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The vacuum region around compact smooth conductors is a bounded smooth domain

Statement

Assume the Axiom of Countable Choice ( ${A C}_{ω}$ ) as carried by the boundary interfaces (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $k \geq 1$ , let $Ω \subset R^{3}$ be a bounded $C^{k}$ domain, and let $D_{1}, \dots, D_{m}$ ( $m \geq 0$ ) be bounded $C^{k}$ domains with pairwise disjoint closures whose closures satisfy $\overline{D_{1}} \cup \dots \cup \overline{D_{m}} \subset Ω$ strictly inside $Ω$ . Put

$Ω_{v} : = Ω ∖ (\overline{D_{1}} \cup \dots \cup \overline{D_{m}}) .$

Then:

(a) [Domain] $Ω_{v}$ is a bounded $C^{k}$ domain in the sense of Bounded C^k domains and boundary charts (in particular nonempty, bounded and open);

(b) [Boundary] $\partial Ω_{v} = \partial Ω \cup \partial D_{1} \cup \dots \cup \partial D_{m}$ , a disjoint union of nonempty compact sets; the compact embedded $C^{k}$ hypersurfaces $\partial Ω, \partial D_{1}, \dots, \partial D_{m}$ are pairwise disjoint (The boundary of a bounded smooth domain is a compact embedded hypersurface);

(c) [Normals and sign] the outward unit normal of $Ω_{v}$ at a point of $\partial Ω$ equals the outward unit normal of $Ω$ , and at a point of $\partial D_{j}$ it equals the negative of the outward unit normal of $D_{j}$ defined by Bounded C^k domains and boundary charts and Bounded C1 domains and their outward normals; so the outward normal of $Ω_{v}$ on $\partial D_{j}$ points from the vacuum into the conductor;

(d) [Componentwise statement] $Ω_{v}$ has finitely many connected components. Each component $U$ is an open bounded $C^{k}$ domain; its boundary is the compact restricted subset $\partial U \subseteq \partial Ω_{v}$ , a union of the boundary components incident to $U$ . The normal formulas in (c) restrict to those boundary pieces. It need not have every boundary piece of $Ω_{v}$ , and connectedness of $Ω_{v}$ is not asserted.

No regularity beyond $C^{k}$ , no convexity and no topological simplicity is claimed.

Facts & Assumptions

Given: ${A C}_{ω}$ as in The Axiom of Countable Choice ( ${A C}_{ω}$ ); a bounded $C^{k}$ domain $Ω \subset R^{3}$ , $k \geq 1$ ; bounded $C^{k}$ domains $D_{1}, \dots, D_{m}$ ( $m \geq 0$ ) with pairwise disjoint closures all strictly contained in $Ω$ . All metric notions use the Euclidean metric $d_{2}$ and norm on $R^{3}$ ; write $B (x, r)$ for open balls, $\overline{A}$ for closure and $\partial A$ for the metric boundary.

[F1]

A bounded set lies in some ball; closures of bounded sets are bounded; a subset of $R^{n}$ is compact exactly when it is closed and bounded; a finite union of compact or of closed sets is closed, and a finite union of bounded sets is bounded (Open ball, closed ball and sphere in a metric space, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open cover, subcover, compact metric space, and compact subset of a metric space, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). Ambient open covers of compact subsets have finite subcovers (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

[F2]

Closure is the smallest closed superset: $A \subseteq F$ with $F$ closed implies $\overline{A} \subseteq F$ , and $A$ is closed iff $A = \overline{A}$ . The boundary is $\partial A = \overline{A} ∖ int (A)$ , and for open $A$ the interior is $A$ itself; boundaries are closed; every ball about a point of $\overline{A}$ meets $A$ (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[F3]

Distances: $d (x, A)$ is the infimum of ${d (x, a) : a \in A}$ for nonempty $A$ ; a continuous real function on a nonempty compact metric space attains its minimum; $∣ d (x, a) - d (x, a^{'}) ∣ \leq d (a, a^{'})$ , so $q \mapsto d (x, q)$ is continuous, and if $K$ is compact with $x \notin K$ then $d (x, K) > 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ ).

[F4]

The chart property and the outward normal: at every boundary point $x$ of a bounded $C^{k}$ domain $D$ there are an open $W ∋ x$ , a rigid motion $R = Q \cdot + b$ with orthogonal $Q$ , an open ball $B \subseteq R^{2}$ , $h \in C^{k} (B)$ with $R (D \cap W) = R (W) \cap {(y, s) : s < h (y)}$ , and the prescribed outward normal in ambient coordinates is $Q^{T} (- D h (y), 1) / \sqrt{1 + ∣ D h (y) ∣^{2}}$ , agreeing on overlaps; shrinking $W$ preserves the identity (Bounded C^k domains and boundary charts, Bounded C1 domains and their outward normals, The boundary of a bounded smooth domain is a compact embedded hypersurface).

