Proof. We first recall Harris’s correlation inequality for product Bernoulli measures (Harris 1960), together with its elementary proof. Increasing functions of independent Bernoulli variables have nonnegative covariance. Indeed, induction on the number of variables and conditioning on the last variable reduce the additional covariance term to \(p(1-p)(f_1-f_0)(g_1-g_0)\ge0\), where \(f_b,g_b\) are the corresponding conditional means. Thus an increasing event \(B\) and a decreasing event \(F\) satisfy \[ \mathbb{P}(B\mid F)\le\mathbb{P}(B) \quad\text{whenever }\mathbb{P}(F)>0. \tag{50}\]

Write \(B_i=\{I_i=1\}\), \(p_i=\mathbb{P}(B_i)\), and \(p_{ij}=\mathbb{P}(B_i\cap B_j)\). For each \(i\), let \(F_i\) be the event that none of the earlier \(B_j\) with \(H_i\cap H_j=\varnothing\) occurs, and let \(G_i\) exclude the remaining earlier events. If some earlier avoidance event has probability zero, the conclusion is immediate. Otherwise all the conditional probabilities below are defined. Since \(F_i\) uses only coordinates outside \(H_i\), it is independent of \(B_i\). It is also decreasing. A conditional union bound and (50) therefore give \[\begin{aligned} \mathbb{P}(B_i\mid F_i\cap G_i) &\ge\mathbb{P}(B_i\cap G_i\mid F_i)\\ &\ge p_i- \sum_{\substack{j<i\\H_i\cap H_j\ne\varnothing}}p_{ij}. \end{aligned}\] The first inequality uses \(\mathbb{P}(G_i\mid F_i)\le1\). Multiplying the successive avoidance probabilities and using \(1-x\le e^{-x}\) yields \[ \mathbb{P}(Y=0)\le\exp(-m+\Delta_{\mathrm{off}}/2), \qquad \Delta_{\mathrm{off}}= \sum_{\substack{i\ne j\\H_i\cap H_j\ne\varnothing}}p_{ij}. \tag{51}\] This remains valid when an intermediate lower bound above is negative.

Attach to each index \(i\) a private independent Bernoulli variable of parameter \(q\), and require it as part of \(B_i\). The new mean is \(qm\), and its ordered off-diagonal sum is \(q^2\Delta_{\mathrm{off}}\). The original zero event implies the new zero event, so (51) bounds its probability by \(\exp(-qm+q^2\Delta_{\mathrm{off}}/2)\). Since \(\Delta^*=m+\Delta_{\mathrm{off}}\), choosing \(q=m/\Delta^*\in(0,1]\) proves (49). The proof retains every index even when two supports coincide. ◻

Proof. Write \(\tau=\mu(\Omega)>0\). Conditional on \(N=\Pi(\Omega)\), the points are independent with law \(\mu/\tau\) and are almost surely distinct. If \(k=|I_\alpha|\), the Poisson factorial moment \(\mathbb{E}(N)_k=\tau^k\) gives \[\mathbb{E}Z_\alpha =\mathbb{E}(N)_k\,\tau^{-k}\prod_i\lambda_{\alpha i} =\nu_\alpha.\] In particular, no factorial divides the activity, even for repeated slot sets.

We apply Lemma 7 to cell-occupation indicators. Fix \(\delta>0\). Form the Boolean atoms generated by all the slot sets and subdivide each positive-measure atom into finitely many measurable cells of measure at most \(\delta\), using nonatomicity. After omitting null cells, this gives a finite partition \(\mathcal P_\delta\) that respects every slot. Modifications on the common null set do not affect the Poisson process. For \(C\in\mathcal P_\delta\), set \[a_C=\mu(C),\qquad X_C=\mathbf1_{\{\Pi(C)>0\}},\qquad p_C=\mathbb{E}X_C=1-e^{-a_C}.\] These variables are independent and \[ 0\le a_C-p_C\le\tfrac12a_C^2\le\tfrac\delta2a_C. \tag{54}\]

For each \(\alpha\), retain every injective assignment of cells \(f:I_\alpha\to\mathcal P_\delta\) with \(f(i)\subseteq E_{\alpha i}\). Its indicator is \(\prod_iX_{f(i)}\). Let \(Y_\delta\) be the sum over these indexed assignments. Different assignments remain different indices even when their underlying cell sets agree.

