A description/proof of that products of sets are associative in 'sets - map morphisms' isomorphism sense
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
- The reader knows a definition of product of sets.
- The reader knows a definition of %category name% isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that the nested product of any sets are associative in the 'sets - map morphisms' isomorphism sense.
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: Description
For any sets, \(S_1, S_2, S_3\), the nested product, \((S_1 \times S_2) \times S_3\) is 'sets - map morphisms' isomorphic to \(S_1 \times (S_2 \times S_3)\).
2: Proof
\((S_1 \times S_2) \times S_3 = \{s\vert \exists s_1 \in S_1, \exists s_2 \in S_2, \exists s_3 \in S_3, s = \langle\langle s_1, s_2 \rangle, s_3 \rangle\}\) is obviously not exactly \(S_1 \times (S_2 \times S_3) = \{s\vert \exists s_1 \in S_1, \exists s_2 \in S_2, \exists s_3 \in S_3, s = \langle s_1, \langle s_2, s_3 \rangle \rangle\}\), but \(f: (S_1 \times S_2) \times S_3 \rightarrow S_1 \times (S_2 \times S_3), \langle \langle s_1, s_2 \rangle, s_3 \rangle \mapsto \langle s_1, \langle s_2, s_3 \rangle \rangle\) is a bijection.
3: Note
Although sloppy expressions like "\((S_1 \times S_2) \times S_3 = S_1 \times (S_2 \times S_3)\)" are prevalently seen, the 2 are not the same, but are isomorphic to each other.
The expression, \(S_1 \times S_2 \times S_3\), is allowed because it is defined as \((S_1 \times S_2) \times S_3\), not because "\((S_1 \times S_2) \times S_3 = S_1 \times (S_2 \times S_3)\)" holds.
