A set [itex]x[/itex] is well-ordered by [itex]<[/itex] if every subset of [itex]x[/itex] has a least element. Here [itex]<[/itex] is assumed a linear ordering, meaning that all members of a set can be compared, unlike with partial ordering.(adsbygoogle = window.adsbygoogle || []).push({});

A set [itex]x[/itex] is transitive if it has property [itex]\forall y\;(y\in x\to y\subset x)[/itex].

A set [itex]\alpha[/itex] is ordinal, if it is transitive and well-ordered by [itex]\in[/itex].

The claim: If [itex]\alpha[/itex] is an ordinal, and [itex]\beta\in\alpha[/itex], then [itex]\beta[/itex] is ordinal too.

A book says that this claim is clear "by definition", however I see only half of the proof by definition.

We have [itex]\beta\in\alpha\to\beta\subset\alpha[/itex], and a subset of a well-ordered set is also well-ordered, so that part is clear by definition.

We should also prove a claim [itex]\forall\gamma\;(\gamma\in\beta\to\gamma\subset\beta)[/itex]. How is this supposed to come from the definition? I only see [itex]\gamma\in\beta\to\gamma\in\alpha\to\gamma\subset\alpha[/itex].

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update: Oh I understood this now! No need for help. But I would like to complain that the book is playing fool on the reader. I wouldn't call that "by definition".

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second update: We assume [itex]\gamma\in\beta[/itex] and then

[tex]

\neg(\gamma\subset\beta)\to \exists\delta\;(\delta\in\gamma\land\delta\notin\beta)

[/tex]

[tex]

\to\exists\delta\;\big(\delta\in\gamma\land(\beta\in\delta\lor \beta=\delta)\big)

[/tex]

[tex]

\to\exists\delta\big(\underbrace{(\delta\in\gamma\land\beta\in\delta)}_{\to 0=1}\lor\underbrace{(\delta\in\gamma\land \beta=\delta)}_{\to 0=1}\big)\to 0=1

[/tex]

Does that look like "by definition"?

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# Ordinal by definition

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