259: Formula That Uniquely Maps Each Element of Set into Set Constitutes Function

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A description/proof of that formula that uniquely maps each element of set into set constitutes function

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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: Note
• 2: Description
• 3: Proof

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Starting Context

• The reader knows a definition of set.

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Target Context

• The reader will have a description and a proof of the proposition that any formula that uniquely maps each element of any set into another set constitutes a function.

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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: Note

This proposition may seem obvious, but its motivation is that the map has to be proved to be a set to be called a function, in the ZFC set theory. Being a part of the product set is not enough, because the subset axiom requires a legitimate formula. So, the formula is required: just knowing that each element maps to the image uniquely somehow is not enough.

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2: Description

For any sets, $S_1,S_2$, and any formula, $\varphi (s_1,s_2,...)$, where $s_1\in S_1$ and $s_2\in S_2$, that maps each element of $S_1$ uniquely to an element of $S_2$, $f:S_1\to S_2,s_1\mapsto s_2\text{ if and only if }\varphi (s_1,s_2,...)$ is a function.

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3: Proof

$f\subseteq S_1\times S_2$, but we need a legitimate formula to declare $f$ to be a set by the subset axiom. $f=\{p\in S_1\times S_2|\mathrm{\exists }{s}_1\in S_1,\mathrm{\exists }{s}_2\in S_2,\varphi (s_1,s_2,...),p=⟨s_1,s_2⟩\}$. There is a unique element, $p=⟨s_1,s_2⟩\in f$ for each $s_1\in S_1$, by the supposition for $\varphi$. So, $f$ is a function.

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References

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