The following estimate controls all cell collisions. For a fixed list \(F_1,\ldots,F_r\) of unions of cells, let \(Q_\delta(F_1,\ldots,F_r)\) be the sum of \(\prod_{v=1}^rp_{C_v}\) over distinct cells \(C_v\subseteq F_v\). Then \[ \begin{split} 0\le\prod_v\mu(F_v)-Q_\delta(F_1,\ldots,F_r) \le\delta\bigg[&\frac r2\prod_v\mu(F_v)\\ &+\sum_{v<w}\mu(F_v\cap F_w) \prod_{u\ne v,w}\mu(F_u)\bigg]. \end{split} \tag{55}\] Indeed, the unconstrained sum factors over the slots, and (54), followed by telescoping the product, bounds its difference from \(\prod_v\mu(F_v)\) by the first term. The weight of assignments with \(C_v=C_w\) is at most \[\sum_{C\subseteq F_v\cap F_w}p_C^2 \prod_{u\ne v,w}\mu(F_u) \le\delta\mu(F_v\cap F_w)\prod_{u\ne v,w}\mu(F_u).\] A union bound proves (55). Empty products are 1, and the estimate also holds if a set has measure zero. Applying it to each slot list gives \[ m_\delta:=\mathbb{E}Y_\delta\longrightarrow\sum_\alpha\nu_\alpha. \tag{56}\]

It remains to identify the finite dependency sum and the limiting event. For two internally injective assignments \(f\) and \(g\), the complete set of equalities \(f(i)=g(j)\) is a unique partial matching \(M\). Their Bernoulli supports intersect precisely when \(M\ne\varnothing\). For fixed \(\alpha,\beta,M\), identify the slots paired by \(M\) and leave the other slots as singletons. Assign to each resulting class its slot set, or the intersection of its two slot sets. Denote these sets by \((F_v)_v\). Requiring distinct cells for these classes excludes exactly the unintended equalities. Thus the contribution with exact matching \(M\) is \(Q_\delta((F_v)_v)\), and its limit is \[\prod_v\mu(F_v)=\nu_\alpha\nu_\beta \prod_{(i,j)\in M} \frac{\mu(E_{\alpha i}\cap E_{\beta j})} {\lambda_{\alpha i}\lambda_{\beta j}}.\] All sums are finite. Estimate (55) therefore shows that the ordered dependency sum, including identical assignment pairs, satisfies \[ \Delta^*_\delta\longrightarrow\sum_{\alpha,\beta}D_{\alpha\beta}. \tag{57}\] Identical assignments give the full identity matching; other full matchings caused by repeated slots remain separate terms.

Finally, couple the cell variables with the original process. The probability that any cell contains two or more points is at most \[\frac12\sum_C\mathbb{E}[\Pi(C)(\Pi(C)-1)] =\frac12\sum_Ca_C^2\le\frac{\delta\tau}{2}.\] Outside this event, point injections and occupied-cell injections are in bijection, simultaneously for every witness type. Hence \[\left|\mathbb{P}(Y_\delta=0)- \mathbb{P}(Z_\alpha=0\text{ for all }\alpha)\right| \le\frac{\delta\tau}{2}.\] Apply Lemma 7 and let \(\delta\downarrow0\), using (56) and (57). The identity-matching contribution makes the limiting denominator positive. This proves (53). ◻

