Tao Analysis I - 3.4.10

Exercise 3.4.10

Suppose that $I$ and $J$ are two sets, and for all $\alpha \in I \cup J$ let $A_\alpha$ be a set. Show that

$$ \left( \bigcup_{\alpha \in I} A_\alpha \right) \cup \left( \bigcup_{\alpha \in J} A_\alpha \right)  = \bigcup_{\alpha \in I \cup J} A_\alpha $$

If $I$ and $I$ are non-empty, show that

$$ \left( \bigcap_{\alpha \in I} A_\alpha \right) \cap \left( \bigcap_{\alpha \in J} A_\alpha \right)  = \bigcap_{\alpha \in I \cup J} A_\alpha $$

Union of Unions

Let's first consider the set $ \left( \bigcup_{\alpha \in I} A_\alpha \right)$. By definition (3.2), we have

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

Similarly, 

$$ x \in \left( \bigcup_{\alpha \in J} A_\alpha \right) \iff x \in A_\alpha \text{ for some } \alpha \in J $$

Now, by definition of pairwise union, if $x \in X \cup Y \iff (x \in Y) \lor (x \in Y)$. So we have,

$$ x \in  \left( \bigcup_{\alpha \in I} A_\alpha \right) \cup \left( \bigcup_{\alpha \in J} A_\alpha \right) \iff (x \in A_\alpha \text{ for some } \alpha \in I) \lor (x \in A_\alpha \text{ for some } \alpha \in J) $$

The two membership conditions are combined in disjunction.

The RHS is equivalent to $(x \in A_\alpha \text{ for some } \alpha \in I \lor J)$. This is the definition of $\bigcup_{\alpha \in I \cup J} A_\alpha$.

Thus, we have shown

$$ \left( \bigcup_{\alpha \in I} A_\alpha \right) \cup \left( \bigcup_{\alpha \in J} A_\alpha \right)  = \bigcup_{\alpha \in I \cup J} A_\alpha \; \square $$

Intersection of Intersections

Let's consider the set $\left( \bigcap_{\alpha \in I} A_\alpha \right)$. By definition (3.4), we have

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

Similarly,

$$ x \in  \left( \bigcap_{\alpha \in J} A_\alpha \right) \iff x \in A_\alpha \text{ for all } \alpha \in J  $$

Now, by definition of pairwise intersection, $x \in X \cap Y \iff (x \in Y) \land (x \in Y)$. So we have,

$$ x \in  \left( \bigcap_{\alpha \in I} A_\alpha \right) \cap \left( \bigcap_{\alpha \in J} A_\alpha \right) \iff (x \in A_\alpha \text{ for all } \alpha \in I) \land (x \in A_\alpha \text{ for all } \alpha \in J) $$

The two membership conditions are combined in conjunction.

The RHS is equivalent to $(x \in A_\alpha \text{ for all } \alpha \in I \land J)$. This is the definition of $\bigcup_{\alpha \in I \cap J} A_\alpha$.

Thus, we have shown

$$ \left( \bigcap_{\alpha \in I} A_\alpha \right) \cap \left( \bigcap_{\alpha \in J} A_\alpha \right)  = \bigcap_{\alpha \in I \cup J} A_\alpha \; \square $$