Analysis Solutions: Tao Analysis I

Let $X$ be a set, let $I$ be an on-empty set, and for all $\alpha \in I$ let $A_\alpha$ be a subset of $X$.

Show that

$$ X \setminus \bigcup_{\alpha \in I}A_\alpha = \bigcap_{\alpha \in I}( X \setminus A_\alpha) $$

and

$$ X \setminus \bigcap_{\alpha \in I}A_\alpha = \bigcup_{\alpha \in I}( X \setminus A_\alpha) $$

Part One

Let's start by considering $\bigcup_{\alpha \in I}A_\alpha$. By definition we have

$$ x \in \bigcup_{\alpha \in I}A_\alpha \iff (x \in A_\alpha \text{ for some } \alpha \in I) $$

If $x$ is not in $\bigcup_{\alpha \in I}A_\alpha$, then we have

$$ x \notin \bigcup_{\alpha \in I}A_\alpha \iff (x \notin A_\alpha \text{ for all } \alpha \in I) $$

Note the quantifier changes from "for some" to "for all" when negated.

So, $x \in X \setminus \bigcup_{\alpha \in I}A_\alpha $ means

$$(x \in X) \land (x \notin \bigcup_{\alpha \in I}A_\alpha )$$

That is,

$$(x \in X) \land (x \notin A_\alpha \text{ for all } \alpha \in I) $$

That means $x \in X$ and also $x \notin A_\alpha$ for each and every $A_\alpha$. This is equivalent to

$$x \in \bigcap_{\alpha \in I} (X \setminus A_\alpha ) $$

We have shown,

$$X \setminus \bigcup_{\alpha \in I}A_\alpha = \bigcap_{\alpha \in I}( X \setminus A_\alpha) \; \square$$

Part Two

The approach is similar to Part One.

Let's start by considering $\bigcap_{\alpha \in I}A_\alpha$. By definition we have

$$ x \in \bigcap_{\alpha \in I}A_\alpha \iff (x \in A_\alpha \text{ for all } \alpha \in I) $$

If $x$ is not in $\bigcap_{\alpha \in I}A_\alpha$, then we have

$$ x \notin \bigcap_{\alpha \in I}A_\alpha \iff (x \notin A_\alpha \text{ for some } \alpha \in I) $$

Note the quantifier changes from "for all" to "for some" when negated. It might be easier to read "for some" as "at least one" in this example.

So, $x \in X \setminus \bigcap_{\alpha \in I}A_\alpha $ means

$$(x \in X) \land (x \notin \bigcap_{\alpha \in I}A_\alpha )$$

That is,

$$(x \in X) \land (x \notin A_\alpha \text{ for some } \alpha \in I) $$

That means $x \in X$ and also $x \notin A_\alpha$ for some $A_\alpha$. This is equivalent to

$$x \in \bigcup_{\alpha \in I} (X \setminus A_\alpha ) $$

We have shown,

$$ X \setminus \bigcap_{\alpha \in I}A_\alpha = \bigcup_{\alpha \in I}( X \setminus A_\alpha) \; \square$$

Analysis Solutions