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An affine minorant of a function $f$ on ${\mathbb{R}}^n$ is an affine function $x\mapsto a\cdot x+b$ satisfying $a\cdot x+b\le f(x)$ everywhere. Every proper convex function that is lower semicontinuous on ${\mathbb{R}}^n$ has an affine minorant: separate a point strictly below its closed epigraph by a hyperplane. The finiteness of $f$ somewhere makes the separating vertical coefficient strictly positive, so the separator can be solved for height.

This supplies a useful existence estimate. For $\tau >0$ and $v\in {\mathbb{R}}^n$,The proximal objective is therefore coercive and lower semicontinuous; in finite dimensions it attains a minimum. Its strong convexity makes that minimizer unique, giving a well-defined proximal operator.^[ref] The finite-dimensional attainment argument should not be applied to arbitrary infinite-dimensional normed spaces merely from boundedness of a minimizing sequence.

For the cubic polynomial, an oriented, normalized tangent vector isIts Euclidean norm is one and $DF(z_k)t_k=0$. The sign here follows increasing $x$; for a general solution branch, choose the sign of each new tangent vector so its dot product with the previous one is positive. First predict $z_{\mathrm{p}}=z_k+h{t}_k$, then apply Newton method starting at $z_{\mathrm{p}}$ toThe second equation keeps the correction on the transverse line in this two-dimensional example; in higher dimensions it is a hyperplane. Solving the linear system $DH(z^{(j)})\delta z=-H(z^{(j)})$ and updating $z^{(j+1)}=z^{(j)}+\delta z$ corrects both coordinates together.

This step gives $z_{\mathrm{p}}\approx (0.617556,0.424641)$ and $z_{k+1}\approx (0.635532,0.378840)$. Check both equations: the solution must satisfy $x^3-x+\lambda =0$ and $t_k^{\mathsf{T}}(z-z_{\mathrm{p}})=0$. Checking only the first could accept any point on the solution branch. The new $x$ exceeds $1/\sqrt{3}$, so the step crosses the continuation fold, even though the predictor's $\lambda$ exceeds $2/(3\sqrt{3})$.

At the seed, set $a=3x_k^2-1$. The determinant of the correction Jacobian matrix with the exact unit vector above isAt either continuation fold, $a=0$, $t_k=(1,0)$ and $DH(z_k)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: the correction remains invertible. More generally, if $DF$ has full rank and a one-dimensional kernel spanned by $t_k$, adding $t_k^{\mathsf{T}}$ gives an invertible matrix, since a vector annihilated by both rows must be zero. This is a local argument; the old transverse hyperplane need not work arbitrarily far along the curve. Reduce $h$ and retry when correction fails, and check the equations and orientation after successful steps. Scaling state and parameter coordinates also matters, because it changes the Euclidean norm used to normalize the tangent vector.

// Target: analysis.bigb

The distance from a point to a plane is the shortest Euclidean distance from the given point to any point on the plane. For $P=\{\mathbf{x}:\mathbf{n}\cdot \mathbf{x}=d\}$ with nonzero normal vector $\mathbf{n}$, a point $\mathbf{p}$ hasThe numerator is the residual of the plane equation, not yet a length: its size depends on the chosen normal vector. Dividing by that vector's Euclidean norm removes this arbitrary scale.

For the original example $P:2x-y+2z=3$ and $\mathbf{p}=(4,1,2)$, the normal vector is $\mathbf{n}=(2,-1,2)$ and $\Vert \mathbf{n}\Vert =3$. The residual is $8-1+4-3=8$, so the distance is $8/3$. Using the equivalent plane equation $-6x+3y-6z=-9$ instead gives numerator $24$ and Euclidean norm $9$, hence the same $8/3$. A point already on the plane has zero residual and zero distance.

