Some questions about the Jech's book (Generalized De Morgan's law and distributive law ) I finished the first chapter of the book Introduction to Set Theory by Jech (I started to love). And I have questions of some exercises where I'm not totally sure if my attempt  was complete or even correct. The two last are where I have more doubts. 
Here we go: 
(A) Generalized distributive law: 
Let $S \not= \emptyset$ and $A$ be a set.  Set, $\;T_{\,1}:= \left\{\,y\in \wp (A): \exists x\in S\, (\, y = A\cap x \,)\,   \right\}$ and prove that: $\, A\cap \bigcup S =\bigcup T_{\,1}.$
Proof: 
($\,\Rightarrow\,$) If $\,z \in A\cap \bigcup S$, $z\in A$ and  $\,z\in \bigcup S$. For $\,z\in \bigcup S,\,$ that means $z \in x_{0}$ for some $\, x_{0}\in S.\, $ Then, $\,z\in A$ and $z \in x_{0}$, i.e.,  $\,z\in A\cap x_0\,$   (where $\,x_0 \in S\,$).
On the other hand, for $z$ be in the union of $T_{\,1}$, $\,z\in \bigcup T_{\,1}$, it is a sufficient condition that $z\in y_{0}$ for some $\,y_0 \in T_{\,1}.\,$ We define $y_{0} = A \cap x_{0}$. It follows immediately that $y_{0} \in T_{\,1}$ $($as $\,y_{0} \in \wp(A)$ and $\,x_0 \in S\,)$. So as $\,z\in A\cap x_0\,$ and we defined $\,y_{0} = A \cap x_{0}$, then $\,z\in y_{0}$ for some $\,y_0 \in T_{\,1}$, as desired. That is, $\, A\cap \bigcup S \subseteq \, \bigcup T_{\,1}$.
($\,\Leftarrow\,$) If  $z\in \bigcup T_{\,1},\, z\in y_{0}$ for some $y_{0} \in T_{\,1}.\,$ Then, by the definition of $T_{\,1},\,\,y_{0} = A\cap x_{0}\, $ for some $x_{0} \in S.\;$ It follows that, if $z\in y_{0}$, then $z\in x_{0}$ ( this is because $\,y_{0} \subseteq x_{0}\,).\,$  And as there a $x_{0} \in S\,$ for which $\,z\in x_{0},$ by the union axiom we can conclude that $\,z\in \bigcup S$. Hence, $z \in A$ and $\,z\in \bigcup S$, $\,z \in A\cap \bigcup S.\,$ That is, $\bigcup T_{\,1} \subseteq\, A\cap \bigcup S. $  $\;\Box$
(B) Generalized De Morgan's laws: 
Set,  $\;T_{\,2}:= \left\{\,y\in \wp (A): \exists x\in S\, (\, y = A - x \,)\,   \right\}\,$ and prove  that: ($\,i\,$) $A - \bigcup S = \bigcap T_{2}\,$ and  ($\,ii\,$) $\,A - \bigcap S = \bigcup T_{2}.$
Proof: 
($\,i\,$)
($\,\Rightarrow\,$) If $z \in A - \bigcup S,\, z\in A\, $ and $ z\notin \bigcup S$ ( which means that,  for each $x \in S,\, z \notin x$ ). For $z$ be a member of the right-hand side, $z \in \bigcap T_{2},\, $ it is necessary that: for every $y\in T_{2}\,$ ( which assume is nonempty ) $z \in y: = A-x.\,$ Then, $z \in A$ and $z \notin x$  as in our assumption has that properties, it follows that $z \in \bigcap T_{2},\, $ i.e., $  A - \bigcup S \subseteq \bigcap T_{2}. $
($\,\Leftarrow\,$) If  $z \in \bigcap T_{2},\, $ where assume that $T_{2}$ is nonempty. So, for each $y\in T_{2},\,z \in y: = A-x\,$. Therefore, $z \in A$ and $z\notin x\,$ and by definition of the set $ T_{2},\, x\in S $; which  holds for each $y \in T_{2}. $ For all $y, \,$ we have that $x\in S\,$ and $ z \notin x,\, $  $ z\notin \bigcup S\,$*??* (* **

How do we know that $S$ cannot have some element out of the elements
