262: Finite Product of Sets Is Set

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A description/proof of that finite product of sets is set

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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

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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.

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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.

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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 sets, $S_1,S_2,...,S_n$, the product of the sets, $S_1\times S_2\times ...\times S_n$, is a set.

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2: Proof

1st, let us think of the case in which $n=2$. In fact, $S_1\times S_2=\{s\in Pow(Pow(S_1\cup S_2))|\mathrm{\exists }{s}_1\in S_1,\mathrm{\exists }{s}_2\in S_2,s=⟨s_1,s_2⟩\}$, because $s_i\in S_1\cup S_2,\{s_1\}\subseteq S_1\cup S_2,\text{ so},\{s_1\}\in Pow(S_1\cup S_2),\{s_1,s_2\}\subseteq S_1\cup S_2,\text{ so},\{s_1,s_2\}\in Pow(S_1\cup S_2),\{\{s_1\},\{s_1,s_2\}\}=⟨s_1,s_2⟩\subseteq Pow(S_1\cup S_2),\text{ so},⟨s_1,s_2⟩\in Pow(Pow(S_1\cup S_2))$. 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.

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References

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