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Each well-orderable set $X$ is equipotent to a unique initial ordinal.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!


My attempt:

Lemma: Every well-ordered set is isomorphic to a unique ordinal.

Existence

By Lemma and Axiom of Choice, $X$ is equipotent to some ordinal $\alpha$. Let $\alpha_0$ be the least ordinal equipotent to $X$. Then $\alpha_0$ is an initial ordinal. If not, $|\alpha_0|=|\beta|=|X|$ for some $\beta<\alpha_0$. This contradicts the minimality of $\alpha_0$.

Uniqueness

Assume the contrary that $X$ is equipotent to initial ordinals $\alpha_0,\alpha_1$ such that $\alpha_0\neq\alpha_1$. WLOG, we assume $\alpha_0<\alpha_1$. Furthermore, $|\alpha_0|=|\alpha_1|$. This contradicts the fact that $\alpha_1$ is an initial ordinal.

Then the cardinal number of $X$, denoted by $|X|$, is defined as the unique initial ordinal equipotent to $X$.

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  • $\begingroup$ Whether "It's clear that $X$ is equipotent to ..." is really clear depends a lot on your assumptions. For instance, if you have assumed the axiom of choice, then it's clear. Otherwise it looks good to me. $\endgroup$ – Arthur Dec 2 '18 at 15:26
  • $\begingroup$ It is not clear that there is such an ordinal. Refer back to the result that allows you to conclude that. Every time you post, the same problem is pointed out to you, and you keep ignoring it. $\endgroup$ – Andrés E. Caicedo Dec 2 '18 at 15:54
  • $\begingroup$ Hi @AndrésE.Caicedo! I have added a lemma as you suggested. Could you please more specific on my same problem? Is it that I appeal to a result without clearly mentioning it? $\endgroup$ – Le Anh Dung Dec 2 '18 at 16:20
  • $\begingroup$ Yes, that's precisely what I meant. $\endgroup$ – Andrés E. Caicedo Dec 2 '18 at 16:36
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Assuming you indeed know that the class of ordinals is linearly well-ordered, this proof looks fine to me. I would say: $\alpha_0$ is initial, because if $\beta < \alpha_0$ it cannot be equipotent to $X$ by minimality of $\alpha_0$ and as $X \simeq \alpha_0$, $\beta \not\simeq \alpha_0$ as otherwise $\beta \simeq X$; and so $\alpha_0$ is initial. It comes down to the same thing, namely using that $\alpha$ is initial iff $\forall \beta < \alpha: \beta \not\simeq \alpha$.

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