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Reference entropy supremum is finite and attained

Statement

Assume Comparison and stability for compounds on compounds over a nonempty state set $X$ , a strict reference pair $x_{0} ⪯ x_{1}$ with not $x_{1} ⪯ x_{0}$ , and the complete ordered field $R$ . For $x \in X$ define

$L_{x} : = {λ \in R : R_{λ} ⪯_{s} e_{x}} .$

Then $L_{x}$ is nonempty, bounded above, and downward closed. Its supremum $S (x) : = \sup L_{x}$ is a real number and satisfies $R_{S (x)} \sim_{s} e_{x}$ . Stability is used to prove attainment; no continuity of states or order is assumed.

Facts & Assumptions

[A1]

Comparison and stability for compounds: equal-mass compounds form a preorder with consistency, positive scaling, stability for fixed positive spectators, and CH.

[F1]

Tagged scalable compounds and Compound mass is well-defined and additive: $e_{x}$ has mass one, compound masses are finite sums, and composition/scaling preserve the stated masses.

[F2]

Signed comparison of finite coefficient maps and Spectator cancellation and signed-order algebra: signed comparisons are well-defined, translation-invariant, and agree with ordinary comparison on nonnegative compounds. For a finite-support signed map $u$ , write $u^{+} (y) : = \max {u (y), 0}$ and $u^{-} (y) : = \max {- u (y), 0}$ pointwise; these are finite-support nonnegative maps with $u + u^{-} = u^{+}$ , and this notation extends the positive and negative parts used in Signed comparison of finite coefficient maps to each signed reference map $R_{λ}$ itself.

[F3]

Ordering the two-state reference mixtures: $R_{λ} ⪯_{s} R_{μ}$ exactly when $λ \leq μ$ .

[F4]

Reciprocals of the positive integers decrease to zero: for $n \in N$ , $ϵ_{n} : = 1 / q_{n}$ with $q_{n} : = ι_{N} (σ (n))$ is positive, strictly decreasing, and converges to zero. Its canonical-natural convention gives $ι_{N} (n) \geq 0$ and $q_{n} = ι_{N} (n) + 1 \geq 1$ ; natural indices and their real images are distinguished.

[F5]

Complete ordered field (least-upper-bound property): every nonempty bounded-above subset of $R$ has a supremum.

Proof

Given: [A1]–[F5], with $x \in X$ fixed.

1.1A1F1F2F4

Write $q_{n} : = ι_{N} (σ (n)) > 0$ , so $ϵ_{n} = 1 / q_{n}$ by [F4]. Suppose no $n \in N$ satisfies $(q_{n} + 1) e_{x_{0}} ⪯ q_{n} e_{x_{1}} + e_{x}$ . Both sides have mass $q_{n} + 1$ , so CH gives $q_{n} e_{x_{1}} + e_{x} ⪯ (q_{n} + 1) e_{x_{0}}$ . Scale by $1 / q_{n} = ϵ_{n}$ to obtain $e_{x_{1}} + ϵ_{n} e_{x} ⪯ e_{x_{0}} + ϵ_{n} e_{x_{0}}$ . Apply stability to the sequence $ϵ_{n}$ , with bases $e_{x_{1}}, e_{x_{0}}$ and positive unit-mass spectators $e_{x}, e_{x_{0}}$ . It yields $x_{1} ⪯ x_{0}$ , contradicting strictness. Thus some $n$ has $(q_{n} + 1) e_{x_{0}} ⪯ q_{n} e_{x_{1}} + e_{x}$ , which by signed translation and cancellation in [F2] is $R_{- q_{n}} ⪯_{s} e_{x}$ ; hence $- q_{n} \in L_{x}$ .

1.2A1F1F2F3F4

Suppose $L_{x}$ is unbounded above. For each $n \in N$ , unboundedness supplies $λ \in L_{x}$ with $λ > q_{n}$ ; [F3] gives $R_{q_{n}} ⪯_{s} R_{λ}$ , and transitivity gives $q_{n} \in L_{x}$ . Since $q_{n} \geq 1$ by [F4], adding the common compound $q_{n} e_{x_{0}}$ to $R_{q_{n}}$ and $e_{x}$ makes them $e_{x_{0}} + q_{n} e_{x_{1}}$ and $e_{x} + q_{n} e_{x_{0}}$ . Thus [F2] identifies $R_{q_{n}} ⪯_{s} e_{x}$ with the compound comparison between those maps. Scaling by $ϵ_{n} = 1 / q_{n}$ gives $e_{x_{1}} + ϵ_{n} e_{x_{0}} ⪯ e_{x_{0}} + ϵ_{n} e_{x}$ . Stability with bases $e_{x_{1}}, e_{x_{0}}$ and positive unit-mass spectators $e_{x_{0}}, e_{x}$ now yields $x_{1} ⪯ x_{0}$ , again a contradiction. Therefore $L_{x}$ is bounded above.

