Cartesian Product of Family/Examples/1 and 2
Jump to navigation
Jump to search
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:
\(\ds A_\O \times A_{\set \O}\) | \(=\) | \(\ds \set \O \times \set {\O, \set \O}\) | Definition of $A_\O$ and $A_{\set \O}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {\tuple {\O, \O}, \tuple {\O, \set \O} }\) | Definition of Cartesian Product |
Then:
\(\ds \prod_{i \mathop \in A_{\set \O} } A_i\) | \(=\) | \(\ds \prod_{i \mathop \in \set {\O, \set \O} } A_i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {f \in \paren {\bigcup_{i \mathop \in A_{\set \O} } A_i}^{A_{\set \O} }: \forall i \in A_{\set \O}: \paren {\map f i \in A_i} }\) | Definition 2 of Cartesian Product of Family |
The above is deconstructed as follows.
We have that:
\(\ds \bigcup_{i \mathop \in A_{\set \O} } A_i\) | \(=\) | \(\ds \bigcup_{i \mathop \in \set {\O, \set \O} } A_i\) | Definition of $A_{\set \O}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup \set {A_\O, A_{\set \O} }\) | Definition of Union of Family | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup \set {\set \O, \set {\O, \set \O} }\) | Definition of $A_\O$ and $A_{\set \O}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \set \O \cup \set {\O, \set \O}\) | Union of Doubleton | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \set {\O, \set \O}\) | Definition of Set Union |
Hence we have that:
\(\ds \paren {\bigcup_{i \mathop \in A_{\set \O} } A_i}^{A_{\set \O} }\) | \(=\) | \(\ds \paren {\set {\O, \set \O} }^{A_{\set \O} }\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\set {\O, \set \O} }^{\set {\O, \set \O} }\) | Definition of $A_{\set \O}$ | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \set {\set {\tuple {\O, \O}, \tuple {\set \O, \O} }, \set {\tuple {\O, \O}, \tuple {\set \O, \set \O} }, \set {\tuple {\O, \set \O}, \tuple {\set \O, \O} }, \set {\tuple {\O, \set \O}, \tuple {\set \O, \set \O} } }\) | Definition of Set of All Mappings |
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:
\(\ds \map f \O\) | \(\in\) | \(\ds A_\O\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f \O\) | \(\in\) | \(\ds \set \O\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f \O\) | \(=\) | \(\ds \O\) |
Then:
\(\ds \map f {\set \O}\) | \(\in\) | \(\ds A_{\set \O}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f {\set \O}\) | \(\in\) | \(\ds \set {\O, \set \O}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f {\set \O}\) | \(=\) | \(\ds \O\) | |||||||||||
\(\, \ds \text {or} \, \) | \(\ds \map f {\set \O}\) | \(=\) | \(\ds \set \O\) |
Hence:
- $\ds \prod_{i \mathop \in A_{\set \O} } A_i = \set {\set {\tuple {\O, \O}, \tuple {\set \O, \O} }, \set {\tuple {\O, \O}, \tuple {\set \O, \set \O} } }$
$\blacksquare$
Sources
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $8$