Terence Tao Analysis 1 Exercise 3.4.11

Question

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.

Answer 1

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.

Answer 2

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! :)