Why is intersection of all sets with indices from T an empty set? I am reading now "Topics in Algebra" of Herstein. Here is said that:
If S is the set of real numbers and T is the set of rational numbers, let for $\alpha \in T$, $A_\alpha=\{ x \in S | x \ge \alpha \}$. Here $\bigcap_{\alpha \in T} A_\alpha$ is the null(empty) set.But the sets $A_\alpha$ are not mutually disjoint.
Could you please explain two last sentences? Why the intersection is empty and at the same time they are not mutually disjoint? Or is it typo in the book?
 A: $A_\alpha=[\alpha,\infty)$
$\bigcap_{\alpha\in T}[\alpha,\infty)$ is empty since there is no $x\in\mathbb{R}$ that is in all the $[\alpha,\infty)$. Concrete example: $\pi$ is not in $[4,\infty)$.
For any $x\in\mathbb{R}$, take $\alpha>x$, where $\alpha\in\mathbb{Q}$. Then $x\notin[\alpha,\infty)$.
However, clearly the $A_\alpha$ are not mutually disjoint: $[1,\infty)$ and $[2,\infty)$ have common intersection $[2,\infty)$.
A: To see that the sets are not mutually disjoint: Pick $\alpha$, $\beta$, with $\alpha$ < $\beta$, then every $x \ge \beta$ is also $\ge \alpha$, hence is in both $A_{\alpha}$ and $A_{\beta}$; this generalizes to any finite number of $A$'s.
To see that the intersection of ALL is empty: For any $x$, there is a rational $\alpha \gt x$, hence NOT $x \in A_{\alpha}$, hence NOT in the intersection.  (intuitively: the more $A$'s you intersect, the 'farther to the right' the intersection result is; intersecting all of them 'moves' the result 'all the way past infinity' so nothing is left)
A: Perhaps you are making a mistaken analogy with taking intersections of two sets: Two sets are mutually disjoint if and only if their intersection is empty. The same is not true for intersections of more sets. A simple example would be intersecting the following three subsets of the real numbers:
$$\{0,1\},\qquad\{1,2\},\qquad\{0,2\}.$$
No two of them are mutually disjoint, but no number is contained in all three sets, so their intersection is empty. 
Something similar is going in in your question. For every pair of rational numbers $\alpha,\beta\in T$, the two sets $A_{\alpha}$ and $A_{\beta}$ are not disjoint. In fact if $\alpha\leq\beta$ then $A_{\beta}\subset A_{\alpha}$. But their intersection is empty; there is no real number $x\in S$ that is contained in $A_{\alpha}$ for every $\alpha\in T$. This would mean that $x>\alpha$ holds for every $\alpha\in T$, i.e. that $x$ is greater than every rational number, which is of course impossible.
A: The intersection is empty: Suppose not for a contradiction, then there is some  $x\in\bigcap_{\alpha\in T}A_\alpha $, i.e. $x\in A_\alpha $ for all $\alpha\in T $. But clearly $x <\lceil x\rceil +1$, so $x\notin A_{\lceil x\rceil +1} $.
The sets are not mutually disjoint: Take any two distinct $A_\alpha $, $A_\beta $ from the collection. Without loss of generality $\alpha>\beta $, then $\lceil\alpha\rceil\in A_\alpha $ and $\lceil\alpha\rceil\in A_\beta $ (notice that $A_\alpha\subset A_\beta $).
