Cartesian Product of Family/Examples/1 and 2

Example of Cartesian Product of Family
Let $A_\O := \set \O$ and $A_{\set \O} := \set {\O, \set \O}$.

Thus $A_\O$ and $A_{\set \O}$ are the numbers $1$ and $2$ as defined by the Von Neumann construction.

Then:
 * $A_\O \times A_{\set \O} = \set {\tuple {\O, \O}, \tuple {\O, \set \O} }$

while:
 * $\ds \prod_{i \mathop \in A_{\set \O} } A_i = \set {\set {\tuple {\O, \O}, \tuple {\set \O, \O} }, \set {\tuple {\O, \O}, \tuple {\set \O, \set \O} } }$

Proof
First we have:

Then:

The above is deconstructed as follows.

We have that:

Hence we have that:

Note that $(2)$ above is the set of all mappings from $\set {\O, \set \O}$ to $\set {\O, \set \O}$ as follows:


 * Each such mapping is a set of $2$ ordered pairs of which the first coordinates are the elements of $\set {\O, \set \O}$


 * From Cardinality of Set of All Mappings there are $2^2 = 4$ set of $2$ such ordered pairs.

Now we have to select the elements $f$ of $\ds \paren {\bigcup_{i \mathop \in A_{\set \O} } A_i}^{A_{\set \O} }$ such that:
 * $\map f i \in A_i$

for all $i \in \set {\O, \set \O}$.

We have that:

Then:

Hence:
 * $\ds \prod_{i \mathop \in A_{\set \O} } A_i = \set {\set {\tuple {\O, \O}, \tuple {\set \O, \O} }, \set {\tuple {\O, \O}, \tuple {\set \O, \set \O} } }$