A problem with transitivity of ordering of stationary sets In chapter 8 of Jech's Set Theory in the Third Millenium, he defines the following ordering on stationary sets: $S<T$ if the set $\{\alpha\in T:\alpha\cap S\text{ is stationary}\}$ is stationary. He goes on to state that 


*

*$S<\text{Tr}(S)$,

*$S<T$ and $T<Q$ implies that $S<Q$, and 

*that if $S\sim S',T\sim T'$ modulo $I_{NS}$ and $S<T$, then $S'<T'$. 


The first seems to be obvious, but I'm having trouble seeing why (2) and (3) are true.
 A: I think that you’ve misinterpreted Definition $\mathbf{8.18}$. I’ll quote it and the sentence that precedes it.

In the context of closed unbounded and stationary sets we use the phrase for almost all $\alpha\in S$ to mean that the set of all contrary $\alpha\in S$ is nonstationary.
Definition $\mathbf{8.18.}$ Let $S$ and $T$ be stationary subsets of $\kappa$.
$$S<T\qquad\text{if and only if}\qquad S\cap\alpha\text{ is stationary for almost all }\alpha\in T\;.$$

The given set of $\alpha\in T$ is $\{\alpha\in T:S\cap\alpha\text{ is stationary}\}$. To say that almost all $\alpha\in T$ are in this set is to say that set of $\alpha\in T$ that are not in this set — i.e., the set of contrary $\alpha\in T$ — is nonstationary. In other words, $S<T$ iff $S\cap\alpha$ is stationary for almost all $\alpha\in T$ iff the set $\{\alpha\in T:S\cap\alpha\text{ is not stationary}\}$ is nonstationary.
Think of non-stationary sets as small, their complements as large, and stationary sets as not small (but not necessarily large). When we say that $\varphi(\alpha)$ holds for almost all $\alpha\in S$, we mean that the set of $\alpha\in S$ for which $\varphi(\alpha)$ fails is small (is non-stationary); this is stronger than saying that the set of $\alpha\in S$ for which $\varphi(\alpha)$ holds is not small (is stationary).
