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

$$\neg(\gamma\subset\beta)\to \exists\delta\;(\delta\in\gamma\land\delta\notin\beta)$$
$$\to\exists\delta\;\big(\delta\in\gamma\land(\beta\in\delta\lor \beta=\delta)\big)$$
$$\to\exists\delta\big(\underbrace{(\delta\in\gamma\land\beta\in\delta)}_{\to 0=1}\lor\underbrace{(\delta\in\gamma\land \beta=\delta)}_{\to 0=1}\big)\to 0=1$$

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.