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Spectator cancellation and signed-order algebra

Statement

Under Comparison and stability for compounds, for compounds A,B with m(A)=m(B) and any compound C, A+C⪯B+C if and only if A⪯B. The signed relation of Signed comparison of finite coefficient maps is well-defined; on nonnegative equal-mass maps it agrees with ⪯, is invariant under adding the same finite-support signed map to both sides, and is transitive and closed under adding two comparisons.

Facts & Assumptions

[F1]

Tagged scalable compounds and Compound mass is well-defined and additive: compounds are finite-support nonnegative maps; their masses are finite sums, additive under composition and homogeneous under positive scaling.

[A1]

Comparison and stability for compounds: ⪯ is a preorder, composition-consistent, positively scale-invariant, admits splitting/recombination, satisfies the stated positive-spectator stability axiom, and has CH.

[F2]

Signed comparison of finite coefficient maps: for equal-total finite signed maps u,v, the signed comparison is defined by comparing (v−u)− with (v−u)+.

[F3]

The canonical natural ι(n)=n⋅1F of a field and Canonical naturals are positive and strictly increasing give the canonical embedding ιN:N→R, whose successor values are positive and are finite sums of 1. Reciprocals of the positive integers decrease to zero gives the real sequence ϵk=1/ιN(σ(k))>0, k∈N, with ϵk→0.

Proof

Given: [F1], [A1], [F2], and [F3].

1.1F1F2

For equal-total u,v, d=v−u has finite support and total sum zero. Pointwise d=d+−d−, so finite-sum linearity gives m(d+)−m(d−)=0; both maps are finite-support and nonnegative. Thus the definition in [F2] compares equal-mass compounds. If u,v are nonnegative, set k(x)=min⁡{u(x),v(x)}; then u=k+(v−u)− and v=k+(v−u)+.

1.2A1F1F3algebrainduction

Suppose A+C⪯B+C and m(C)>0. For each k∈N, set n=σ(k) and r=ιN(n)>0, and write A/r=(1/r)A. The embedding's successor rule and finite induction show that the sum of n copies of A/r is r(A/r)=A, and likewise for B. Scaling the given comparison by 1/r gives A/r+C/r⪯B/r+C/r. By composition consistency, replace the n copies of A/r one at a time by B/r, keeping the already replaced copies, the remaining copies, and C/r as spectators; after these finitely many replacements, A+ϵkC⪯B+ϵkC. This holds for every natural index k. By [F3], ϵk>0 and ϵk→0, so stability with Z0=Z1=C gives A⪯B. If m(C)=0, nonnegativity implies C=0 and the conclusion is the premise itself. Conversely, A⪯B gives A+C⪯B+C by consistency.

2.1F2step 1.1step 1.2

For nonnegative u,v, step 1.1 gives u=k+(v−u)− and v=k+(v−u)+. Cancellation from step 1.2 shows u⪯v exactly when (v−u)−⪯(v−u)+, which is u⪯sv. For general signed maps, adding a common map leaves v−u unchanged, so it leaves signed comparison unchanged.

3.1A1F2step 1.2∎

If u⪯sv and v⪯sw, consistency gives N⪯P, where N:=(v−u)−+(w−v)− and P:=(v−u)++(w−v)+. Pointwise P−N=w−u. Set K(x):=min⁡{P(x),N(x)}. If P(x)≥N(x) then P(x)−K(x)=(w−u)+(x) and N(x)−K(x)=(w−u)−(x); if P(x)<N(x) the same identities follow with the two sides exchanged. Thus P=K+(w−u)+ and N=K+(w−u)−, with all four maps finite-support and nonnegative. Cancellation from step 1.2 gives (w−u)−⪯(w−u)+, hence u⪯sw. For two independent signed comparisons, the same pointwise decomposition of the sums of their positive and negative parts proves closure under adding comparisons. Reflexivity follows because 0⪯0.

Depends on

Dependency tree · 2 levels

Sources