254: 'Natural Number'-th Power of Cardinality of Set Is That Times Multiplication of Cardinality

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A description/proof of that 'natural number'-th power of cardinality of set is that times multiplication of cardinality

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of set.
• The reader knows a definition of cardinality of set.

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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.

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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.

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Main Body

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1: Description

For any set, $S$, and any natural number, $n$, $(cardS{)}^n=cardScardS...cardS$, which is the $n$ times multiplication.

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2: Proof

For $n=1$, $(cardS{)}^n=cardS$, because it is the cardinality of the set of the functions, $F:=\{f\in Pow(1\times S)|f:1\to S\}$ where $1=\{0\}$.

Let us suppose that for an $n$, $(cardS{)}^n=cardScardS...cardS$, which is the $n$ times multiplication. $(cardS{)}^{n+1}$ is the cardinality of the set of the functions, $F:=\{f\in Pow((n+1)\times S)|f:(n+1)\to S\}$. Any $f=f^{\prime }\cup \{⟨n,s⟩\}$ where $f^{\prime }$ is any $f^{\prime }\in F^{\prime }:=\{f^{\prime }\in Pow(n\times S)|f^{\prime }:n\to S\}$ and $s$ is any $s\in S$. On the other hand, $(cardS{)}^{n}cardS=card{F}^{\prime }cardS=card(F^{\prime }\times S)$ by the definition of arithmetic of cardinalities. Any element of $F^{\prime }\times S$ is $⟨f^{\prime },s⟩$ where $f^{\prime }$ is any $f^{\prime }\in F^{\prime }$ and $s$ is any $s\in S$. There is the bijection, $g:F\to F^{\prime }\times S,f^{\prime }\cup \{⟨n,s⟩\}\mapsto ⟨f^{\prime },s⟩$, so, $cardF=card(F^{\prime }\times S)=card{F}^{\prime }cardS=cardScardS...cardS$, which is the $n+1$ times multiplication.

So, by the mathematical induction principle, for any natural number, $n$, $(cardS{)}^n=cardScardS...cardS$, which is the $n$ times multiplication.

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3: Note

The proposition is not so obvious, because $(cardS{)}^n$ is not defined as the $n$ times multiplication, but as the cardinality of $\{f\in Pow(n\times S)|f:n\to S\}$.

Only because of this proposition, $(cardS{)}^n$ can be regarded to be the $n$ times multiplication.

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References

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