Prove that if R^*=(R_1,...,R_n) is a sequence of sets such…

Prove that if $R^*=(R_1,\dots,R_n)$ is a sequence of sets such that property $b_{ijk}$ holds for $R_i,R_j,R_k$ for every $i\lt j\lt k$, then $R^*$ is a linearly ordered sequence of sets (see Problem 66).

Step 1: Each membership sequence is monotone

For an outcome $\omega$ let $x_t=1$ if $\omega\in R_t$, else $0$. Property $b_{ijk}$ for all $i\lt j\lt k$ says: there are no $i\lt j\lt k$ with $x_i=x_k\ne x_j$.

If the sequence $x_1,\dots,x_n$ changed value twice, say at positions after $i$ and after $j$, we would have $x_i\ne x_{i+1}=\dots=x_j\ne x_{j+1}$, i.e. $x_i = x_{j+1} \ne x_j$ with $i\lt j\lt j+1$ — forbidden. So each sequence changes value at most once, i.e. it is monotone.

Step 2: Four types of outcomes

A monotone $0/1$ sequence is determined by its first and last values:

Step 3: Define $C$, $A_t$, $B_t$

\[ C = R_1R_n, \qquad A_t = \overline{R_1}\,R_t, \qquad B_t = \overline{R_n}\,R_{n-t+1} \qquad (t=1,\dots,n). \]

So $A_t$ collects the "switch on" outcomes that are already in $R_t$, and $B_t$ collects the "switch off" outcomes that are still in $R_{n-t+1}$.

Step 4: Check the conditions of Problem 66

• $A_t\subseteq A_{t+1}$: if $\omega\in A_t$, then $x_1=0$ and $x_t=1$; a monotone sequence that went from $0$ to $1$ stays $1$, so $x_{t+1}=1$ and $\omega\in A_{t+1}$.
• $B_t\subseteq B_{t+1}$: if $\omega\in B_t$, then $x_n=0$ and $x_{n-t+1}=1$; a sequence ending in $0$ that is $1$ at position $n-t+1$ is decreasing, so it is also $1$ at the earlier position $n-t$: $\omega\in B_{t+1}$.
• Disjointness: $A_n=\overline{R_1}R_n$, $B_n=\overline{R_n}R_1$ and $C=R_1R_n$ are three different rows of the table in Step 2, so they are pairwise disjoint.

Step 5: Check $R_i = A_i + B_{n-i+1} + C$

Note $B_{n-i+1}=\overline{R_n}R_i$. Take any outcome and use its type:

In every row, "$\omega\in R_i$" agrees with "$\omega$ is in $A_i$ or $B_{n-i+1}$ or $C$". Hence

\[ \boxed{R_i = A_i + B_{n-i+1} + C \quad (i=1,\dots,n),} \]

with $A^*$, $B^*$ increasing and $A_n,B_n,C$ pairwise disjoint: $R^*$ is linearly ordered. $\blacksquare$

Step 6: Summary of Problems 64–68

For a sequence of events $R_1,\dots,R_n$ the following are equivalent:

1. $R^*$ is linearly ordered (Problem 66 definition);
2. property $b_{ijk}$ holds for all $i\lt j\lt k$ (Problems 66 and 68);

and they imply that the distances add up like points on a line (Problem 67); the converse also holds when only the impossible event has probability $0$ (Problem 64).