Terence Tao Analysis 1 Exercise 3.4.11 My real analysis class is using Terence Tao's "Analysis 1" this semester, which unfortunately doesn't have solutions available anywhere. Part of the current homework is exercise 3.4.11:
"Let $X$ be a set, let $I$ be a non-empty set, and for all $\alpha \in I$ let $A_\alpha$ be a subset of $X$. Show that
$$X - \bigcup_{\alpha \in I} A_\alpha = \bigcap_{\alpha \in I} (X - A_\alpha)$$
and
$$X - \bigcap_{\alpha \in I} A_\alpha = \bigcup_{\alpha \in I} (X - A_\alpha).$$"
The book suggests comparing them with de Morgan's laws, and I see the similarities, but it also says that one cannot derive these identities from de Morgan's laws since $I$ could be infinite.
I'm not sure where to go with this. Since we're comparing sets, I know that the place to start is "Let $x \in X - \bigcup_{\alpha \in I} A_\alpha$." I also think from this we can deduce that $x \in X$ and $x \notin \bigcup_{\alpha \in I} A_\alpha$, but I'm not sure where to go from there. 
 A: De Morgan's laws are not necessary. We have
\begin{align}
x\in X\setminus\bigcup_{\alpha\in I} A_\alpha &\iff  x\in X, x\notin A_\alpha \text{ for all $\alpha\in I$}\\
&\iff  x\in X\setminus A_\alpha \text{ for all $\alpha\in I$}\\
&\iff x\in \bigcap_{\alpha \in I}(X\setminus A_\alpha),
\end{align}
and similarly for the other statement.
Also, De Morgan's laws hold for sets of arbitrary cardinality.
A: I will second @Math1000's proof but with additional details
Definitions:

*

*$x \in \bigcup_{\alpha\in I} A_\alpha := \exists a \in I (x \in A_a)$

*$x \in \bigcap_{\alpha\in I} A_\alpha := \forall a \in I (x \in A_a)$
Proofs:
\begin{align}
x \in X\setminus\bigcup_{\alpha\in I} A_\alpha &\iff x\in X \ \land \ x \notin \ \bigcup_{\alpha\in I} A_\alpha 
\\
&\iff x\in X \ \land \lnot (\exists a \in I (x \in A_a))
\\
&\iff x \in X \ \land \forall a \in I (x \notin A_a)
\\
&\iff \forall a \in I(x \in X \land x \notin A_a)
\\
&\iff x \in \bigcap_{\alpha\in I} \ (X \setminus A_a)
\end{align}
\begin{align}
x \in X\setminus\bigcap_{\alpha\in I} A_\alpha &\iff x\in X \ \land \ x \notin \ \bigcap_{\alpha\in I} A_\alpha 
\\
&\iff x\in X \ \land \lnot (\forall a \in I (x \in A_a))
\\
&\iff x \in X \ \land \exists a \in I (x \notin A_a)
\\
&\iff \exists a \in I(x \in X \land x \notin A_a)
\\
&\iff x \in \bigcup_{\alpha\in I} \ (X \setminus A_a)
\end{align}
I'm not a Professional Mathematician, if you are, please review the above proofs and comment below. Much appreciated! :)