Proof. Write \(h_{ij}=\mu(E_i\cap E_j)/(\lambda_i\lambda_j)\). For the self sum, fix the off-diagonal edges \(P\) of a partial matching between two copies of the slot list. Regard \(P\) as a directed graph on the common label set. Every vertex has total degree \(d_P(i)\in\{0,1,2\}\), since its in-degree and out-degree are at most one. A diagonal loop is available exactly at the vertices of degree zero. Since \(h_{ii}=\lambda_i^{-1}\), summing the compatible loop choices gives \[\left(\prod_i\delta_i\right) \left(\prod_{(i,j)\in P}h_{ij}\right) \left(\prod_{i:d_P(i)>0}\delta_i^{-1}\right) \le\left(\prod_i\delta_i\right)\prod_{(i,j)\in P}w_{ij}.\] Here \(\delta_i^{-1}\le \delta_i^{-d_P(i)/2}\) at each incident vertex. This includes directed paths and cycles; Figure 2 illustrates the permitted incidences. Summing over \(P\) and dropping the requirement of distinct right endpoints bounds the remaining sum by \(\prod_i(1+\sum_{j\ne i}w_{ij})\). The empty matching contributes 1, proving (58).

For two lists, drop all matching restrictions and sum over nonempty subsets of cross edges. With the analogous nonnegative weights \(h_{ij}\), the result is at most \(\prod_{i,j}(1+h_{ij})-1\le e^{\sum_{i,j}h_{ij}}-1\). This proves (59). ◻

Proof. Apply Lemma 4 with shell parameter \(10\) and perturbation bound \(3\). Its affinity and common-intersection estimates supply the hypotheses of Proposition 6; the common-intersection constant is available before the radius multiplier in that proposition is fixed. Apply the proposition separately to each list, making all choices before sampling \(\Pi\). The budget (36) gives \[-\log\nu_\alpha \le\sum_S\log(1+\lambda_{\alpha S}^{-1})\le AR.\] Together with (37) and (58), it also gives the self bound in (60), after choosing \(C_*\).

To check the cross estimate, write \(\bar c_{\alpha i}=\min\{\mu(C_{\alpha i}),1/2\}\) and let \(i(S)\) be the anchor of a block. Lemma 4 and Proposition 6 provide fixed exponents \(P_0,P_1>0\) such that \[\bar c_{\alpha i}\ge n^{-P_0},\qquad \lambda_{\alpha S}\ge n^{-P_1}\bar c_{\alpha,i(S)},\qquad E_{\alpha S}\subseteq C_{\alpha,i(S)}.\] Apply the affinity estimate to each two-element list of anchor residuals; this does not require the union of the two full lists to have at most \(n^2\) entries. It gives \[\frac{\mu(E_{\alpha S}\cap E_{\beta T})} {\lambda_{\alpha S}\lambda_{\beta T}} \le Cn^{2P_1+P_0} e^{-c n\angle(x_{\alpha,i(S)},x_{\beta,i(T)})^2}.\] There are at most \(n^4\) block pairs. Choose \(H\) after these constants, so that \(cH^2>9+2P_1+P_0\). For sufficiently large \(n\), the angular hypothesis gives \(\rho_{\alpha\beta}\le n^{-4}\). Equation (59) now gives \(D_{\alpha\beta}\le2n^{-4}\nu_\alpha\nu_\beta\), which implies (61). At cross angles greater than \(4\sqrt\varepsilon\), Corollary 5 makes every cross-slot intersection empty, and the diagram is zero.

Finally, the point assigned to a block lies in every cap belonging to that block. A positive injection count thus hits every residual cap. ◻

Proof. Condition on \(Y=y\) with \(0<|y-p|\le L\), and write \(u=(y-p)/|y-p|\). If the event in the statement occurs, the component of \(Z-q\) perpendicular to \(u\) has length at most \(L\sin\xi\). Both orientations allowed by the projective angle therefore lie in the same cylinder around the affine line \(q+\mathbb Ru\). The projection of \(B_D\) onto the direction \(u\) has length \(2D\). Every perpendicular slice of the cylinder has \((n-1)\)-dimensional volume at most \(\omega_{n-1}(L\sin\xi)^{n-1}\). Thus the allowed points \(Z\) have volume at most \(2D\omega_{n-1}(L\sin\xi)^{n-1}\). Divide by \(|G|\) and then average over \(Y\).

