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Let $D$ be a set with cardinality $\aleph_\alpha$, and give an ordinal number $\beta$ between $\omega_\alpha$ and $\omega_{\alpha+1}$, can $D$ be well-ordered with order type $\beta$?

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    Of course! What is the cardinality of $\beta$?2012-11-28

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Note that $\omega_\alpha\leq\beta<\omega_{\alpha+1}$ means that there is a bijection between $\beta$ and $\omega_\alpha$.

If $D$ has a bijection with $\omega_\alpha$ then it has a bijection with $\beta$ as well. Let $f\colon D\to\beta$ be such bijection, we define $\prec$ on $D$: $x\prec y\iff f(x)

It is not hard to see that $f$ is now an order isomorphism, so $(D,\prec)$ has order type $\beta$, as wanted.

Also note that we didn't use the axiom of choice at any point. We assumed that $D$ was well-ordered to begin with, so the axiom of choice was not needed to have that.

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    So $(D,\prec)=img(f^{-1})$ thus isomorphic to $\beta$.2012-11-29
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What does it mean to be equipotent? It means that there exists a bijection $f:D\to\beta$. Then simply define $R:=\{(a,b)\mid f(a)\le f(b) \}$. This is going to give a well-ordering of type $\beta$.

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    Yes, you are right, then $f$ will be a isomorphism and $(D,R)$ will be the image of $f^{-1}$.2012-11-29