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 $X$.
Best Answer
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 $\forall \beta < \alpha: \beta \not\simeq \alpha$.