[F5]

For a bounded $C^{k}$ domain the boundary is nonempty, compact, and a compact embedded $C^{k}$ hypersurface with finitely many connected components; hence also $\partial D \cap W = R^{- 1} ({(y, h (y)) : y \in B} \cap R (W))$ and points of $R (W)$ with $s > h (y)$ lie outside $\overline{D}$ (The boundary of a bounded smooth domain is a compact embedded hypersurface).

[F6]

Let $B_{1} \subseteq R^{2}$ be a ball, $h \in C (B_{1}; R)$ , $c \in R$ and $δ > 0$ , and suppose $∣ h (y) - c ∣ \leq δ / 4$ for all $y \in B_{1}$ . With $H : = c + δ / 2$ one then has $h (y) < H < h (y) + δ$ for every $y \in B_{1}$ . Consequently each of the regions $S_{+} : = {(y, s) : y \in B_{1}, h (y) < s < h (y) + δ}$ and $S_{-} : = {(y, s) : y \in B_{1}, h (y) - δ < s < h (y)}$ is polygonally connected: for $u = (y_{1}, s_{1})$ and $v = (y_{2}, s_{2})$ in $S_{+}$ the segment $[y_{1}, y_{2}]$ lies in the convex ball $B_{1}$ (Open ball, closed ball and sphere in a metric space, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it), every point $(y, H)$ of the middle segment satisfies $h (y) < H < h (y) + δ$ by the displayed inequality, and the vertical segments from $s_{1}$ up to $H$ at $y_{1}$ and from $H$ down to $s_{2}$ at $y_{2}$ have all their intermediate heights between the endpoint heights, hence also lie in $S_{+}$ ; the same with $H_{-} : = c - δ / 2$ and the reversed inequalities applies to $S_{-}$ . These three-segment paths are polygonal paths (Polygonal paths and polygonally connected subsets of $R^{n}$ , A finite concatenation of straight segments in $R^{n}$ is a continuous path), so a polygonally connected set is path-connected and hence connected (Every path-connected space is connected, and every path component lies inside a component).

[F7]

Rigid motions and reflections: $R = Q \cdot + b$ and $ρ (y, s) = (y, - s)$ are continuous bijections with continuous inverses and preserve distances; images of connected sets under them are connected (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, A continuous image of a connected space is connected, and connectedness is a topological property).

[F8]

$C^{k}$ maps and their algebra: $h \in C^{k}$ remains $C^{k}$ under negation; the coordinate functions $(y, s) \mapsto y$ and $(y, s) \mapsto s$ and affine maps are $C^{\infty}$ ; the set ${s > h (y)}$ is open as the preimage of the open set $(0, \infty)$ under the continuous map $s - h (y)$ ( $C^{k}$ Euclidean maps and diffeomorphisms, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Metric continuity characterisations, with countable choice for the sequential converse, Total differentiability gives a local $O (∥ h ∥_{2})$ increment bound and therefore continuity).

[F9]

Every connected component of an open subset of $R^{n}$ is open in $R^{n}$ and polygonally connected; components are nonempty, pairwise disjoint, cover the space, are the maximal connected subsets, and are closed; a union of connected sets with a common point is connected (Every connected component of an open subset of $R^{n}$ is open and polygonally connected, Connected components, quasicomponents, and totally disconnected spaces, The components of a space are its maximal connected subsets, they partition it, and each of them is closed, A union of connected subspaces with a point in common is connected, and so is a union of a family in which every member meets a fixed connected member).

[F10]

A nonempty space is connected exactly when its only clopen subsets are the empty set and the whole space; a connected space is not the disjoint union of two nonempty open subsets; $R^{3}$ is connected (For a topological space the following agree: no separation exists, the only clopen subsets are $\emptyset$ and $X$ , and every continuous map to the two-point discrete space is constant, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, $R^{n}$ is polygonally connected, connected, locally path-connected and locally connected).