2.1A1F2F3F5step 1.1step 1.2

If $λ \in L_{x}$ and $μ \leq λ$ , [F3] gives $R_{μ} ⪯_{s} R_{λ}$ ; signed transitivity and $R_{λ} ⪯_{s} e_{x}$ imply $μ \in L_{x}$ . Thus $L_{x}$ is downward closed. By steps 1.1 and 1.2 it is nonempty and bounded above, so [F5] supplies the real number $s : = \sup L_{x}$ .

3.1A1F1F2F4step 2.1

For each $n \in N$ , $s - ϵ_{n}$ is not an upper bound of $L_{x}$ , so some $λ \in L_{x}$ has $λ > s - ϵ_{n}$ . Downward closure gives $s - ϵ_{n} \in L_{x}$ . Consistency then gives $R_{s} + ϵ_{n} e_{x_{0}} ⪯_{s} e_{x} + ϵ_{n} e_{x_{1}}$ , because $R_{s - ϵ_{n}} + ϵ_{n} e_{x_{1}} = R_{s} + ϵ_{n} e_{x_{0}}$ . Let $K : = R_{s}^{-}$ . Adding the fixed nonnegative map $K$ to both sides converts this signed comparison into the compound comparison $R_{s}^{+} + ϵ_{n} e_{x_{0}} ⪯ e_{x} + K + ϵ_{n} e_{x_{1}}$ . Their bases have equal mass because $m (R_{s}^{+}) - m (K) = 1 = m (e_{x})$ . Stability with positive unit-mass spectators $e_{x_{0}}, e_{x_{1}}$ yields $R_{s}^{+} ⪯ e_{x} + K$ , that is, $R_{s} ⪯_{s} e_{x}$ .

4.1A1F1F2F4step 2.1step 3.1∎

Since $s$ is an upper bound, $s + ϵ_{n} \notin L_{x}$ for every $n \in N$ . To derive the reverse signed comparison from CH, set $H_{n} : = R_{s + ϵ_{n}}^{-}$ . The translated maps $e_{x} + H_{n}$ and $R_{s + ϵ_{n}} + H_{n} = R_{s + ϵ_{n}}^{+}$ are nonnegative compounds. Since $R_{s + ϵ_{n}} = R_{s + ϵ_{n}}^{+} - H_{n}$ has total coefficient one, finite-sum linearity gives $m (R_{s + ϵ_{n}}^{+}) - m (H_{n}) = 1$ ; with $m (e_{x}) = 1$ , the translated compounds have equal mass $1 + m (H_{n})$ . CH applies to this pair. If it gave $R_{s + ϵ_{n}}^{+} ⪯ e_{x} + H_{n}$ , signed translation invariance in [F2] would imply $R_{s + ϵ_{n}} ⪯_{s} e_{x}$ , contrary to $s + ϵ_{n} \notin L_{x}$ . Therefore it gives $e_{x} + H_{n} ⪯ R_{s + ϵ_{n}}^{+}$ , which by [F2] is $e_{x} ⪯_{s} R_{s + ϵ_{n}}$ . Add $ϵ_{n} e_{x_{0}}$ to both signed maps and use $R_{s + ϵ_{n}} + ϵ_{n} e_{x_{0}} = R_{s} + ϵ_{n} e_{x_{1}}$ . Then add $K = R_{s}^{-}$ to both; the resulting equal-mass compound comparison is $e_{x} + K + ϵ_{n} e_{x_{0}} ⪯ R_{s}^{+} + ϵ_{n} e_{x_{1}}$ . Stability with bases $e_{x} + K, R_{s}^{+}$ and positive unit-mass spectators $e_{x_{0}}, e_{x_{1}}$ gives $e_{x} + K ⪯ R_{s}^{+}$ , or $e_{x} ⪯_{s} R_{s}$ . Together with step 3.1 this proves $R_{s} \sim_{s} e_{x}$ and the claimed attainment.

Reference entropy supremum is finite and attained

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources