Ordinal Property of Subsets in Well-Ordered Sets

[jostpuur]

A set x is well-ordered by < if every subset of x has a least element. Here < is assumed a linear ordering, meaning that all members of a set can be compared, unlike with partial ordering.

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

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

The claim: If \alpha is an ordinal, and \beta\in\alpha, then \beta 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 \beta\in\alpha\to\beta\subset\alpha, 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 \forall\gamma\;(\gamma\in\beta\to\gamma\subset\beta). How is this supposed to come from the definition? I only see \gamma\in\beta\to\gamma\in\alpha\to\gamma\subset\alpha.

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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 \gamma\in\beta and then

<br /> \neg(\gamma\subset\beta)\to \exists\delta\;(\delta\in\gamma\land\delta\notin\beta)<br />
<br /> \to\exists\delta\;\big(\delta\in\gamma\land(\beta\in\delta\lor \beta=\delta)\big)<br />
<br /> \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<br />

Does that look like "by definition"?

 

[verty]

I think they call it by definition for this reason. Since $\beta\subset\alpha$, $\beta$ is well ordered by $\in$, as you pointed out. So for $\beta' < \beta$ in $\alpha$, $\beta'\in\beta$, and for $\beta' > \beta$, $\beta\in\beta'$ which precludes $\beta$ containing any of these larger elements. But $\beta\subset\alpha$, therefore $\beta$ is exactly the union of elements of $\alpha$ less than $\beta$. But then $\forall\gamma\in\beta \; (\gamma\subset\beta)$ and $\beta$ is transitive.

So in a sense, $\beta$ is defined in this way by those definitions.