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Let $A$ and $B$ be well-ordered sets such that $A \subseteq B$ and suppose $\alpha$ and $\beta$ are ordinals such that A is order-isomorphic to $\alpha$ and B is order-isomorphic to $\beta$.

Show that $\alpha \le \beta$ (so $\alpha = \beta$ or $\alpha \in \beta$).

This seems like it must be true, but I haven't been able to figure out a proof. My idea was to show that $\beta \nless \alpha$ and use linearity of ordinals, but I'm not sure how to do this.

Also, I'm pretty new to studying ordinals (working through Goldrei's book "Classical Set Theory" on my own), so maybe there's something obvious I'm missing!

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Let's see. If $\alpha\cong A\subseteq B\cong\beta$ then $\alpha$ is isomorphic to a subset of $\beta,$ right? So there is a strictly increasing function $f:\alpha\to\beta.$

We want to show that $\alpha\le\beta,$ so we assume $\beta\lt\alpha$ and try to derive a contradiction.

You can easily show (if you have not already shown) that, if $A$ is a well-ordered set and $f:A\to A$ is a strictly increasing function, that $f(x)\ge x$ for all $x\in A.$

In particular, then, if $\alpha$ is an ordinal and $f:\alpha\to\beta\le\alpha$ is a strictly increasing function, then $f(\xi)\ge\xi$ for all $\xi\lt\alpha.$ If $\beta\lt\alpha$ that means $f(\beta)\ge\beta,$ contradicting our assumption that $f:\alpha\to\beta.$

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