Why is the perpendicular route shortest? Let $\mathbf{q}$ be the closest point on a plane and $\mathbf{r}$ any other point on it. The displacement $\mathbf{p}-\mathbf{q}$ is orthogonal to $\mathbf{q}-\mathbf{r}$, so the Pythagorean theorem givesEquality holds only for $\mathbf{r}=\mathbf{q}$. This is the closest-point principle of orthogonal decomposition, applied after a translation of an affine plane to its direction linear subspace.^[ref]

The formula is useful whenever the deviation from a flat reference should be measured geometrically. A raw linear equation residual cannot be compared across arbitrary equation scalings; a perpendicular distance can. The same reasoning works for a hyperplane in any finite-dimensional Euclidean space, with the appropriate number of coordinates.

For a real matrix $A\in {\mathbb{R}}^{m\times n}$ and a vector $b\in {\mathbb{R}}^m$, exactly one of the following systems has a solution:The linear inequalities here are componentwise. In the first system $z$ is nonnegative; in the second system $y$ is unrestricted in sign. The second system is a certificate that the first has no feasible point. This nonnegative-equations alternative is developed in Boyd and Vandenberghe, Convex Optimization slides, §5.34.

Build the target, or separate it. Write the columns of $A$ as $a_1,\dots ,a_n$. Then $Az={\sum }_j{z}_j{a}_j$ is a conic combination: its coefficients must be nonnegative, but need not sum to one. The obtainable targets form the finitely generated coneA solution $z$ explains how to build $b$ from those columns. A certificate $y$ instead defines a linear functional $v\mapsto y^{T}v$. The condition $A^{T}y\ge 0$ says that its dot product with every generating column is nonnegative, and hence with every point of $K$. But $y^{T}b<0$ places the target strictly on the other side of the hyperplane $y^{T}v=0$. Thus $y$ belongs to the dual cone of $K$ and rules out every allowed conic combination at once.

A certificate that can be checked by eye. Take the identity matrixHere $K$ is the nonnegative orthant. The equation $Az=b$ forces $z=(-1,1{)}^T$, violating $z\ge 0$. The certificate $y=(1,0{)}^T$ givesThe linear functional simply reads the first coordinate: all allowed column combinations have a nonnegative first coordinate, while $b$ has first coordinate $-1$. For comparison, the target $(1,1{)}^T$ is obtained with $z=(1,1{)}^T$.

Why this is useful and striking. A successful search can exhibit one feasible point; an unsuccessful search alone cannot explain why every candidate fails. Farkas' lemma guarantees an equally concrete witness of impossibility for this finite linear problem. Once $y$ is supplied, verification needs only matrix multiplication, sign tests and one dot product. The same idea gives a Farkas certificate for linear inequalities and underlies linear programming duality.

Brief geometric proof. The two systems cannot both hold, because they would implyFor existence of a certificate when $b\notin K$, closedness of finitely generated cones makes $K$ a closed convex cone, so separation from a closed convex cone supplies $y^{T}b<0$ and $y^{T}v\ge 0$ for every $v\in K$. Testing each column gives $A^{T}y\ge 0$. Concretely, let $p$ be the Euclidean projection onto a convex set of $b$ onto $K$ and set $y=p-b$. The projection inequality gives $y^T(v-p)\ge 0$ for $v\in K$. Testing $v=0$ and $v=2p$ gives $y^{T}p=0$, whence $y^{T}b=-\Vert p-b{\Vert }_2^2<0$. In the example, $p=(0,1{)}^T$ and $y=(1,0{)}^T$. Alternatively, the Fenchel-Moreau theorem applied to the cone's indicator functional proves the same separation through its polar cone.

An original historical source is Julius Farkas's ^[ref]. The linked archival record verifies the author, title and publication details.

For a finite group $G$, the Frattini subgroup $\Phi (G)$ is the intersection of all maximal subgroups. Its elements are exactly the non-generators: an element in every maximal subgroup cannot enlarge a proper generated subgroup to $G$, while an element outside a maximal subgroup generates $G$ together with it.

For a finite p-group,where $G^p$ is generated by all $p$th powers and $[G,G]$ is the commutator subgroup. Every maximal subgroup has index $p$ and is normal, so its quotient group is cyclic of order $p$ and it contains $G^p[G,G]$. Conversely $G/G^p[G,G]$ is an elementary abelian p-group; each nonzero vector is excluded by some hyperplane, whose preimage is a maximal subgroup. This proves equality.