Proof

technique · direct

1.1F1F2F3

Put $K : = \overline{D_{1}} \cup \dots \cup \overline{D_{m}}$ (so $K = \emptyset$ when $m = 0$ ). Each $\overline{D_{j}}$ is closed and bounded, hence compact [F1, F2]; $K$ is a finite union of compact sets, hence closed and bounded and therefore compact [F1]. By hypothesis $K \subseteq Ω$ , so $\partial Ω \cap K = \emptyset$ , and the sets $\overline{D_{1}}, \dots, \overline{D_{m}}$ are pairwise disjoint. Put $Ω_{v} : = Ω ∖ K$ ; it is open, since $Ω$ is open and $K$ is closed, and bounded, since $Ω$ is; and, when $m > 0$ , for every $x \in \partial Ω$ we have $d (x, K) > 0$ by [F3]. For $m = 0$ set $ρ_{x} : = 1$ ; for $m > 0$ set $ρ_{x} : = d (x, K) > 0$ . In either case $B (x, ρ_{x} / 2)$ misses $K$ .

2.1F2F3step 1.1

$Ω_{v} \neq \emptyset$ . Since $Ω$ is a bounded $C^{k}$ domain, $\partial Ω \neq \emptyset$ [F5]; fix $x \in \partial Ω$ and put $ρ : = ρ_{x} > 0$ as in step 1.1. Every ball about $x$ meets $Ω$ while $x \notin Ω$ [F2], so there is $y \in B (x, ρ / 2) \cap Ω$ . Then $d (y, q) \geq d (x, q) - d (x, y) > ρ - ρ / 2 > 0$ for every $q \in K$ , so $y \notin K$ ; hence $y \in Ω_{v}$ .

2.2F2F3F4F5step 1.1

$\partial Ω_{v} = \partial Ω \cup \partial D_{1} \cup \dots \cup \partial D_{m}$ as a disjoint union. For $x \in \partial D_{j}$ : $x \in \overline{D_{j}} \subseteq Ω$ , so $x \notin Ω_{v}$ ; every ball about $x$ meets $D_{j} \subseteq K$ [F2], and for a sufficiently small ball it lies in $Ω$ and misses $⋃_{i \neq j} \overline{D_{i}}$ (positive distances between the pairwise disjoint compact sets [F3]) and meets $R^{3} ∖ \overline{D_{j}}$ [F4, F5], hence meets $Ω_{v}$ ; so $x \in \partial Ω_{v}$ . For $x \in \partial Ω$ : every ball about $x$ meets $Ω$ while $x \notin K$ with $B (x, ρ_{x} / 2) \cap K = \emptyset$ [step 1.1], so small balls meet $Ω_{v}$ ; and $x \notin Ω_{v}$ ; so $x \in \partial Ω_{v}$ . Conversely let $x \in \partial Ω_{v}$ ; then $x \in \overline{Ω_{v}} \subseteq \overline{Ω}$ and $x \notin Ω_{v}$ [F2]. If $x \notin Ω$ then $x \in \partial Ω$ . If $x \in Ω$ , then $x \in K$ , say $x \in \overline{D_{j}}$ ; if $x \in D_{j}$ then $D_{j}$ is open and a small ball about $x$ lies in $D_{j} \subseteq K$ , so it misses $Ω_{v}$ , contradicting $x \in \overline{Ω_{v}}$ ; hence $x \in \overline{D_{j}} ∖ D_{j} = \partial D_{j}$ . The displayed pieces are pairwise disjoint because $K \subseteq Ω$ and the $\overline{D_{j}}$ are pairwise disjoint, and each is nonempty and compact [F5].

3.1F4F5F7F8step 1.1step 2.2

At every point of $\partial Ω_{v}$ the vacuum region $Ω_{v}$ is locally exactly the one-sided subgraph of a $C^{k}$ graph. Let $x \in \partial Ω$ . Use the chart of [F4] for $Ω$ at $x$ and let $ρ : = ρ_{x} > 0$ [step 1.1]; replacing $W$ by the open set $W \cap B (x, ρ / 2)$ preserves the chart identity [F4] and makes $W \cap K = \emptyset$ , so $Ω_{v} \cap W = Ω \cap W$ and $R (Ω_{v} \cap W) = R (W) \cap {s < h (y)}$ . Let $x \in \partial D_{j}$ . Use the chart of [F4] for $D_{j}$ at $x$ , and replace $W$ by an open ball about $x$ contained in $Ω$ and avoiding $⋃_{i \neq j} \overline{D_{i}}$ ; this keeps the chart identity [F4] and gives $Ω_{v} \cap W = W ∖ \overline{D_{j}}$ , whose image is $R (W) \cap {s > h (y)}$ : points with $s < h (y)$ lie in $D_{j}$ , points with $s = h (y)$ lie on $\partial D_{j}$ [F4, F5]. Put $ρ_{0} (y, s) : = (y, - s)$ and $R_{0} : = ρ_{0} \circ R$ , $h_{0} : = - h$ ; then $R_{0}$ is a rigid motion, $h_{0} \in C^{k}$ , and $R_{0} (Ω_{v} \cap W) = R_{0} (W) \cap {s < h_{0} (y)}$ [F7, F8].

