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  • Let $\alpha$ be an ordinal. Let $z$ be an initial segment of $\alpha$. Then either $z=\alpha$ or $z\in\alpha$. Thus, an initial segment of an ordinal is an ordinal.

Proof. If $z\neq\alpha$, then $\alpha\setminus z$ does have a minimal element, say $x$. Then

$$z=\left\{y\in\alpha : y<\alpha\right\}=\left\{y\in\alpha : y\in\alpha\right\}=x\in\alpha.$$

I have a question in the proof: How did writer connect from $\left\{y\in\alpha : y<\alpha\right\}$ to the set $\left\{y\in\alpha : y\in\alpha\right\}=x\in\alpha.$ Can you explain?

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    $\begingroup$ I think that the chain of equalities contains two typos: it should be $z=\{y\in\alpha:y<\color{red}{x}\}$ and similarly the second one. After you fix them, you should easily see why the conclusion holds. $\endgroup$
    – Leo163
    Jun 18, 2019 at 14:59
  • $\begingroup$ @Leo163 İ think so there is a typo $\endgroup$
    – user295645
    Jun 18, 2019 at 15:05

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