273: For Transitive Set with Partial Ordering by Membership, Element Is Initial Segment Up to It

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A description/proof of that for transitive set with partial ordering by membership, element is initial segment up to it

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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 transitive set.
• The reader knows a definition of initial segment up to element of set.

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Target Context

• The reader will have a description and a proof of the proposition that for any transitive set with the at least partial ordering by membership (supposing that the ordering by membership is really a partial ordering), any element is the initial segment up to it.

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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 transitive set, $S$, with the at least partial ordering by membership (supposing that the ordering by membership is really a partial ordering), any element, $p\in S$, is the initial segment, $segp$, up to $p$, which is $p=segp$.

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2: Proof

For any element, $p^{\prime }\in segp$, of $segp$, $p^{\prime }\in p$, because $p^{\prime }<p$ if and only if $p^{\prime }\in S$ and $p^{\prime }\in p$, by the definition of ordering by membership. For any element, $p^{\prime }\in p$, of $p$, $p^{\prime }\in segp$, because $p^{\prime }\in S$ as $S$ is transitive and so, $p^{\prime }<p$.

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References

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