Show that property b_(ijm) always follows from properties…
Show that property $b_{ijm}$ always follows from properties $b_{ijk}$ and $b_{jkm}$ (see Problem 64).
Explanation of Method
Everything can be checked outcome by outcome. For each outcome $\omega$ write $x_t=1$ if $\omega\in A_t$ and $x_t=0$ otherwise. Property $b_{ijk}$ says: there is no outcome with $x_i=x_k\ne x_j$ ("the middle one is the odd one out"). We test the claim on all possible patterns $(x_i,x_j,x_k,x_m)$ and find a pattern that satisfies $b_{ijk}$ and $b_{jkm}$ but violates $b_{ijm}$, so the statement as written is false. We then prove the correct rule (Menger's transitivity of betweenness): $b_{ijk}$ and $b_{ikm}$ imply $b_{ijm}$ and $b_{jkm}$.
Step 1: Property $b$ outcome by outcome
With the definition of Problem 64, $b_{ijk}$ means that the atoms $A_i\overline{A_j}A_k$ and $\overline{A_i}A_j\overline{A_k}$ are empty, i.e.
\[ b_{ijk}: \quad \text{no outcome has } x_i = x_k \ne x_j. \]
Think of $A_j$ as lying "between" $A_i$ and $A_k$: where $A_i$ and $A_k$ agree, $A_j$ agrees with them.
Step 2: A counterexample to the statement as written
Let $\Omega=\{\omega_1,\omega_2\}$ with $P(\omega_1)=P(\omega_2)=\tfrac12$ and
\[ A_i=\{\omega_1\},\qquad A_j=\varnothing,\qquad A_k=\{\omega_2\},\qquad A_m=\Omega. \]
| Outcome | $x_i$ | $x_j$ | $x_k$ | $x_m$ | $b_{ijk}$ ok? | $b_{jkm}$ ok? | $b_{ijm}$ ok? |
|---|---|---|---|---|---|---|---|
| $\omega_1$ | 1 | 0 | 0 | 1 | yes ($x_i\ne x_k$) | yes ($x_j\ne x_m$) | no ($x_i=x_m=1\ne x_j$) |
| $\omega_2$ | 0 | 0 | 1 | 1 | yes | yes | yes |
So $b_{ijk}$ and $b_{jkm}$ hold, but $b_{ijm}$ fails. In distances: $\varrho(A_i,A_j)+\varrho(A_j,A_k)=\tfrac12+\tfrac12=1=\varrho(A_i,A_k)$ and $\varrho(A_j,A_k)+\varrho(A_k,A_m)=\tfrac12+\tfrac12=1=\varrho(A_j,A_m)$, yet $\varrho(A_i,A_j)+\varrho(A_j,A_m)=\tfrac32\ne\tfrac12=\varrho(A_i,A_m)$.
The same happens for points on a line only when $j$ and $k$ coincide; for events, $A_j$ and $A_k$ can "coincide" at some outcomes (here $\omega_1$) while differing elsewhere. Hence the claim cannot be proved.
Step 3: The correct rule
\[ \boxed{b_{ijk}\ \text{and}\ b_{ikm}\ \Longrightarrow\ b_{ijm}\ \text{and}\ b_{jkm}.} \]
(For points on a line: if $j$ is between $i$ and $k$, and $k$ is between $i$ and $m$, then $j$ is between $i$ and $m$, and $k$ is between $j$ and $m$.)
Step 4: Proof of $b_{ijm}$
Suppose some outcome had $x_i=x_m=a$ and $x_j=1-a$.
- By $b_{ikm}$ (no $x_i=x_m\ne x_k$), we must have $x_k=a$.
- Then $x_i=x_k=a\ne x_j$, which contradicts $b_{ijk}$.
So no such outcome exists: $b_{ijm}$ holds.
Step 5: Proof of $b_{jkm}$
Suppose some outcome had $x_j=x_m=c$ and $x_k=1-c$.
- By $b_{ikm}$, $x_i=x_m\ne x_k$ is forbidden; since $x_m=c\ne x_k$, we need $x_i\ne c$, i.e. $x_i=1-c=x_k$.
- Then $x_i=x_k=1-c\ne c=x_j$, contradicting $b_{ijk}$.
So $b_{jkm}$ holds. $\blacksquare$
Step 6: When does the original claim hold?
The claim "$b_{ijk}$ and $b_{jkm}\Rightarrow b_{ijm}$" is true inside a linearly ordered sequence (Problems 66–68), where each outcome's pattern $x_1,x_2,\dots,x_n$ changes value at most once. It fails only because of patterns like $1,0,0,1$, which a linearly ordered sequence cannot contain.
Result
| Claim | Status |
|---|---|
| $b_{ijk}\wedge b_{jkm}\Rightarrow b_{ijm}$ | false in general: $A_i=\{\omega_1\}, A_j=\varnothing, A_k=\{\omega_2\}, A_m=\Omega$ |
| $b_{ijk}\wedge b_{ikm}\Rightarrow b_{ijm}\wedge b_{jkm}$ | true (transitivity of betweenness) |