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Source: Imre Leader, Paul A. Russell and Mark Walters, Transitive sets and cyclic quadrilaterals, Journal of Combinatorics 2 (2011), no. 3, 457--462: the unnumbered consequence of Theorem 1 on p. 458. The edition read is identified on the source card.

The paper states that almost every cyclic quadrilateral does not embed into a transitive set, calls the deduction from Theorem 1 routine, and supports it only by noting that many parameter pairs occur, for example every pair with $\alpha$ and $\beta$ sufficiently close to $1$. It names no measure. The measure below and the whole proof are supplied here.

Statement

As made precise here: choose an ordered quadruple of pairwise distinct points of the unit circle with respect to product arc-length measure. For almost every such cyclic quadrilateral, there is no embedding into a finite transitive set in any Euclidean dimension.

Equivalently, the exceptional set has measure zero in every circular-order chamber. The same conclusion holds for any measure on this configuration space that is absolutely continuous in angular coordinates.

Full proof

Let $\mathcal C$ be the open subset of $(S^1)^4$ consisting of ordered quadruples $(x,y,z,w)$ of distinct points. The three points $x,y,z$ are noncollinear, so the parameters are unique and equal to

$$\alpha=\frac{\det(w-z,y-z)}{\det(x-z,y-z)}, \qquad \beta=\frac{\det(x-z,w-z)}{\det(x-z,y-z)}. \tag{1}$$

Thus the parameter map $\Phi:\mathcal C\to\mathbb R^2$ is real analytic.

We check that $\Phi$ has rank two somewhere in every circular-order chamber. Within any such chamber, choose a configuration with

$$z=(-1,0),\qquad y=(1,0),\qquad x=(\cos u,\sin u),\qquad w=(\cos t,\sin t).$$

The points $x,w$ can be placed in the required two semicircles, and in either order when they share a semicircle, so all six circular orders occur. Choose them generically so that

$$\sin u\ne0,\qquad \sin t\ne0,\qquad \sin(t-u)\ne0.$$

Formula (1) becomes

$$\alpha=\frac{\sin t}{\sin u}, \qquad \beta=\frac{1+\cos t-\alpha(1+\cos u)}2.$$

Direct differentiation gives

$$\det\frac{\partial(\alpha,\beta)}{\partial(u,t)} =\frac{\sin t\,\sin(t-u)}{2\sin^2u}\ne0. \tag{2}$$

Hence the parameter map has a locally open image in every chamber.

For a nonzero polynomial $F\in\mathbb Q[X,Y]$, the analytic function $F(\alpha,\beta)$ is not identically zero on any chamber: otherwise its local open image from (2) would force the polynomial $F$ to vanish on an open subset of $\mathbb R^2$. The zero set of a nonzero real-analytic function on a connected real-analytic manifold has measure zero. Therefore each set

$$\{(x,y,z,w)\in\mathcal C:F(\alpha,\beta)=0\}$$

has measure zero, as does the set $\alpha=1$.

There are only countably many polynomials in $\mathbb Q[X,Y]$. If $\beta$ is algebraic over $\mathbb Q(\alpha)$, clearing the denominators of a polynomial relation over $\mathbb Q(\alpha)$ gives some nonzero $F\in\mathbb Q[X,Y]$ with $F(\alpha,\beta)=0$. Thus the configurations for which $\alpha=1$ or $\beta$ is algebraic over $\mathbb Q(\alpha)$ lie in a countable union of null sets. Outside this union, Theorem 1 rules out an embedding into a finite transitive set.

The only external analytic facts used in this expansion are the local submersion theorem and the measure-zero theorem for the zero set of a nonzero real-analytic function.

Depends on. Theorem 1.

Bears on. Problem 174: the statement shows that spherical four-point sets which are not subtransitive are typical among cyclic quadrilaterals, not exceptional. It says nothing about which of them are Ramsey.