description/proof of that finite product set is 'sets - maps' isomorphic to sequential products of sets
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of product set.
- The reader knows a definition of %category name% isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any finite product set is 'sets - maps' isomorphic to the sequential products of the sets.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J\): \(\in \{\text{ the finite index sets }\}\), with \(\vert J \vert = n\) and any ordering
\(\{S_j \in \{\text{ the sets } \} \vert j \in J\}\):
\(\times_{j \in J} S_j\):
\(( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\):
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Statements:
\(\times_{j \in J} S_j \cong_{sets} ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\)
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2: Proof
Whole Strategy: Step 1: define a bijection, \(g: \times_{j \in J} S_j \to ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\).
Step 1:
Let us define the map, \(g: \times_{j \in J} S_j \to ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}, f \mapsto f_{n - 1}\), where \(f_1 \in S_{J_1} \times S_{J_2}\) (which is a map from \(\{1, 2\}\) into \(S_{J_1} \cup S_{J_2}\)) is determined as \(f_1 (1) = f (J_1)\) and \(f_1 (2) = f (J_2)\), \(f_2 \in (S_{J_1} \times S_{J_2}) \times S_{J_3}\) (which is a map from \(\{1, 2\}\) into \((S_{J_1} \times S_{J_2}) \cup S_{J_3}\)) is determined as \(f_2 (1) = f_1\) and \(f_2 (2) = f (J_3)\), ..., and \(f_{n - 1} \in ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\) (which is a map from \(\{1, 2\}\) into \(( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_{n - 1}} \cup S_{J_n}\)) is determined as \(f_{n - 1} (1) = f_{n - 1}\) and \(f_{n - 1} (2) = f (J_n)\).
\(g\) is an injection, because for any \(f, f' \in \times_{j \in J} S_j\) such that \(f \neq f'\), \(f (J_{n'}) \neq f' (J_{n'})\) for an \(n'\), and when \(n' = 1\), \(f_1 \neq f'_1\), because \(f_1 (1) = f (J_1) \neq f' (J_1) = f'_1 (1)\), and otherwise, \(f_{n' - 1} \neq f'_{n' - 1}\), because \(f_{n' - 1} (2) = f (J_{n'}) \neq f' (J_{n'}) = f'_{n' - 1} (2)\), then, \(f_{n - 1} \neq f'_{n - 1}\).
\(g\) is a surjection, because for each \(f_{n - 1} \in ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\), there is the \(f \in \times_{j \in J} S_j\) such that \(f (J_1) = f_1 (1)\), \(f (J_2) = f_1 (2)\), \(f (J_3) = f_2 (2)\), ..., \(f (J_n) = f_{n - 1} (2)\), and \(g (f) = f_{n - 1}\).
So, \(g\) is a bijection.