3.2F5step 2.2

The boundary pieces of [step 2.2] are the compact embedded $C^{k}$ hypersurfaces of the boundary lemma, and they are pairwise disjoint with nonempty compact pieces; this is (b).

4.1F4F8step 1.1step 2.1step 2.2step 3.1

$Ω_{v}$ is a bounded $C^{k}$ domain: it is nonempty [step 2.1], bounded and open [step 1.1], and by [step 2.2] every boundary point is treated by one of the two local charts of [step 3.1], each exhibiting $Ω_{v}$ locally as the one-sided subgraph of a $C^{k}$ function over an open ball after a rigid motion [F4, F8]. This is (a).

4.2F4F7F8step 3.1

The normals. In the $\partial Ω$ case of [step 3.1] the chart data for $Ω_{v}$ and for $Ω$ are identical, so the prescribed outward normal of $Ω_{v}$ , computed by the formula of [F4], equals that of $Ω$ at every point of $\partial Ω$ . In the $\partial D_{j}$ case, the chart of $Ω_{v}$ is $R_{0} = ρ_{0} \circ R$ with graph $h_{0} = - h$ ; the orthogonal part of $R_{0}$ is $ρ_{0} Q$ , whose inverse is $Q^{T} ρ_{0}$ , so [F4] gives the vacuum normal $(Q^{T} ρ_{0}) (- D h_{0} (y), 1) / \sqrt{1 + ∣ D h_{0} (y) ∣^{2}} = (Q^{T} ρ_{0}) (D h (y), 1) / \sqrt{1 + ∣ D h (y) ∣^{2}}$ , and since $ρ_{0} (D h, 1) = (D h, - 1) = - (- D h, 1)$ this equals $- Q^{T} (- D h (y), 1) / \sqrt{1 + ∣ D h (y) ∣^{2}}$ , the negative of the prescribed outward normal of $D_{j}$ [F4, F8]. This is (c).

4.3F3F6F7F8step 3.1

Connected collars and patches. Fix one of the charts of [step 3.1], shrinking its neighborhood to be bounded first. Write $R$ for its rigid motion, $W$ for its domain, $h$ for its graph function and $B^{''}$ for the set of parameters $y$ with $(y, h (y)) \in R (W)$ , an open subset of $B$ containing the parameter of the distinguished boundary point. Choose a ball $B_{1}$ with $\overline{B_{1}} \subseteq B^{''}$ ; the compact graph set $G : = {(y, h (y)) : y \in \overline{B_{1}}} \subseteq R (W)$ (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value) is disjoint from the closed set $R^{3} ∖ R (W)$ , the function $z \mapsto d (z, R^{3} ∖ R (W))$ is positive and continuous on $G$ (its Lipschitz constant is at most one by the triangle inequality), hence has a positive minimum $δ$ by [F3]. Its complement is nonempty because $W$ is bounded. By continuity of $h$ at the parameter $y_{x}$ of the distinguished point choose a ball $B_{1}^{'} \subseteq B_{1}$ around $y_{x}$ with $∣ h (y) - h (y_{x}) ∣ \leq δ / 4$ for $y \in B_{1}^{'}$ [F8]. In the inner case the region $T : = {(y, s) : y \in B_{1}^{'}, h (y) < s < h (y) + δ}$ is contained in $R (W)$ (every point of $T$ is within $δ$ of its graph point, which lies in $G$ ) and in ${s > h (y)}$ , so $R^{- 1} (T) \subseteq Ω_{v} \cap W$ ; in the outer case the same holds for $T : = {(y, s) : y \in B_{1}^{'}, h (y) - δ < s < h (y)}$ and ${s < h (y)}$ , by the descriptions of [step 3.1]. The set $T$ is polygonally connected by [F6], hence connected, so $N : = R^{- 1} (T)$ is connected as its continuous image [F7]; moreover $N$ accumulates at every point $X (y)$ of the patch $P : = X (B_{1}^{'}) \subseteq \partial Ω_{v}$ , since the points $(y, h (y) \pm ε)$ with $0 < ε < δ$ lie in $T$ and converge to $(y, h (y))$ as $ε ↓ 0$ .

5.1F9step 4.1

Each connected component of the open set $Ω_{v}$ is open in $R^{3}$ and polygonally connected [F9]; in particular every component is open.

