Increasing set or decreasing set I know this is probably trivial but I am confused on how we determine a set is increasing or decreasing. For example suppose we have a set $\{E_j\}_{1}^{\infty}\subset M$ where $M$ is a $\sigma$-algebra. Then supposedly $$\left\{\bigcap_{j=k}^{\infty}E_j\right\}_{k=1}^\infty$$ is an increasing sequence of sets and $$\left\{\bigcup_{j=k}^{\infty}E_j\right\}_{k=1}^\infty$$ is an decreasing sequence of sets
Why is that the case?
 A: A sequence $A_1 \supset A_2 \supset A_3 \supset \ldots $ is decreasing,
a sequence $A_1 \subset A_2 \subset A_3 \subset \ldots $ is increasing.
In this case, for $$B_k := \bigcap_{j=k}^\infty E_j$$
we have $B_k = B_{k+1} \cap E_k \subset B_{k+1}$, so the sequence $(B_k)_{k\in\mathbb{N}}$ is increasing.
Analogously, for $$C_k := \bigcup_{j=k}^\infty E_j$$
we get $C_k = C_{k+1} \cup E_k \supset C_{k+1}$, so the sequence $(C_k)$ is decreasing.
For another example, if we say $$B'_k := \bigcap_{j=1}^k E_j$$
we get a decreasing sequence $(B'_k)$ (unlike $(B_k)$), since $B'_{k+1} = B'_k \cap E_{k+1} \subset B'_k$.
A: The sequence index is $k$. So define $$I(k) = \bigcap_{j=k}^\infty E_j$$
Then $x \in I(k)$ iff for all $j \ge k$ we have $x \in E_j$. So it's "easier" to be in $I(k+1)$ because then you only have to be in $E_j$ for all $j \ge k+1$, which is fewer sets. And if $x$ is in $I(k)$, then it is certainly in $I(k+1)$ for that reason, hence the sequence is increasing: $$\forall k: I(k) \subseteq I(k+1)\text{.}$$
For the union $$U(k) = \bigcup_{j=k}^\infty E_j$$ the oppossite holds: if $x \in U(k+1)$ then $x \in E_j$ for some $j \ge k+1$. The $j$ that works is also $\ge k$ so $x \in U(k)$. So this sequence decreases:
$$ \forall k :U(k+1) \subseteq U(k)\text{.}$$
