sets and functions proof help I started to get excited about mathematical analysis. So I bought a mathematical analysis book and started to study. But because of the reason that, book do not have solutions I do not have an idea where to start and how to prove the following:
Let $A_t$, $t \in T$, be a family of sets, and let $X$ be a set. Prove the identities:
$$X \setminus \bigcup A_t = \bigcap (X\setminus A_t)$$
$$X \setminus \bigcap A_t = \bigcup(X\setminus A_t) $$
Could you please help me? Also I need a book recommendation about mathematical analysis which goes like a theorem and its proof, a theorem and its proof.. Do you have any suggestions?
Regards
 A: For example:
$$x\in X\setminus \bigcup A_t\Longleftrightarrow \,\,\,\text{for no}\,\,t\,\,,\,\,x\in A_t\Longleftrightarrow \,\,\forall\,t\,\,,\,x\in X\setminus A_t\Longleftrightarrow x\in\bigcap(X\setminus A_t)$$
The other equality is very similar.
A: Let's look at the case there $|T| = 1$, i.e. that the family of sets has only one element.
The $X\setminus A$ is trivially $X \setminus A$.
What about when $|T| = 2$?
Then $X \setminus (A_1 \cup A_2)$ is the set of $x \in X$ such that $x \not \in A_1, A_2$. 
$X \setminus A_1$ is the set of $x \in X$ such that $x\not \in A_1$, and $X \setminus A_2$ is the set of $x \in X$ such that $x \not \in A_2$. This means that $\bigcap (X \setminus A_t)$ is the set of $x \in X$ such that $x \not \in A_1$ and such that $x \not \in A_2$. Equivalently, this is the set of $x \in X$ such that $x \not \in A_1, A_2$. My gosh, this is the same as the set above!
Now you might look back at this and say, where did we use our ability to enumerate the sets? 
Do you see how to finish from here?
