It's commonly known that the cardinality of the set of all well-orders on $\aleph_0$ is the continuum (correct me if I'm wrong plz). What about that of all well-orders on $\mathbb{R}$? Is there a pattern here resembling the GCH? In general, what results do we have regarding the set of all well-orders on a particular cardinal?
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$\begingroup$ It may help you to know that "the number of well orders of $X$, up to isomorphism" is the definition of $|X|^{+}$, and always exists (in ZF). Using the axiom of choice, we can prove that "the next cardinal above $|X|$" exists and is equal to $|X|^{+}$, but without the axiom of choice we cannot prove that there is a "next", so this isn't suitable as a general definition of $|X|^{+}$. $\endgroup$– Robert FurberCommented Oct 29 at 12:37
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$\begingroup$ @RobertFurber Thank you a lot for this response! Since GCH allows $|X|^+$ to be defined by $P(X)$ (correct me if this is a misunderstanding), this got me to think about the relation between power sets and well orders, about which I posted another question. Please check it out if you have time! $\endgroup$– Edouard JiCommented Nov 2 at 7:47
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Every well order of the real numbers has some order type between $\mathfrak{c}$ and $\mathfrak{c}^+$, and for any given order type arising, every permutation of $\mathbb{R}$ induces another well order of the same order type and conversely. So the total number of well-orders of $\mathbb{R}$ is exactly $$\mathfrak{c}^+\cdot\mathfrak{c}^{\mathfrak{c}}=\mathfrak{c}^{\mathfrak{c}}=2^{\mathfrak{c}}=\beth_2.$$
Similarly, for any infinite cardinal $\kappa$, there will be $\kappa^+$ many possible order types, and again for each order type we have $\kappa^\kappa$ many rearrangements of it, making $\kappa^+\cdot\kappa^\kappa=\kappa^\kappa=2^\kappa$ many well orders in all.
answered Oct 28 at 16:11