Proof of Cyclically Monotone Subsets of the Doubled Real Line: Two-Point Monotonicity, Ordering of the Sections, and Countability of the Multi-Valued Abscissae

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here.

Step 1 (Claim 1). Let $z,z'\in\Gamma$ and write $x=\mathrm{pr}_{1}(z)$, $y=\mathrm{pr}_{2}(z)$, $x'=\mathrm{pr}_{1}(z')$, $y'=\mathrm{pr}_{2}(z')$. Apply Cyclically Monotone Subset of a Doubled Euclidean Space (i) with $N=2$, $z_{1}=z$ and $z_{2}=z'$, so that $x_{1}=x$, $y_{1}=y$, $x_{2}=x'$, $y_{2}=y'$ and $x_{3}=x_{1}=x$. Since $u\cdot v=uv$ in dimension one by claim 1 of One-Dimensional Test Functions: Scalars, Derivatives, and the Difference Quotient of the Derivative, the defining inequality reads

$$y(x'-x)+y'(x-x')\le0 .$$

Since $x-x'=-(x'-x)$ by claim 6 of Additive Cancellation and Elementary Additive Identities in a Field, claim 2 of Zero Products and Elementary Identities in a Field and axiom 8 of Field give $y'(x-x')=-y'(x'-x)$, so the left-hand side equals $(y-y')(x'-x)$, again by claim 2 of that lemma and the distributive law of Field. Hence $(y-y')(x'-x)\le0$, and multiplying by $-1$, which reverses the inequality by claim 4 of Elementary Order Arithmetic in an Ordered Field, gives

$$0\le-(y-y')(x'-x)=(y-y')(x-x'),$$

the last identity by claim 6 of Additive Cancellation and Elementary Additive Identities in a Field, claim 2 of Zero Products and Elementary Identities in a Field and axiom 8 of Field. This is claim 1.

Step 2 (Claim 2). Let $x<x'$, $y\in\Gamma_{x}$ and $y'\in\Gamma_{x'}$. By the definition of the sections there are $z,z'\in\Gamma$ with $\mathrm{pr}_{1}(z)=x$, $\mathrm{pr}_{2}(z)=y$, $\mathrm{pr}_{1}(z')=x'$ and $\mathrm{pr}_{2}(z')=y'$, so claim 1 gives $0\le(y-y')(x-x')$. Suppose $y'<y$. By claim 3 of Elementary Arithmetic in an Ordered Field one has $0\le y-y'$, and $y-y'\ne0$ since $y'\ne y$, so $0<y-y'$; likewise $0\le x'-x$ with $x'-x\ne0$, so $0<x'-x$. Hence $0<(y-y')(x'-x)$ by claim 5 of Elementary Order Arithmetic in an Ordered Field. By claim 6 of Additive Cancellation and Elementary Additive Identities in a Field, claim 2 of Zero Products and Elementary Identities in a Field and axiom 8 of Field, $(y-y')(x-x')=-\bigl((y-y')(x'-x)\bigr)$, so $(y-y')(x-x')<0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, contradicting $0\le(y-y')(x-x')$. Hence $y'<y$ is impossible. Since $\le$ is a total order on $\mathbb{R}$ by Ordered Field, either $y\le y'$ or $y'\le y$; in the latter case $y=y'$, because $y'<y$ is excluded. In both cases $y\le y'$.

Step 3 (Claim 3). The set $\mathbb{Q}$ of rational numbers is countable by claim 2 of The Integers and the Rational Numbers are Countable, and it is nonempty: $0$ and $1$ lie in $\mathbb{Z}$, $1$ being the image of the natural number $1$ under the canonical map by claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, and $1^{-1}$ exists by axioms 6 and 7 of Field, as $1\ne0$; so $0=0\cdot1^{-1}\in\mathbb{Q}$ by claim 1 of Zero Products and Elementary Identities in a Field. Hence Countable Set itself supplies a sequence $(q_{k})_{k\in\mathbb{N}}$ whose set of terms is $\mathbb{Q}$. For $k\in\mathbb{N}$ put

$$M_{k}=\{x\in M_{\Gamma}:\ \text{there are }u,v\in\Gamma_{x}\text{ with }u<q_{k}<v\}.$$

Each $M_{k}$ has at most one element. Suppose $x,x'\in M_{k}$ with $x\ne x'$; since $\le$ is a total order on $\mathbb{R}$, either $x<x'$ or $x'<x$, and as the two play symmetric roles we may assume $x<x'$. Choose $u,v\in\Gamma_{x}$ with $u<q_{k}<v$ and $u',v'\in\Gamma_{x'}$ with $u'<q_{k}<v'$. By claim 2, applied to $x<x'$ with $v\in\Gamma_{x}$ and $u'\in\Gamma_{x'}$, one has $v\le u'$. Then $q_{k}<v\le u'<q_{k}$, so $q_{k}<q_{k}$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, contradicting the irreflexivity of the strict order. Hence $x=x'$.

$M_{\Gamma}$ is the union of the $M_{k}$. Each $M_{k}$ is contained in $M_{\Gamma}$. Conversely let $x\in M_{\Gamma}$, so $\Gamma_{x}$ has two distinct elements $y_{1},y_{2}$; since $\le$ is a total order on $\mathbb{R}$ and $y_{1}\ne y_{2}$, we may assume $y_{1}<y_{2}$. By claim 1 of The Rational Numbers are Dense in the Real Numbers there is a rational number $q$ with $y_{1}<q<y_{2}$, and $q=q_{k}$ for some $k\in\mathbb{N}$; then $x\in M_{k}$.

Therefore $M_{\Gamma}=\bigcup_{k\in\mathbb{N}}M_{k}$ is a countable union of sets each of which is countable, being finite with at most one element and hence countable by claim 2 of Basic Properties of Countable Sets; so $M_{\Gamma}$ is countable by A Countable Union of Countable Sets is Countable.

Finally, let $x\in\mathbb{R}\setminus M_{\Gamma}$. Then $\Gamma_{x}$ does not have two distinct elements, so it is empty or has exactly one element.