Thus $\Phi (C_{p^2})$ is its order-$p$ subgroup, whereas $\Phi (C_p\times C_p)=1$. The Frattini quotient records the independent generators in the Burnside basis theorem.

A closed half-space consists of a hyperplane and one of its two sides in a real affine space. Replacing the weak inequality by a strict inequality gives an open half-space. Half-spaces are convex sets, and finite intersections of closed half-spaces are linear polyhedra.

A Householder transformation is a reflection that leaves every direction in a chosen hyperplane fixed and reverses its normal direction. For a real unit vector $u$, its matrix isTo understand the formula, decompose any vector into its orthogonal projection onto the normal and a perpendicular component:Then $Hx=x_{\perp }-x_{\parallel }$. Subtracting twice the normal component changes its sign while retaining everything parallel to the mirror. A nonzero normal $v$ gives the equivalent formula $H=I-2v{v}^T/(v^{T}v)$, so normalization can be deferred.

For example, take $u=(2,1{)}^T/\sqrt{5}$. Its mirror is the straight line $2x_1+x_2=0$, andThe midpoint of these two vectors is $(-1,2{)}^T$, which lies on the mirror; their connecting segment is parallel to its normal. Both vectors have Euclidean norm five.

Several useful properties follow from this decomposition. The Householder reflection is a symmetric matrix and an orthogonal matrix:Reflecting twice restores the original vector; Euclidean norms and inner products are preserved. The eigenvalue is $-1$ along $u$ and $+1$ on $u^{\perp }$, so $\det H=-1$. A reflection reverses orientation even though it preserves lengths.

The computational payoff is that the mirror can be chosen to send a whole vector onto one coordinate axis, eliminating several entries at once. Householder QR decomposition repeats this operation on the remaining parts of successive columns. For a real symmetric matrix, applying the same Householder reflection on both sides gives a matrix similarity $HAH$ that preserves symmetry and eigenvalues; this is the basis of Householder tridiagonalization.

Alston S. Householder's 1958 paper Unitary Triangularization of a Nonsymmetric Matrix develops this elimination method. Its motivation includes reducing the arithmetic needed for triangularization while retaining the conditioning advantages of orthogonal transformations. ^[ref] Over the complex numbers, $I-2u{u}^{\ast }$ for a unit $u$ is Hermitian and unitary, with the star denoting the conjugate transpose; the coordinate-axis construction below is stated for real vectors.

For a linear map $T:V\to V$, an invariant subspace $U$ has an annihilator of a vector subspace $U^0$ invariant under the dual map $T^{\prime }$. Indeed, for $\phi \in U^0$ and $u\in U$,For a finite-dimensional vector space, the converse also holds. If $T^{\prime }(U^0)\subseteq U^0$, every $\phi \in U^0$ vanishes on $Tu$ for each $u\in U$. The dual basis description of $U^0$ shows that the common zero set of its linear functionals is exactly $U$, so $Tu\in U$. ThusThe dimension identity $\dim U^0=\dim V-\dim U$ exchanges a one-dimensional invariant subspace with an invariant hyperplane in the dual space.

Classical Laguerre geometry studies oriented hyperplanes, oriented spheres and their oriented tangency in Euclidean space. Its transformations preserve that contact relation. In the plane the objects are oriented straight lines and oriented circles; points can be represented as radius-zero circles. The orientation and signed radius are part of the data, so ordinary unoriented tangency alone loses information.^[ref]

For example, fix a unit normal $u$ and an oriented line $u\cdot x=h$. Under a consistent signed-radius convention, a circle with centre $c$ and signed radius $r$ is in oriented contact when $u\cdot c-h=r$. The absolute-distance equation $|u\cdot c-h|=|r|$ forgets which side and orientation participate. This linear signed-contact relation motivates coordinate models for tangency and the use of Laguerre geometry in studying incircular nets.

