Exercise 8.1.11

Proof. First let

so that we must show that $L=R$.

$\text{(⊆)}$ Consider any $x\in L$. For any $t\in T$, define the set $S_t=\left\{s\in S\mid x\in A_{t,s}\right\}$. Since $x\in L$ we have that $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{s\in S}{A}_{t,s}$ for all $t\in T$. Hence, for all $t\in T$, there is an $s\in S$ such that $x\in A_{t,s}$. From this it follows that $S_t\ne \varnothing$ for any $t\in T$. Now, by the Axiom of Choice, the system of sets $\left\{S_t\mid t\in T\right\}$ has a choice function $g$. We then define a function $f(t)=g(S_t)$ for any $t\in T$, noting that this is defined for all $t\in T$ since $S_t$ is nonempty. Clearly then we have, for any such $t\in T$, that $f(t)=g(S_t)\in S_t\subseteq S$ so that $f(t)\in S$. Hence $f$ is a function from $T$ into $S$ so that $f\in S^T$.

Now consider any specific $t\in T$ so that $f(t)\in S_t$ and hence $x\in A_{t,f(t)}$ by the definition of $S_t$. Thus since $t$ was arbitrary this shows that $x\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{t\in T}{A}_{t,f(t)}$. Moreover, since $f\in S^T$ we clearly have that $x\in =R$. Thus, since $x$ was arbitrary, this shows that $L\subseteq R$.

$\text{(⊇)}$ Now consider any $x\in R$ so that there is an $f\in S^T$ such that $x\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{t\in T}{A}_{t,f(t)}$. Thus we have that $x\in A_{t,f(t)}$ for all $t\in T$. So consider any $t\in T$ and let $s=f(t)$, noting that $s=f(t)\in S$ since $f\in S^T$. Thus we have that $x\in A_{t,f(t)}=A_{t,s}$. Since we have shown that there is an $s\in S$ such that $x\in A_{t,s}$, it follows that $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{s\in S}{A}_{t,s}$. Since $t\in T$ was arbitrary we have $x\in =L$. Hence $R\subseteq L$ since $x$ was arbitrary. □

Proof. First let

so that we must show that $L=R$.

$\text{(⊆)}$ Consider any $x\in L$ so that there is a $t\in T$ such that $x\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{s\in S}{A}_{t,s}$. Hence we have that $x\in A_{t,s}$ for all $s\in S$. Now consider any $f\in S^T$. Then we have that $f(t)\in S$ so that $x\in A_{t,f(t)}$. Therefore there is a $t\in T$ such that $x\in A_{t,f(t)}$ so that $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{t\in T}{A}_{t,f(t)}$. Moreover, since $f$ was arbitrary, we have that $x\in =R$. This shows that $L\subseteq R$ since $x$ was arbitrary.

$\text{(⊇)}$ We show this by contrapositive. So suppose that $x\notin L$. Hence we have

Now, let $S_t=\left\{s\in S\mid x\notin A_{t,s}\right\}$ for each $t\in T$, noting that $S_t\ne \varnothing$ by the above for any $t\in T$. Then, by the Axiom of Choice, the system of sets $\left\{S_t\mid t\in T\right\}$ has a choice function $g$. We then set $f(t)=g(S_t)$ for any $t\in T$. Then clearly $f(t)=g(S_t)\in S_t\subseteq S$ for any $t\in T$ so that $f(t)\in S$ since $g$ is a choice function. It then follows that $f$ is a function from $T$ into $S$ so that $f\in S^T$. Now consider any $t\in T$ so that $f(t)=g(S_t)\in S_t$. Then, by the definition of $S_t$, we have that $x\notin A_{t,f(t)}$. Since $t\in T$ was arbitrary, we have thus shown that

Since $x$ was arbitrary, this shows by contrapositive that $x\in R$ implies $x\in L$ so that $R\subseteq L$. □