half-closed intervals and Lebesgue measures

Question

I am reading Bartle's book.

define $$K=\{ a \in \mathbb{Q}\,|\, 0 < a \le 1\}$$

and define $A$ by the family of all finite unions of half-closed intervals in the form of

$$\{a \in K\, |\, x < a \le y\}$$ in which $0 \le x \le y \le 1$ and $x, y \in K$.

1. Is $A$ an algebra of subsets of $K$ ? And how is that concluded?

2. It leads to another conclusion that every non-empty set in $A$ is infinite, can someone show me why?

3. Why does $\sigma(A)$ contain all subsets of $K$?

Answer 1

Notation: $(x,y]=\{z\in\Bbb Q\,:\,x< z\le y\}$

1. $A$ is clearly non-empty ( $\emptyset =(x,x]$ ) and closed under finite union. It's also closed under finite inteserction, by the identity $$\left(\bigcup_{i=1}^n L_i\right)\cap\left(\bigcup_{j=1}^m N_j\right)=\bigcup_{i,j} L_i\cap N_j$$ and that the intersection of two left-open, right-closed intervals is left-open and right-closed.

   Since $K\setminus(x,y]=(0,x]\cup (y,1]$, using De-Morgan's law and the previous observation you get closure under complementation.

2. Every non-empty set in $A$ must contain an interval in the form $(x,y]$ with $x<y$, which contains countably many elements.

3. Because $\Bbb Q$ is countable and $\sigma(A)$ contains each singleton: if $y>0$, then $$\{y\}=\bigcap_{0<x<y,\ x\in\Bbb Q } (x,y]$$