For a fixed nonzero $a\in {\mathbb{R}}^n$, a nonzero level set $f(x)=h$ is the sphereThis follows by completing the square after multiplying by the positive denominator. Nonzero levels exist only for $|h|\le \Vert a\Vert /2$. The zero level is the hyperplane $a\cdot x=0$; the extreme levels shrink to points $x=\pm a/\Vert a\Vert$. The Cauchy-Schwarz inequality gives $|f(x)|\le \Vert a\Vert \Vert x\Vert /(1+\Vert x{\Vert }^2)\le \Vert a\Vert /2$, so these extrema are global. At them the Hessian matrix is respectively $-\Vert a\Vert I/2$ and $+\Vert a\Vert I/2$. In two dimensions the spheres are circles whose centres all lie on the axis spanned by $a$.

Two subsets of a Euclidean space are weakly linearly separable when a hyperplane with nonzero normal vector places them in opposite closed half-spaces. Points on the separating hyperplane may belong to either subset. Strict separation instead requires strictly opposite signs, excluding all boundary ties.

In Minkowski spacetime of metric signature $(-,+,+,+)$, let $u$ be a constant future-directed four-velocity with $\eta (u,u)=-c^2$. The time difference assigned to two spacetime events is $\Delta t_u=-\eta (u,Q-P)/c^2$. Thus the observer's simultaneous events form an affine hyperplane orthogonal to $u$. This is an invariant relation of the three arguments $u,P,Q$; the Minkowski metric alone does not single out one $u$. For fixed $u$, it defines an equivalence relation and the hyperplanes form a foliation. Changing $u$ changes this foliation.

Let $C\subset {\mathbb{P}}_{\mathbb{C}}^r$, $r\ge 2$, be a smooth irreducible nondegenerate projective variety of dimension one and degree $d\ge 2$. Let $U$ be the open subset of the dual projective space consisting of hyperplanes meeting $C$ in $d$ distinct points. The incidence projectionis a $d$-sheeted covering space in the complex analytic topology. Its monodromy is the full symmetric group $S_d$.^[ref]

The incidence space choosing two distinct ordered intersection points is irreducible, so the monodromy action of a covering space is two-transitive. A small loop around an ordinary tangent hyperplane exchanges its two nearby intersection points and fixes the others. Conjugating this transposition by double transitivity gives every transposition, and these generate $S_d$. In particular the action on $k$-element subsets is transitive for each $0\le k\le d$. For a degree-one projective line, a transverse section has one point and the corresponding group is the trivial $S_1$.

For two distinct centroids in a Euclidean space, equal squared Euclidean distances expand to the displayed affine hyperplane. It is their perpendicular bisector; each closed half-space consists of points at least as close to the centroid on that side. This proves weak linear separability after a two-centroid nearest-assignment step, allowing ties on the boundary. Coincident centroids are an essential exception: every point ties and the distance comparison supplies no nonzero normal vector.

An extreme point $p\in P$ cannot be written as a nontrivial convex combination of two distinct points of $P$: if $p=\lambda u+(1-\lambda )v$ with $u,v\in P$ and $0<\lambda <1$, then $u=v=p$.

A vertex of a linear polyhedron is a feasible point at which the active constraints include $n$ linearly independent row normals. Equivalently, some $n$ independent bounding hyperplanes intersect uniquely at that point. This definition does not require $P$ to be bounded or full-dimensional. For a linear polyhedron, it is also equivalent to the point uniquely maximizing a suitable linear function: sum the selected active inequalities to construct one such linear function.

Choose representatives $v_1,\dots ,v_{n+1}$ of the first $n+1$ points. The general-position condition says that these vectors form a basis of ${\mathbb{C}}^{n+1}$. Write a representative of the last point as $v_{n+2}={\sum }_i{c}_i{v}_i$. Every $c_i$ is nonzero; otherwise the last point and the other $n$ basis points would lie in a hyperplane. Rescaling $v_i$ by $c_i$ gives representatives with $v_{n+2}={\sum }_i{v}_i$. Normalize representatives $w_i$ of the second frame in the same way.

