Reference-strip gluing proves comparison for scaled copies

By T4, the open strips $\mathrm{\Sigma }(a,b):=\{x:a\prec \prec x\prec \prec b\}$ with $a\prec \prec b$ and $a{\sim }_{T}b$ cover ${\mathrm{\Gamma }}_i$. They are open: by [F2], $a\prec \prec x$ is $x\in \mathrm{int}{A}_a$, and $x\prec \prec b$ is $x\mathrm{\notin }{A}_b$; the latter set is open because $A_b$ is relatively closed.

Let $\mathrm{\Sigma }(a,b)$ and $\mathrm{\Sigma }(c,d)$ overlap. Their intersection is open. Choose two states $e\prec \prec f$ in a small vertical segment contained in the overlap, using [F4]. Let $s_e,s_f$ be their coordinates in the first strip and $t_e,t_f$ in the second, supplied by [F1]. If $s_f<s_e$, the signed difference $R_{s_e}-R_{s_f}=(s_e-s_f)(\mathsf{J}(b)-\mathsf{J}(a))$ and positive scaling and signed compatibility [A1, F3] give $R_{s_f}\prec R_{s_e}$, so $f\prec e$ by the two reference equivalences; if $s_f=s_e$, those references coincide and give $f\sim e$. Both contradict $e\prec \prec f$. Hence $s_f>s_e$, and the same argument gives $t_f>t_e$.

The four reference equivalences give, in signed comparison, $f-e{\sim }_s(s_f-s_e)(b-a),f-e{\sim }_s(t_f-t_e)(d-c)\mathrm{.}$ Put $r=(s_f-s_e)\mathrm{/}(t_f-t_e)>0$. Additivity and positive scaling in [F3] give $d-c{\sim }_{s}r(b-a)$. Applying the same algebra to the equivalences for $e$ gives $c{\sim }_{s}a+(s_e-r{t}_e)(b-a)$. Consequently a state in the second strip with coordinate $t$ has first-chart coordinate $s_e+r(t-t_e)$. The change of chart is affine with strictly positive slope.

The same calculation covers a finite chain of overlapping strips: after each overlap, express the new strip's endpoint states as signed-equivalent affine maps in the existing reference pair. The local strip lemma gives every state $z$ of the added strip a coordinate $s_z\in \mathbb{R}$ with $\mathsf{J}(z){\sim }_s(1-s_z)\mathsf{J}(a)+s_z\mathsf{J}(b)$ in the existing chart. Signed addition, positive scaling, translation, and equivalence under negative scaling are valid by [F3]; hence induction along a finite overlap chain gives one such common chart for every state in the union.

Take any finite list of target states in ${\mathrm{\Gamma }}_i$ and join them in order by finitely many straight segments, which lie in ${\mathrm{\Gamma }}_i$ by convexity. The resulting connected path is a finite union of closed bounded segments, hence compact by [F5]. The open strips from step 1.1 cover it, so compactness gives a finite subcover. The intersection graph of those strips meeting the path is connected: otherwise their unions would partition the connected path into two disjoint nonempty relatively open sets. Order a spanning tree from one root strip; each later strip overlaps an earlier one in an open set and can be glued by step 3.1. Thus every target state has a common signed chart $\mathsf{J}(z){\sim }_s(1-s_z)\mathsf{J}(a)+s_z\mathsf{J}(b)$ with $a\prec \prec b$.

Write the arbitrary scaled-copy compounds as $A={\sum }_j{\alpha }_j{\mathsf{J}}_{r_j{\mathrm{\Gamma }}_i}(r_j{x}_j),B={\sum }_k{\beta }_k{\mathsf{J}}_{h_k{\mathrm{\Gamma }}_i}(h_k{y}_k),$ with $r_j,h_k>0$ and nonnegative coefficients, omitting zero terms. The declared reversible transport in [A1] gives ${\mathsf{J}}_{r_j{\mathrm{\Gamma }}_i}(r_j{x}_j)\sim r_j{\mathsf{J}}_{{\mathrm{\Gamma }}_i}(x_j)$ and similarly on the other side; ordinary positive scaling and composition therefore reduce them to the unscaled base-type states with effective amounts $a_j={\alpha }_j{r}_j$ and $b_k={\beta }_k{h}_k$. Their masses are respectively $r_j{w}_{{\mathrm{\Gamma }}_i}$ and $h_k{w}_{{\mathrm{\Gamma }}_i}$ before this transport, so equal total mass $M$ implies $L=M\mathrm{/}{w}_{{\mathrm{\Gamma }}_i}={\sum }_j{a}_j={\sum }_k{b}_k\mathrm{.}$ If $M=0$, both maps are empty and reflexivity applies. Otherwise use step 4.1 for the finite list of base states $x_j,y_k$, with common-chart coordinates $s_j,t_k$. Signed addition and splitting [F3] make the transported compounds equivalent to $R_{q_A}=(L-q_A)\mathsf{J}(a)+q_A\mathsf{J}(b)$ and $R_{q_B}=(L-q_B)\mathsf{J}(a)+q_B\mathsf{J}(b)$, where $q_A={\sum }_j{a}_j{s}_j$ and $q_B={\sum }_k{b}_k{t}_k$. If $q_A<q_B$, their signed difference has positive part $(q_B-q_A)\mathsf{J}(b)$ and negative part $(q_B-q_A)\mathsf{J}(a)$, so $a\prec b$ and positive scaling compare them; equality gives reflexivity, and $q_B<q_A$ gives the reverse comparison. Transitivity transfers this order through the reversible transports to the original distinct scaled tags. Hence CH holds for every allowed scaled-copy pair, without identifying those tags or dropping their scale factors.