Let $c_1:=0$, $c_2:=4e_1$ and $D_0:=B_1(c_1)\cup B_1(c_2)$. Each ball is a bounded $C^1$ domain whose local graph function is $C^{\mathrm{\infty }}$ [F20]; since $d_2(c_1,c_2)=4>2$, the closed balls are disjoint, and at a boundary point of $B_1(c_1)$ the chart of [F20] shrunk to the open complement of the closed ball around $c_2$ exhibits $D_0$ locally as the same subgraph, and symmetrically, so $D_0$ is a bounded $C^k$ domain for every $k\ge 1$ [F1, F20]. Its boundary is $\mathrm{\partial }{D}_0=S_1(c_1)\cup S_1(c_2)$: a point of either sphere lies in $\overline{D_0}$ but, every ball about it meeting the exterior of that ball, outside $\mathrm{int}(D_0)$ [F6, F20], while a point of $B_1(c_1)$ or $B_1(c_2)$ lies in the interior of $D_0$, and a point outside both closed balls has a ball avoiding $D_0$ entirely [F6, F7, F20]. Now put $U:=B_1(c_1)$ and $V:=B_1(c_2)\cup ({\mathbb{R}}^3\setminus ({\bar{B}}_1(c_1)\cup {\bar{B}}_1(c_2)))$. Both are open [F7, F20], nonempty, and disjoint, because $B_1(c_2)\subseteq {\bar{B}}_1(c_2)$ is disjoint from the exterior term; and $U\cup V={\mathbb{R}}^3\setminus \mathrm{\partial }{D}_0$, since the removed set $\mathrm{\partial }{D}_0$ is exactly the union of the two spheres. Thus $U$ and $V$, open in the subspace ${\mathbb{R}}^3\setminus \mathrm{\partial }{D}_0$, nonempty, disjoint and covering it, form a separation of the complement [F6, F7]. This witnesses the asserted non-claim for every $k\ge 1$.