There is a unique invertible linear map $L$ with $L(v_i)=w_i$ for $i\le n+1$. It also maps $v_{n+2}$ to $w_{n+2}$, so its projectivization has the required effect. For uniqueness, the quotient of two candidate projective transformations fixes the normalized frame. A representing matrix must be diagonal in the $v_i$ basis, with entries $a_i$. Fixing $[{\sum }_i{v}_i]$ forces all $a_i$ to be equal. Such a scalar represents the identity projective transformation. Thus the ordered projective frame determines the transformation uniquely.

Expanding the squared Euclidean distances givesFor distinct centroids, chooseA point assigned to $C_1$ has the displayed distance difference at most zero and therefore $w^{\mathsf{T}}x+b\ge 0$; assignment to $C_2$ gives the reverse inequality. The separating hyperplane is the perpendicular bisector of the line segment joining the two centroids. Points tied between them lie on that hyperplane. This is the nearest-centroid decision hyperplane.

The requested nonzero $w$ needs the additional assumption $z_1\ne z_2$. It does not follow from the stated algorithm alone. In one dimension, take $C_1=\{-1,1\}$ and $C_2=\{0\}$, whose centroids are both zero. A tie rule that retains previous assignments can leave this configuration after an iteration. The inequalities at $-1,1$ imply $b\ge |w|$, while the inequality at zero implies $b\le 0$, forcing $w=0$. Thus the unconditional nonzero-separator claim has this degenerate exception.

For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.

A vector space over the real numbers is a set $V$ with an addition making it an abelian group and a scalar multiplication by $\mathbb{R}$ satisfying, for scalars $a,b$ and vectors $u,v$, the identities $a(u+v)=au+av$, $(a+b)v=av+bv$, $a(bv)=(ab)v$ and $1v=v$. A vector subspace is a nonempty subset closed under addition and scalar multiplication; a proper vector subspace is one unequal to $V$. A spanning set is a set whose finite linear combinations give every vector of $V$. A basis is a linearly independent spanning set. The dimension is the number of vectors in a basis, or its cardinality when the basis is infinite.

The sum of vector subspaces and their intersection of vector subspaces areBoth are vector subspaces: addition and scalar multiplication preserve the displayed forms or simultaneous membership. The intersection of vector subspaces is never empty because $0\in U\cap W$.

For the two given hyperplanes, adding and subtracting their defining equations gives $x_1=x_2$ and $x_3=x_4$. HenceThese two nonzero vectors are orthogonal, and consequently form an orthogonal basis of the intersection of vector subspaces. Put $u=(1,-1,-1,1)$ and $w=(1,-1,1,-1)$. Direct substitution gives $u\in U$, $w\in W$. Their dot products with $b_1,b_2$ vanish, and $u\cdot w=0$. Each defining equation of $U$ or $W$ is one nonzero linear constraint in four variables, so each vector subspace has dimension three. Thus each space has the indicated orthogonal basis:The last four vectors are nonzero and mutually orthogonal, so they are a basis of ${\mathbb{R}}^4$. They lie in $U+W$, proving $U+W=V$. The dimension identity is therefore verified numerically:

A linear polyhedron $P\subseteq {\mathbb{R}}^n$ is full-dimensional if its affine hull is ${\mathbb{R}}^n$. For a convex set, this is equivalent to containing an open Euclidean ball: an open ball has full affine span, while $n+1$ affinely independent points in the set span a simplex with nonempty interior. In particular, a nonempty linear polyhedron need not be full-dimensional; a singleton or a hyperplane is an example.

Assume $U\ge 1$, increasing the coefficient bound to one if necessary. If $x_0\in P$ and $A_i$ denotes the $i$th row, then $\Vert A_i{\Vert }_2\le \sqrt{n}U$. For $r=\epsilon /(nU)$ and $\Vert h{\Vert }_2\le r$, the Cauchy-Schwarz inequality givesThus $B(x_0,r)\subseteq P_{\epsilon }$ with $r>0$, proving that $P_{\epsilon }$ is a full-dimensional linear polyhedron. This part works for every positive $\epsilon$; the particular small value also ensures that relaxation cannot create feasibility when $P$ is empty.

