Below is a statement from the book 'Topics in Algebra' by I.N.Herstein.
For instance, if $S$ is the set of real numbers, and if $T$ is the set of rational numbers, let, for $\alpha \in T$, $A_\alpha =\{x\in S | x\ge\alpha\}$. It is an easy exercise to see that $\cup_{\alpha\in T}A_\alpha= S$ whereas $\cap_{\alpha\in T}A_\alpha $ is the null set. The sets $A_\alpha$ are not mutually disjoint.
My question is, if $\cap_{\alpha\in T}A_\alpha $ is a null set, then $A_\alpha$'s are mutually disjoint sets. But it is stated that it is not a mutually disjoint sets. This may be due to my lack of understanding the statement properly. Can someone explain it?
I will expand here what i have written in the comment. Let $A_\alpha=\{ x \in \mathbb{R} : x \ge \alpha\}$ and consider $$ \bigcap_{\alpha \in \mathbb{Q}} A_\alpha $$ suppose $ x \in \bigcap_{\alpha \in \mathbb{Q}} A_\alpha$ So $ x \in A_\alpha$ for any $\alpha \in \mathbb{Q}$, but this implies that $$ x \ge \alpha $$ for all the rational numbers. This is absurd, so $$\bigcap_{\alpha \in \mathbb{Q}} A_\alpha= \emptyset$$ but, for any $\alpha \le \beta$ you have $$ A_\alpha \cap A_\beta = A_\beta \ne \emptyset $$ The reason this is possible is that $\bigcap_n A_n = \emptyset$ does not imply that there exists $n,m$ such that $A_n \cap A_m = \emptyset$ (think e.g. to the sets $\{1,2\},\{2,3\}, \{1,3\}$)
Moreover being mutually disjoint means that for all $n \ne m$ $A_n \cap A_m =\emptyset$ that is a stronger property and, in general it is not true (unless of course there are only 2 sets)