Exercise 1.4.5

Proof. $\text{(⊆)}$ Consider any $x\in A\cap \bigcup <!--\; FUNCTION\; APPLICATION\; -->S$ so that $x\in A$ and there exists an $X\in S$ such that $x\in X$. Set $Y=A\cap X$ so that $x\in Y$ since $x\in A$ and $x\in X$. Since clearly $Y=A\cap X\subseteq A$ so that $Y\in 𝒫\left(A\right)$, and $X\in S$, it follows that $Y\in T_1$. Hence, $x\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->T_1$ since $x\in Y$ and $Y\in T_1$. This shows that $A\cap \bigcup <!--\; FUNCTION\; APPLICATION\; -->S\subseteq \bigcup T_1$.

$\text{(⊇)}$ Now consider any $x\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->T_1$ so that there is a $Y\in T_1$ such that $x\in Y$. Since $Y\in T_1$, it must be that $Y=A\cap X$ for some $X\in S$. And since $x\in Y=A\cap X$, we have that $x\in A$ and $x\in X$. Then, because $X\in S$ and $x\in X$, it must be that $x\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->S$. Since also $x\in A$ we have that $x\in A\cap \bigcup <!--\; FUNCTION\; APPLICATION\; -->S$, which suffices to show that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->T_1\subseteq A\cap \bigcup S$ as well. □

Proof. $\text{(⊆)}$ Consider any $x\in A-\bigcup <!--\; FUNCTION\; APPLICATION\; -->S$ so that $x\in A$ but $x\notin \bigcup <!--\; FUNCTION\; APPLICATION\; -->S$. It then has to be that $x\notin X$ for every $X\in S$ since otherwise $x$would be a member of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->S$. Now consider any $Y\in T_2$ so that $Y=A-X$ for some $X\in S$. Since $X\in S$ we know that $x\notin X$ from what has already been established. Since also $x\in A$, it follows that $x\in A-X=Y$. Then, since $Y$ was an arbitrary member of $T_2$, it follows that $x\in \bigcap <!--\; FUNCTION\; APPLICATION\; -->T_2$, which proves that $A-\bigcup <!--\; FUNCTION\; APPLICATION\; -->S\subseteq \bigcap T_2$.

$\text{(⊇)}$ Next, consider any $x\in \bigcap <!--\; FUNCTION\; APPLICATION\; -->T_2$ so that $x\in Y$ for every $Y\in T_2$. To digress for a moment, consider any $X\in S$ and set $Y=A-X$. Since any $y\in Y$ is also in $A$ we have $Y\subseteq A$ and thus $Y\in 𝒫\left(A\right)$. Since also of course $Y=A-X$ where $X\in S$, we have that $Y\in T_2$ by definition so that $x\in Y$ because $x\in \bigcap <!--\; FUNCTION\; APPLICATION\; -->T_2$. Hence, $x\in A$ but not in $X$ since $Y=A-X$. Then, since $X\in S$ was arbitrary and $x\notin X$, it follows that $x\notin \bigcup <!--\; FUNCTION\; APPLICATION\; -->S$. As it has been established that also $x\in A$, we have that $x\in A-\bigcup <!--\; FUNCTION\; APPLICATION\; -->S$. This suffices to show that $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T_2\subseteq A-\bigcup S$ since $x$ was arbitrary. □

Proof. First we note that $S$ is not empty so that $\bigcap <!--\; FUNCTION\; APPLICATION\; -->S$ exists (refer to the next exercise).

$\text{(⊆)}$ First, suppose any $x\in A-\bigcap <!--\; FUNCTION\; APPLICATION\; -->S$, implying that $x\in A$ but $x\notin \bigcap <!--\; FUNCTION\; APPLICATION\; -->S$. It then has to be that there is an $X\in S$ such that $x\notin X$ since otherwise $x$ would be in $\bigcap <!--\; FUNCTION\; APPLICATION\; -->S$. Set $Y=A-X$, noting that $x\in Y$ since $x\in A$ but $x\notin X$. As before, clearly $Y\in 𝒫\left(A\right)$. Since also $Y=A-X$ and $X\in S$, it is necessary that $Y\in T_2$. From this it follows that $x\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->T_2$ since we have already shown the $x\in Y$. Therefore, $A-\bigcap <!--\; FUNCTION\; APPLICATION\; -->S\subseteq \bigcup T_2$ since $x$ was arbitrary.

$\text{(⊇)}$ Now consider any $x\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->T_2$ so that there is a $Y\in T_2$ such that $x\in Y$. Then, by the definition of $T_2$, $Y=A-X$ for some $X\in S$. Then, since $x\in Y=A-X$, we have that $x\in A$ and $x\notin X$. Since $X\in S$ and $x\notin X$, it cannot be that $x$ is a member of $\bigcap <!--\; FUNCTION\; APPLICATION\; -->S$. Since also $x\in A$, clearly $x\in A-\bigcap <!--\; FUNCTION\; APPLICATION\; -->S$, and hence $\bigcup <!--\; FUNCTION\; APPLICATION\; -->T_2\subseteq A-\bigcap S$ as desired. □