What does $\wedge$ mean for measurable partitions?

Question

Sorry for this silly question, but I'm reading a textbook on Ergodic theory by Cornfeld, Sinai & Fomin and suddenly this operation on measurable partitions appears, and I didn't find definition in this textbook. Note, that $\xi\vee\eta$ is also used in text for partition that consists of intersections of sets of partitions $\xi$ and $\eta$. So, what $\xi\wedge\eta$ could possibly mean? Thanks!

Answer 1

Recall that $\xi(x)$ is the element of the partition $\xi$ containing $x$. There is an order on partitions: $\xi \geq \eta$ if $\xi(x) \subset \eta(x)$ for all $x$, or for almost all $x$ if we are on a measure space.

Then $\xi \vee \eta$ is the maximum of the two partitions $\xi$ and $\eta$, it satisfies $\xi \vee \eta \geq \xi$, $\xi \vee \eta \geq \eta$ and it is the smallest partition satisfying these two inequalities. It is often defined by the equality $\xi \vee \eta (x) = \xi(x) \cap \eta(x)$. When the partitions are finite, $\xi = \{A_1,...,A_n\}$, $\eta = \{B_1,...,B_m\}$, $\xi \vee \eta = \{A_i \cap B_j\}_{i,j}$.

The partition $\xi \wedge \eta$ is the minimum of the two partitions. it satisfies $\xi \wedge \eta \leq \xi$, $\xi \wedge \eta \leq \eta$ and it is the largest partition satisfying these two inequalities. A set $A$ is a union of elements of $\xi \wedge \eta$ if it satisfies $A = \bigcup_{x\in A} \xi(x) \bigcup_{y\in A} \eta(y)$ i.e. it is saturated both by $\xi$ and $\eta$.