281: No 2 Sets Have Each Other as Members

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A description/proof of that no 2 sets have each other as members

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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 set.

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Target Context

• The reader will have a description and a proof of the proposition that there are no 2 sets such that each of them has the other as a member.

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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

There are no sets, $S_1,S_2$, such that $S_1\in S_2$ and $S_2\in S_1$.

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2: Proof

This proposition stems from the regularity axiom.

Suppose that $S_1\in S_2$ and $S_2\in S_1$. By the subset axiom, $S_1^{\prime }:=\{s\in S_1|s=S_2\}=\{S_2\}$ and $S_2^{\prime }:=\{s\in S_2|s=S_1\}=\{S_1\}$ would be sets. By the union axiom, $S:=S_1^{\prime }\cup S_2^{\prime }=\{S_1,S_2\}$ would be a set. $S\cap S_1\ne \mathrm{\varnothing }$, because $S_2\in S$ and $S_2\in S_1$, and $S\cap S_2\ne \mathrm{\varnothing }$, because $S_1\in S$ and $S_1\in S_2$, a contradiction, being against the regularity axiom.

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References

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