Exercise 8.2.4

Proof. For any $x$ we have

Proof. Consider any $x\in S-B$ so that $x\in S$ and $x\notin B$. Then it has to be that also $x\notin A$ since otherwise it would not be that $A\subseteq B$. Hence $x\in S-A$, which shows the result since $x$ was arbitrary. □

Proof. We must show that the above definition of $$ satisfies the three parts of the definition of a $\sigma$-algebra:

For (a) clearly $\left|\text{∅}\right|=0\le {\aleph }_0$ so that $\varnothing \in$, and $S-S=\varnothing$ so that $\left|S-S\right|=\left|\text{∅}\right|=0\le {\aleph }_0$. Hence $S\in$ as well.

For (b) suppose that $X\subseteq S$ and $X\in$. Then either $\left|X\right|\le {\aleph }_0$ or $\left|S-X\right|\le {\aleph }_0$. If $\left|X\right|\le {\aleph }_0$ then by Lemma 1 we have

since $X\subseteq S$. Therefore $\left|S-(S-X)\right|=\left|X\right|\le {\aleph }_0$ so that $S-X\in$. On the other hand, if $\left|S-X\right|\le {\aleph }_0$, then obviously $S-X\in$ by definition.

Lastly, regarding (c), suppose that ${\left\{X_n\right\}}_{n=0}^{\infty }$ is a system of sets where each $X_n$ is in $$. Thus $\left|X_n\right|\le {\aleph }_0$ or $\left|S-X_n\right|\le {\aleph }_0$ for each natural $n$.

Now we show that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\in$. First, if $\left|X_n\right|\le {\aleph }_0$ for all natural $n$, then it follows from Theorem 8.1.7 that $\left|{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|\le {\aleph }_0$, which of course uses the Axiom of Choice. If, on the other hand, there is a natural $m$ such that $\left|X_m\right|\nleqq {\aleph }_0$, then it has to be that $\left|S-X_m\right|\le {\aleph }_0$ since $X_m\in$. Then, since clearly $X_m\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n$, it follows from Lemma 2 that $S-{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\subseteq S-X_m$ so that clearly $\left|S-{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|\le \left|S-X_m\right|\le {\aleph }_0$. Thus in all cases we have that either $\left|{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|\le {\aleph }_0$ or $\left|S-{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|\le {\aleph }_0$ so that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\in$.

Lastly we show that ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\in$ as well. If it is the case $\left|S-X_n\right|\le {\aleph }_0$ for all natural $n$, then clearly we have that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }(S-X_n)\le {\aleph }_0$, again by Theorem 8.1.7. It also follows from Exercise 2.3.11 that $S-{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n={\bigcup }_{n=0}^{\infty }(S-X_n)$ so that we have $\left|S-{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|=\left|{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }(S-X_n)\right|\le {\aleph }_0$. Now, on the other hand, if there is a natural $m$ such that $\left|S-X_m\right|\nleqq {\aleph }_0$, then it has to be that $\left|X_m\right|\le {\aleph }_0$ since $X_m\in$. Since clearly ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\subseteq X_m$, we then have $\left|{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|\le \left|X_m\right|\le {\aleph }_0$. Hence in either case we have that $\left|{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|\le {\aleph }_0$ or $\left|S-{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right|\le {\aleph }_0$ so that ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\in$ by definition.

We have therefore shown parts (a), (b), and (c) so that $$ is a $\sigma$-algebra as desired. □