A description/proof of that 'natural number'-th power of cardinality of set is that times multiplication of cardinality
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 cardinality of set.
Target Context
- The reader will have a description and a proof of the proposition that for the cardinality of any set, any 'natural number'-th power of the cardinality is the natural number times multiplication of the cardinality.
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 set, \(S\), and any natural number, \(n\), \((card S)^n = card S card S . . . card S\), which is the \(n\) times multiplication.
2: Proof
For \(n = 1\), \((card S)^n = card S\), because it is the cardinality of the set of the functions, \(F:= \{f \in Pow (1 \times S)\vert f: 1 \rightarrow S\}\) where \(1 = \{0\}\).
Let us suppose that for an \(n\), \((card S)^n = card S card S . . . card S\), which is the \(n\) times multiplication. \((card S)^{n + 1}\) is the cardinality of the set of the functions, \(F:= \{f \in Pow ((n + 1) \times S)\vert f: (n + 1) \rightarrow S\}\). Any \(f = f' \cup \{\langle n, s\rangle\}\) where \(f'\) is any \(f' \in F':= \{f' \in Pow (n \times S)\vert f': n \rightarrow S\}\) and \(s\) is any \(s \in S\). On the other hand, \((card S)^n card S = card F' card S = card (F' \times S)\) by the definition of arithmetic of cardinalities. Any element of \(F' \times S\) is \(\langle f', s\rangle\) where \(f'\) is any \(f' \in F'\) and \(s\) is any \(s \in S\). There is the bijection, \(g: F \rightarrow F' \times S, f' \cup \{\langle n, s\rangle\} \mapsto \langle f', s\rangle\), so, \(card F = card (F' \times S) = card F' card S = card S card S . . . card S\), which is the \(n + 1\) times multiplication.
So, by the mathematical induction principle, for any natural number, \(n\), \((card S)^n = card S card S . . . card S\), which is the \(n\) times multiplication.
3: Note
The proposition is not so obvious, because \((card S)^n\) is not defined as the \(n\) times multiplication, but as the cardinality of \(\{f \in Pow (n \times S)\vert f: n \rightarrow S\}\).
Only because of this proposition, \((card S)^n\) can be regarded to be the \(n\) times multiplication.