Here is that exact-feasibility point. Put $D=[(n+1)U{]}^{n+1}$. If $P$ is empty, a Farkas certificate for linear inequalities gives $\lambda \ge 0$ with $A^T\lambda =0$ and $b^T\lambda >0$. Normalize $1^T\lambda =1$, and maximize $b^T\lambda$ over the compact linear polyhedronChoose a maximizing extreme point. Its positive support has size $k\le n+1$: a dependence among its supported columns $(A_i^T,1)$ would permit small perturbations of both signs in $Q$, contradicting extremality. Select $k$ independent equations for those $k$ coordinates. Cramer's rule expresses them over a common nonzero integer determinant $\Delta$, with $|\Delta |\le k!U^k\le D$. Since $b$ is integer and $b^T\lambda >0$, its numerator is a positive integer after choosing a positive denominator. Hence $b^T\lambda \ge 1/D$. This is the normalized integer Farkas infeasibility gap. A point in $P_{\epsilon }$ would implya contradiction. Consequently $P=\varnothing$ if and only if $P_{\epsilon }=\varnothing$.

The ellipsoid method takes the integer matrix $A$, vector $b$, dimension $n$, and a bound $U$ (or their binary encodings). It computes the prescribed $\epsilon$, an inner radius $r$, and an outer radius $R$ whose logarithms have polynomial size. For an explicit outer bound, let $M=(nU{)}^n$. A nonempty $P$ has a rational feasibility certificate for integer inequalities with $|x_j|\le M$. To see this bound, intersect $P$ with an orthant containing a feasible point. This intersection is a nonempty linear polyhedron containing no whole nontrivial affine line, and it has an extreme point. Indeed, if its active rows do not yet span the ambient space, move along a nonzero direction annihilating them. The orthant prevents motion in both directions from being feasible forever, so in at least one direction a new independent row becomes active. Repeating at most $n$ times produces an extreme point. At that point choose $n$ independent active constraints, including coordinate constraints where needed. Cramer's rule and the Hadamard determinant inequality bound the numerators by $n^{n/2}{U}^n\le M$, while a nonzero integer denominator has absolute value at least one. Thus the point has norm at most $nM$. Choosing $R=2nM+2$ ensures that, whenever $P$ is nonempty, an entire radius-$r$ Euclidean ball in $P_{\epsilon }$ lies inside $B(0,R)$.

Start with the ellipsoid $E_0=B(0,R)$ and retain $K=P_{\epsilon }\cap B(0,R)$. At each iteration test the centre $c$ against every relaxed inequality. If it passes all tests, $P_{\epsilon }$ is nonempty and hence so is $P$. Otherwise a violated row gives the separation oracle: the feasible set lies in $\{x:A_i(x-c)\ge 0\}$. Cut the current ellipsoid by this half-space through its centre and replace it by the standard containing ellipsoid. Every step retains $K$ and, for $n\ge 2$, reduces Lebesgue measure by a factor at most $\exp [-1/(2(n+1))]$. In dimension one the corresponding step bisects an interval.

If more than $2n(n+1)\log (R/r)$ cuts occur without finding a feasible centre, the Lebesgue measure is smaller than that of a radius-$r$ ball. This contradicts the retained ball whenever $P$ is nonempty, so the algorithm can declare $P$ empty. Both $\log R$ and $\log (1/r)$ are polynomial in $n$ and $\log U$, and each inequality test uses $O(mn)$ arithmetic operations for $m$ rows. The bit implementation uses rational rounding with outward enlargement of the containing ellipsoids; the encoding bounds allow polynomial precision while keeping a constant fraction of the volume decrease. Detailed update and rounding formulae are not needed for this account. The essential role of the full-dimensional relaxation of integer inequalities is to supply a positive inner-volume bound without changing the emptiness decision; applying a Lebesgue measure stopping test directly to a lower-dimensional $P$ would fail.