To obtain the last inequality, use the slicing identity \[\frac{\omega_n}{\omega_{n-1}} =\int_{-1}^1(1-t^2)^{(n-1)/2}\,dt \ge\frac{2\exp(-1/2)}{\sqrt n}.\] Indeed, restrict the integral to \(|t|\le n^{-1/2}\) and use \(\log(1-1/n)\ge-1/(n-1)\). In particular, \(\omega_{n-1}/\omega_n\le\sqrt n\). ◻

Proof. Sample \(M_n\) points independently and uniformly from \(G\). For independent sampled points \(Y,Z\), take the union in Lemma 11 over all \(N^2\) ordered pairs of center labels. Since \(b+3\eta\le2\) for sufficiently large \(n\), this gives \[ \mathbb P\bigl((Y,Z)\text{ is $\xi$-bad}\bigr) \le8\sqrt n\,N^2 \left(\frac{2\sin\xi}{D}\right)^{n-1}. \tag{66}\] All relevant labels are included even though the relevant lists vary with \(Y\) and \(Z\). Repeated centers only repeat events in this union.

Set \(t_n=H\sqrt{\log n/n}\). Multiply the right side of (66), with \(\sin t_n\le t_n\), by \(\binom{M_n}{2}\). The logarithm of the resulting bound is at most \[-\frac38n\log n+\frac12n\log\log n+O_{H,k_0}(n),\] which tends to \(-\infty\). Thus the probability that any pair of sampled points is \(t_n\)-bad tends to zero, uniformly in the pattern. For \(\xi=4\sqrt\varepsilon\), the inequality \(\sin\xi\le4\sqrt\varepsilon\) instead bounds (66) by \(f_n\). The expected fraction of these bad unordered pairs is at most \(f_n\). Markov’s inequality shows that this fraction exceeds \(10f_n\) with probability at most \(1/10\). Consequently both conclusions hold with probability at least \(9/10-o(1)>0\). The sampled points are distinct almost surely because uniform measure on \(G\) has no atoms. Fix a successful list. No independence between the two separation properties is needed. ◻

Proof. Fix \(P\in\mathcal F_n\) and its deterministic good set \(G_P\). At each \(y\in G_P\), Proposition 3 gives \[ r_--2\eta\le |x_i|\le b+3\eta,\qquad m_y\le2C_{\mathrm{loc}}R,\qquad \sum_{i=1}^{m_y}w_a(|x_i|)\le2C_{\mathrm{loc}}R \tag{68}\] for its relevant residuals \(x_i=y-p_i\). If this list is empty at any point of \(G_P\), that point lies in none of the balls \(p+(b+3\eta)B\), so the probability in (67) is zero. We may assume all these lists are nonempty. For the range of \(R\) in the proposition, \(m_y\le n^2\) once \(n\) is sufficiently large.

For each relevant residual, define the cutting cap \[C_i(y)=\{u\in S^{n-1}:|u\cdot x_i|>1+3\eta\}.\] Lemma 4, applied with radial loss \(10\log n/n\) and a fixed perturbation bound, gives \[s_i=-\log\min\{\mu(C_i(y)),1/2\} \le C_a w_a(|x_i|).\] Consequently \[ m_y+\sum_{i=1}^{m_y}s_i\le A_0R, \qquad A_0=2C_{\mathrm{loc}}(1+C_a). \tag{69}\] All assumptions of Corollary 10 therefore hold uniformly in \(y\) and \(P\). Fix its constants \(A,C_*\) and \(H\) for this value of \(A_0\).

