A description/proof of that finite product of sets is set
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of product set.
- The reader admits the proposition that some expressions can be parts of legitimate formulas for the ZFC set theory.
Target Context
- The reader will have a description and a proof of the proposition that the product of any finite number of sets is a set.
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, \(S_1, S_2, . . ., S_n\), the product of the sets, \(S_1 \times S_2 \times . . . \times S_n\), is a set.
2: Proof
1st, let us think of the case in which \(n = 2\). In fact, \(S_1 \times S_2 = \{s \in Pow (Pow (S_1 \cup S_2))\vert \exists s_1 \in S_1, \exists s_2 \in S_2, s = \langle s_1, s_2 \rangle\}\), because \(s_i \in S_1 \cup S_2, \{s_1\} \subseteq S_1 \cup S_2, \text{ so}, \{s_1\} \in Pow (S_1 \cup S_2), \{s_1, s_2\} \subseteq S_1 \cup S_2, \text{ so}, \{s_1, s_2\} \in Pow (S_1 \cup S_2), \{ \{s_1\}, \{s_1, s_2\}\} = \langle s_1, s_2 \rangle \subseteq Pow (S_1 \cup S_2), \text{ so}, \langle s_1, s_2 \rangle \in Pow (Pow (S_1 \cup S_2))\). By the subset axiom and the proposition that some expressions can be parts of legitimate formulas for the ZFC set theory, \(S_1 \times S_2\) is a set.
\(S_1 \times S_2, \times . . . \times S_n\) is really \(( . . . (S_1 \times S_2) \times . . .) \times S_n\). As \(S_1 \times S_2\) is a set, \((S_1 \times S_2) \times S_3\) is a set; as \((S_1 \times S_2) \times S_3\) is a set, \(((S_1 \times S_2) \times S_3) \times S_4\) is a set; . . .; as \(( . . . (S_1 \times S_2) \times . . .) \times S_{n - 1}\) is a set, \(( . . . (S_1 \times S_2) \times . . .) \times S_n\) is a set. Of course, we should use the induction principle to be exact.
