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Polymer analyticity and dilute continuum correlations — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Hard-core spheres in the explicit dilute domain

Example

Assume Countable Choice. Fix $d \in N$ with $d \geq 1$ , a hard-core diameter $a \in R$ with $a > 0$ , and $β \in R$ with $β > 0$ . For $x = (x_{j})_{j < d} \in R^{d}$ , write $∥ x ∥_{2} : = \sqrt{\sum_{j < d} x_{j}^{2}}$ , and let $B (0, a) : = {x \in R^{d} : ∥ x ∥_{2} < a}$ . Let $λ_{d}$ denote Lebesgue measure on $R^{d}$ . Define the hard-core pair potential $v_{a} : R^{d} \to R \cup {+ \infty}$ by $v_{a} (x) = + \infty$ when $∥ x ∥_{2} < a$ and $v_{a} (x) = 0$ when $∥ x ∥_{2} \geq a$ . Its stability constant may be chosen as $B = 0$ , the function $x \mapsto e^{- β v_{a} (x)} - 1$ is $- 1$ on $B (0, a)$ and $0$ elsewhere, and its absolute integral is $C_{β} = λ_{d} (B (0, a))$ . For a complex activity $z \in C$ , with modulus $∣ z ∣$ as defined by $∣ x + i y ∣ = \sqrt{x^{2} + y^{2}}$ , the sufficient analytic domain is $e ∣ z ∣ λ_{d} (B (0, a)) < 1$ .

Statement references

The stable continuum Mayer pressure series converges in an explicit dilute disk

Facts & Assumptions

Given: Countable Choice, $d \in N$ with $d \geq 1$ , $a \in R$ with $a > 0$ , $β \in R$ with $β > 0$ , and the hard-core potential in the Example.

[A1]

Assume Countable Choice. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

Stability, superstability and regular tails: stability means that every $N$ -particle pair-energy sum is at least $- B N$ ; the potential must be even Borel.

[F2]

The stable continuum Mayer pressure series converges in an explicit dilute disk: under Countable Choice, for a stable even Borel potential with finite $C_{β}$ , the centered-cube pressure series is absolutely convergent whenever $∣ z ∣ e^{2 β B + 1} C_{β} < 1$ .

[F3]

Euclidean balls have positive finite Lebesgue measure: under Countable Choice, the Euclidean ball $B (0, a)$ is Lebesgue measurable and has positive finite measure.

[F4]

Open ball, closed ball and sphere in a metric space, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, The Borel sigma-algebra of a topological space, and The Borel sigma-algebra on the extended real line: $B (0, a) = {x : ∥ x ∥_{2} < a}$ is an open Euclidean ball, every open set is Borel, and $+ \infty$ is Borel in the extended real line.

[F5]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ x ∥_{2} = \sqrt{\sum_{j < d} x_{j}^{2}}$ , so $∥ - x ∥_{2} = ∥ x ∥_{2}$ .

[F6]

The nonnegative integral agrees with the simple integral on simple functions and The integral of a nonnegative simple function: the integral of an indicator of a measurable set equals that set's measure.

[F7]

Real and imaginary parts, complex conjugation, and modulus: if $z = x + i y$ with $x, y \in R$ , its complex modulus is $∣ z ∣ = \sqrt{x^{2} + y^{2}}$ .

[F8]

Absolute value in an ordered field, Basic properties of the absolute value: real absolute value is defined by sign, is nonnegative, and satisfies $∣ - 1 ∣ = 1$ and $∣ 0 ∣ = 0$ .

Proof

Given: The model and assumptions in the Example, [A1], [F1], [F2], [F3], [F4], [F5], [F6], [F7], and [F8].

1.1A1F1F4F5algebra

The Euclidean norm is even by [F5], so $v_{a} (- x) = v_{a} (x)$ . The ball $B (0, a)$ is open, hence Borel, by [F4]; its complement is Borel as well. The target singleton ${+ \infty}$ is Borel by [F4], and ${0} = ⋂_{m \geq 1} (- 1 / m, 1 / m)$ is Borel by the extended-real topology and sigma-algebra closure in [F4]. Since $v_{a}$ takes the constant value $+ \infty$ on the ball and $0$ on its complement, the preimage of any target Borel set is among the empty set, the ball, its complement, and all of $R^{d}$ ; hence $v_{a}$ is Borel. Every pair energy is either $0$ or $+ \infty$ , so every finite total energy is at least zero; therefore the stability inequality holds with $B = 0$ .

1.2A1F3F6F8givenalgebra

Under the stated convention, $e^{- β v_{a} (x)} - 1 = - 1$ on $B (0, a)$ and $0$ elsewhere, so $∣ f_{β} ∣ = 1_{B (0, a)}$ by [F8]. The ball lemma [F3] gives measurability and finite measure, and [F6] gives $C_{β} = \int 1_{B (0, a)} d x = λ_{d} (B (0, a)) \in (0, \infty)$ .

2.1A1F2F7step 1.1step 1.2algebra∎

The hypotheses of [F2] hold by steps 1.1–1.2 and [A1]. Substituting $B = 0$ and $C_{β} = λ_{d} (B (0, a))$ into its sufficient disk gives $e ∣ z ∣ λ_{d} (B (0, a)) < 1$ . This is a sufficient analytic domain; it is not asserted to be necessary and implies no freezing or coexistence point.