The list-length estimate in Proposition 3 gives \(N\le n^{k_0}\exp(\lambda_0n)\) for a fixed \(k_0\), uniformly in the present range of \(R\). Apply Lemma 12 to \(G_P\) with the just-fixed \(H\). At each of the resulting \(M=M_n\) targets, choose the block partition and measurable witness sets from Corollary 10. Denote its injection count by \(Z_y\), its activity by \(\nu_y\), and the diagram sums by \(D_{yz}\). The corollary gives \[ \exp(-AR)\le\nu_y\le1, \qquad D_{yy}\le\nu_y^2\exp(C_*R). \tag{70}\] Every pair of distinct selected targets satisfies \(D_{yz}\le n^{-3}\nu_y\nu_z\). Moreover, \(D_{yz}=0\) unless the pair is \(4\sqrt\varepsilon\)-bad. Our use of “at most” in the definition of badness makes both angular interfaces strict on its complement.

All choices so far depend only on \(P\) and its targets. In particular, the witness sets are chosen before the Poisson process; their activities are deterministic intensities. There is no need to choose witness sets measurably at every point of \(G_P\), since only the finite selected list is used. If \(Z_y>0\), every relevant residual is cut by a slab of the process, and the other centers are too far from \(y\) to contain it in a translate of \(K^+\). Thus \(Z_y>0\) certifies a hole.

Divide the possible values of \(-\log\nu_y\) in \([0,AR]\) into intervals of length \(\log2\). At most \[Q=1+\left\lceil\frac{AR}{\log2}\right\rceil\] bins suffice. A largest bin has \(m\ge M/Q\) targets and, for some \(\nu>0\), all its activities belong to \([\nu,2\nu]\). There are at most \(10f_n\binom M2\) bad unordered pairs in the full list. Hence the chosen bin contains at most \[ 10f_nM(M-1)\le10f_nQ^2m^2 \tag{71}\] bad ordered pairs with distinct indices. This bound allows the bin to contain every bad pair; no independence between bin membership and angles is asserted.

Apply Proposition 8 to the witnesses in this bin. Its numerator is at least \(m^2\nu^2\). By (70) and (71), its denominator before the factor \(2\) is at most \[\sum_{y,z}D_{yz} \le4m\nu^2\exp(C_*R) +40m^2\nu^2n^{-3}Q^2f_n.\] The first term counts diagonal pairs. Among distinct pairs, only the coarse bad pairs contribute. Coverage forces \(Z_y=0\) at every target of the bin, so \[ \mathbb P\left(B_D\subseteq\bigcup_{p\in P}(p+K^+)\right) \le\exp\left[-\frac1{ 8\bigl(\exp(C_*R)/m+10n^{-3}Q^2f_n\bigr)}\right]. \tag{72}\]

Now choose \[c_0=\frac1{64\max\{C_*,1\}}.\] Uniformly for \(1\le R\le c_0n\log n\), we have \(\log Q=O(\log n)\) and, for sufficiently large \(n\), \[m\exp(-C_*R) \ge\frac{M_n}{Q}\exp(-C_*R) \ge\exp\left(\frac{n\log n}{32}\right).\] The first term inside the parentheses in (72) is therefore at most \(\exp(-n\log n/32)\). By (65), the second is \(\exp(-\gamma n+O(\log n))\). Since \(\gamma\) is fixed, their sum is at most \(\exp(-\gamma n+O(\log n))\); absorbing the fixed factor \(8\) gives (67).

The order of the choices is fixed throughout: the geometric parameters come first, then the cap and block constants and \(H\), then \(c_0\), and finally the dimension threshold. Every estimate is uniform over the retained finite class. In particular, the auxiliary target selection introduces no dependence on the realization of \(\Pi\). ◻

Proof. Theorem 1 gives the lower bound already among centrally symmetric bodies, and \(\theta_T\le\theta_L\) transfers it to lattice coverings. Rogers’ bound (3) supplies the translative upper bound. The companion single-lattice theorem (OpenAI 2026, Theorem 1.1) gives \(\theta_L(K)\le Cn\log n\) for every convex body and every \(n\ge2\), which supplies the two lattice upper bounds. ◻