5.2F2F9step 3.1step 4.3

Each patch is incident to exactly one component. The connected nonempty collar $N$ of step 4.3 lies in one component $U$ of $Ω_{v}$ [F9]. For $z = X (y_{0}) \in P$ , continuity of $h$ and openness of $B_{1}^{'}$ give a ball $B (z, r)$ so small that its chart coordinates satisfy $y \in B_{1}^{'}$ and $∣ s - h (y) ∣ < δ$ , and that $B (z, r) \subset W$ . By step 3.1 the vacuum portion of this ball lies in the one-sided collar, hence $B (z, r) \cap Ω_{v} \subset N \subset U$ . If $z \in \partial U_{1}$ for another component, the ball meets $U_{1}$ by [F2], so $U_{1}$ meets $N$ and equals $U$ by [F9]. Conversely, every $z \in P$ is in $\partial U$ because the collar accumulates there and $z \notin Ω_{v}$ . Thus every point of $P$ is incident to exactly $U$ .

6.1F2F9F10step 2.2step 3.2step 5.2

Every component is incident to a patch. Let $U$ be a component of $Ω_{v}$ . It is nonempty, open and bounded, so $\partial U \neq \emptyset$ : otherwise $\overline{U} = U$ , making $U$ a nonempty clopen proper subset of the connected space $R^{3}$ (proper since $U$ is bounded), contradicting [F10]. Also $\partial U \subseteq \partial Ω_{v}$ : if $x \in \partial U$ then $x \in \overline{U} \subseteq \overline{Ω_{v}}$ and $x \notin Ω_{v}$ , because $Ω_{v}$ is open and $x$ lies neither in the open set $U$ nor in its open complement in $Ω_{v}$ [F2, F9]. By [step 2.2] and [step 3.2] every point of $\partial Ω_{v}$ lies on one of the finitely many boundary pieces, and the patches of the kind constructed in [step 4.3] cover each piece; so some such patch $P$ meets $\partial U$ , and [step 5.2] identifies $U$ as the unique component incident to $P$ .

7.1step 3.2step 5.2step 6.1

Finitely many components. The patches of the kind constructed in [step 4.3], for all boundary points, cover $\partial Ω_{v}$ [step 6.1], and $\partial Ω_{v}$ is compact [step 3.2]; hence finitely many of them already cover $\partial Ω_{v}$ . Each determines exactly one component [step 5.2], and every component is determined by one of them [step 6.1]; therefore the components of $Ω_{v}$ are among these finitely many, so $Ω_{v}$ has finitely many connected components.

8.1F4F8F9F10step 1.1step 3.1step 4.2step 5.1step 3.2step 4.3step 5.2step 6.1step 7.1∎

Each component is a bounded $C^{k}$ domain with the stated boundary. Let $U$ be a component of $Ω_{v}$ and let $P$ be a patch incident to $U$ [step 6.1]. The local description of [step 3.1] shows that, in the chart of $P$ , $U$ coincides with the one-sided region described there, hence is locally exactly the one-sided subgraph of the $C^{k}$ function $h$ (outer case) or of the reflected $C^{k}$ function $h_{0} = - h$ (inner case) over a ball [F4, F8, step 3.1, step 5.2]. So $U$ is a bounded $C^{k}$ domain: it is open [step 5.1], bounded (a subset of bounded $Ω_{v}$ ) and nonempty [F9]. Every point of a patch incident to $U$ is a boundary point of $U$ (it lies in $\partial Ω_{v}$ and in the closure of the collar $N \subseteq U$ [step 4.3, step 5.2]), and conversely every boundary point of $U$ lies in a patch incident to $U$ [step 6.1]; hence $\partial U$ equals the union of those patches in a finite covering that are incident to $U$ . It is compact because it is closed in $R^{3}$ [F2] and contained in the compact $\partial Ω_{v}$ [step 3.2], by [F1]. Finally, let $C$ be a connected component of $\partial Ω_{v}$ ; the assignment to $x \in C$ of the component of $Ω_{v}$ incident to a patch containing $x$ takes values in a finite set and is locally constant on $C$ (near a point of a patch it is given by the single component of that patch [step 5.2]), so it is constant on the connected set $C$ [F10]; hence each boundary component of $\partial Ω_{v}$ is incident to exactly one component of $Ω_{v}$ , and $\partial U$ is a union of such boundary components. The normal formulas of [step 4.2] apply verbatim to each piece of $\partial U$ , since they are the same local charts. This is (d); connectedness of $Ω_{v}$ itself is neither assumed nor asserted.

The vacuum region around compact smooth conductors is a bounded smooth domain

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources