Set-Indexed operations

Definition 23   Let $\Lambda$ and $A$ be sets, and for each $\lambda\in \Lambda$ let $A_\lambda$ be a subset of $A$. Then

$$\bigcap_{\lambda\in\Lambda}A_\lambda=\{a\vert a\in A_\lambda \mbox{ for every }\lambda\in \Lambda\}$$

Definition 24   Let $\Lambda$ and $A$ be sets, and for each $\lambda\in \Lambda$ let $A_\lambda$ be a set. Then

$$\bigcup_{\lambda\in\Lambda}A_\lambda=\{a\vert a\in A_\lambda \mbox{ for some }\lambda\in \Lambda\}$$

Notice that if $\Lambda$ is a two element set these give the usual intersection and union and that we have made no restrictions on the size of $\Lambda$ other than that it be a set.

Proposition 31   Suppose $\Lambda=\emptyset$, then both $${\bigcap_{\lambda\in\Lambda}A_\lambda \mbox{ and }\bigcup_{\lambda\in\Lambda}A_\lambda}$$ can be determined. Find what they are, and then prove the result.

(4 Points )

Notice that the difference between the definitions is needed precisely for this case.

Definition 25   Let $\Lambda$ be a set, and for each $\lambda\in \Lambda$ let $A_\lambda$ be a set. Then

$$\prod_{\lambda \in \Lambda}A_\lambda=\{f:\Lambda\to \bigcup_{\lambda\in \Lambda}A_\lambda\vert f(\lambda)\in A_\lambda\}$$

Such a function $f$ simultaneously chooses an element of $A_\lambda$ for each $\lambda\in \Lambda$ and may be thought of as a sort of $\Lambda$-tuple.

Axiom: The axiom of choice says that if $A_\lambda \neq \emptyset$ for all $\lambda\in \Lambda$ then

$$\prod_{\lambda\in \Lambda}A_\lambda\neq \emptyset$$

While this may seem obvious (and innocuous) it has some paradoxical consequences and is independent of the rest of set theory.

Proposition 32   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$ and $B\subseteq A$. Then

$$B\cup \bigcup_{\lambda\in \Lambda}A_\lambda = \bigcup_{\lambda\in \Lambda}(B\cup A_\lambda).$$

(2 Points )

Proposition 33   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$ and $B\subseteq A$. Then

$$B\cap \bigcup_{\lambda\in \Lambda}A_\lambda = \bigcup_{\lambda\in \Lambda}(B\cap A_\lambda).$$

(2 Points )

Proposition 34   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$ and $B\subseteq A$. Then

$$B\cup \bigcap_{\lambda\in \Lambda}A_\lambda = \bigcap_{\lambda\in \Lambda}(B\cup A_\lambda).$$

(2 Points )

Proposition 35   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$ and $B\subseteq A$. Then

$$B\cap \bigcap_{\lambda\in \Lambda}A_\lambda = \bigcap_{\lambda\in \Lambda}(B\cap A_\lambda).$$

(2 Points )

Proposition 36   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$ and $B$ is any set. Then

$$B\times \bigcap_{\lambda\in \Lambda}A_\lambda = \bigcap_{\lambda\in \Lambda}(B\times A_\lambda).$$

(2 Points )

Proposition 37   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$ and $B$ is any set. Then

$$B\times \bigcup_{\lambda\in \Lambda}A_\lambda = \bigcup_{\lambda\in \Lambda}(B\times A_\lambda).$$

(2 Points )

Proposition 38   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$. Then

$$A\setminus \bigcap_{\lambda\in \Lambda}A_\lambda = \bigcup_{\lambda\in \Lambda}(A\setminus A_\lambda).$$

(2 Points )

Proposition 39   Suppose that $A_\lambda\subseteq A$ for each $\lambda\in \Lambda$. Then

$$A\setminus \bigcup_{\lambda\in \Lambda}A_\lambda = \bigcap_{\lambda\in \Lambda}(A \setminus A_\lambda).$$

(2 Points )