  that we use by the definition of the set $T_{2}$ which could be in? I
  don't know maybe I misunderstood this part

*)
Hence, $z\in A$ and $ z\notin \bigcup S,\,$ $z \in A - \bigcup S\, $, i.e, $\bigcap T_{2} \subseteq  A- \bigcup S.$
$(\, ii \,)$
$(\, \Rightarrow \,)$ If $z\in A - \bigcap S,\, z\in A\,$ and $z \notin \bigcap S.\,$ For  $z \notin \bigcap S\,$, means that there exist some $x\in S$ for which $z\notin x.$  Then, $z\in A- x_{0}\,$ for some $x_{0} \in S.\,$ We set, $\,y_{0}:= A- x_{0}.\,$ So, $\,y_{0} \in T_{2}\,$ because $\,y_{0} \in \wp(A)$ and $\, x_{0}\in S.\,$ As $\, y_{0} \in T_{2}\,$ and $\, z\in y_0,\,$ it follows that $z \in \bigcup T_{2},\,$ i.e., $A-\bigcap S \subseteq \,\bigcup T_{2}.\, $
($\,\Leftarrow\,$) if $\,z \in \bigcup T_{2}\,$, then there exist a $\,y_{0} \in T_{2}\,$ for which $\,z\in y_{0}: = A-x_{0}\,$ ( for some $\,x_{0} \in S\,).\,$ Then $z \in A\,$ and for some $\,x_{0} \in S,\, z\notin x_{0}.\,$ Therefore, $z \in A\,$ and $\,z\notin \bigcap S,\, $ $z\in A - \bigcap S,\, $ i.e., $\, \,\bigcup T_{2} \subseteq\, A-\bigcap S .\, $
Claim 1: The set $T_{2}$ is non empty
We'll show that the set $T_{2}$ is empty iff the set $S$ is empty.
Proof: 
Suppose $S = \emptyset$, we need to show that $T_{2}$ is empty. Assume for the sake of the contradiction that $y$ is in $T_{2},$  $\,y\in T_{2} \leftrightarrow y = A-x$ for some $x\in S,\,$ but since $x\notin S:=\emptyset$ we have a contradiction, it follows that $y$ cannot be in $T_{2},\,$ i.e., $y \notin  T_{2}.\,$ Therefore $T_{2} = \emptyset,\,$ as desired. 
On the other hand, if we assume that $\,T_{2} = \emptyset,\,$ we need to seek if this assumption implies the emptiness of $S.\,$  By contradiction, suppose $S \not= \emptyset,\,$ then $x\in S,\,$ and the set $A-x \in T_{2},\,$ which is a contradiction, because is empty by hypothesis. Therefore, $S = \emptyset.\,$ 
Then, if we assume that $S \not= \emptyset$ it follows that $T_{2} \not= \emptyset,\,$ as desired.  $\;\Box$
**
I have problems to understand what's going on in that parts where I put the question mark in boldface.... I don't know maybe I'm tired. I need a coffee. 
As usual, thanks in advance :)
 A: In more everyday notation, $T_2=\{A\setminus x:x\in S\}$, so $T_2=\varnothing$ iff $S=\varnothing$. If $S=\varnothing$, then $\bigcup S=\varnothing$, so $A\setminus\bigcup S=A$. Now $\bigcap T_2=\bigcap\{y\in\wp(A):\exists x\in S(y=A\setminus x)\}$; it’s an intersection of subsets of $A$, so it must be a subset of $A$. What elements of $A$ are not in it? Let $a\in A$ be arbitrary; $a\notin\bigcap T_2$ iff there is a $y\in T_2$ such that $a\notin y$. But $T_2=\varnothing$, so there is no such $y$, and therefore it’s not the case that $a\notin\bigcap T_2$, i.e., $a\in\bigcap T_2$. And $a$ was arbitrary, so $\bigcap T_2=A$ in this case, just as we wanted. (If you’re familiar with the expression vacuously true, you can say that for each $a\in A$ it’s vacuously true that $a\in\bigcap T_2$.)
Now I’ll go back to the point in the proof of $(i,\Leftarrow)$ where you had trouble. You’ve assumed that $z\in\bigcap T_2$, and you want to show that $z\in A\setminus\bigcup S$. Let $x\in S$ be arbitrary. Then $A\setminus x\in T_2$, so $z\in A\setminus x$, and in particular $z\notin x$. Thus, for each $x\in S$ we have $z\notin x$, so $z\notin\bigcup S$. And $z\in\bigcap T_2\subseteq A$, so certainly $z\in A$, and it follows that $z\in A\setminus\bigcup S$, as desired.
A: (A) 
Proof. ($\rightarrow$) Suppose $x \in A \cap \bigcup S$. Then $x \in A$ and $x \in \bigcup S$. Since $x \in \bigcup S$, we can choose some $B \in S$ such that $x \in B$. Since $x \in A$ and $x \in B$, $x \in A \cap B$. We have shown that $B \in S$ and $x \in A \cap B$, so since $A \cap B \in \mathcal P \left({A}\right)$, we can conclude that $x \in \bigcup T_1$. 
($\leftarrow$) Suppose $x \in \bigcup T_1$. Then we can choose some $B \in T_1$ such that $x \in B$. Since $B \in T_1$, $B \in \mathcal P \left({A}\right)$ and $B = A \cap X$ for some $X \in S$. But then since $x \in B$ and $B = A \cap X$, $x \in A \cap X$, so $x \in A$ and $x \in X$. Since $X \in S$ and $x \in X$, it follows that $x \in \bigcup S$. We have shown that $x \in A$ and $x \in \bigcup S$, so $x \in A \cap \bigcup S$. 
(B) 
i) Proof. ($\rightarrow$) Suppose $x \in A \setminus \bigcup S$. Then $x \in A$ and $x \notin \bigcup S$. Let $B \in T_2$ be arbitrary. Then $B \in \mathcal P \left({A}\right)$ and $B = A \setminus X$ for some $X \in S$. Since $x \notin \bigcup S$ and $X \in S$, $x \notin X$. Since $x \in A$ and $x \notin X$, $x \in A\setminus X$, so $x \in B$. But then since $B \in T_2$ was arbitrary, $x \in \bigcap T_2$. 
($\leftarrow$) Suppose $x \in \bigcap T_2$. Let $B \in S$ be arbitrary. Clearly $A \setminus B \in \mathcal P \left({A}\right)$, so since $x \in \bigcap T_2$ and $B \in S$, it follows that $x \in A \setminus B$. This means that $x \in A$ and $x \notin B$. Since $B \in S$ was arbitrary, $x \notin \bigcup S$. Since $x \in A$ and $x \notin \bigcup S$, we can conclude that $x \in A \setminus \bigcup S$. 
ii) Proof. ($\rightarrow$) Suppose $x \in A \setminus \bigcap S$. Then $x \in A$ and $x \notin \bigcap S$. Since $x \notin \bigcap S$, we can choose some set $B \in S$ such that $x \notin B$. Since $x \in A$ and $x \notin B$, $x \in A \setminus B$. Clearly $A \setminus B \in \mathcal P \left({A}\right)$, so since $B \in S$ and $x \in A \setminus B$, it follows that $x \in \bigcup T_2$. 
($\leftarrow$) Suppose $x \in \bigcup T_2$. Then we can choose some set $B \in T_2$ such that $x \in B$. Since $B \in T_2$, it follows that $B \in \mathcal P \left({A}\right)$ and $B = A \setminus X$ for some $X \in S$. But then since $x \in B$ and $B = A \setminus X$, $x \in A \setminus X$, so $x \in A$ and $x \notin X$. Since $X \in S$ and $x \notin X$, $x \notin \bigcap S$. We have shown that $x \in A$ and $x \notin \bigcap S$, so $x \in A \setminus \bigcap S$.
