Every well-ordered set is isomorphic to a unique ordinal number.
This is a well-known and important theorem, so i would like to give it a shot by myself. Does my proof look fine, or contain logical gaps and flaws? Thank you for your verification!
My attempt:
Lemma 1: Two different ordinals are not isomorphic.
Lemma 2: The set of ordinals is well-ordered with respect to $\in$.
Lemma 3: An initial segment of an ordinal is an ordinal.
Lemma 4: No well-ordered set is isomorphic to an initial segment of itself.
Let $(W,<)$ be a well-ordered set and $W[a]:=\{x\in W\mid x<a\}$ be an initial segment of $W$ given by $a\in W$.
Let $A:=\{a\in W\mid W[a]\text{ is isomorphic to some ordinal}\}$.
It follows from Lemma 1 that for each $a\in W$, the ordinal, which is isomorphic to $a$, is unique. As a result, we denote it by $\alpha_a$.
Let $S:=\{\alpha_a\mid a\in A\}$. The existence of set $S$ is asserted by Axiom Schema of Replacement. It follows that $S$ is a set of ordinals and thus is well-ordered by Lemma 2.
I claim that $S$ is transitive. Suppose that $\gamma\in\alpha_a\in S$. It follows that $\gamma\subsetneq\alpha_a$ and that $\gamma$ is an initial segement of $\alpha_a$. Let $\psi$ be an isomorphism between $W[a]$ and $\alpha_a$. Then $\psi^{-1}\restriction\gamma$ is an isomorphism between $\gamma$ and an initial segment $W[a']$ of $W[a]$ where $a'<a$. Then $\alpha_{a'}=\gamma$ and thus $\gamma\in S$.
As a result, $S$ is an ordinal. Let $\alpha:=S$.
I claim that $A$ is an initial segment of $W$. Suppose that $a\in A$ and $b\in W$ such that $b<a$. It follows that $W[a]$ is isomorphic to an ordinal $\alpha_a$. Let $\varphi$ be an isomorphism between $W[a]$ and $\alpha_a$. Then $\varphi\restriction W[b]$ is an isomorphism between an initial segment $W[b]$ of $W[a]$ and $\varphi [W[b]]$. It follows that $\varphi [W[b]]$ is an initial segment of $\alpha_a$ and thus is an ordinal by Lemma 3. Then $b\in A$.
As a result, either $A=W$ or $A=W[c]$ for some $c\in W$.
Existence
We define a mapping $f:A\to S=\alpha$ by $f(a)=\alpha_a$.
- $f$ is surjective
Suppose that $\gamma\in\alpha=S$.
From the definition of $S$: $\gamma=\alpha_a$ for some $a\in A$. Then $f(a)=\alpha_a=\gamma$.
- $\forall a,b\in A:a<b\implies f(a)\in f(b)$
It's clear that $a<b\implies f(a)\neq f(b)$. If not, $\alpha_a=\alpha_b$, then $W[a]$ and $W[b]$ are isomorphic, whereas $W[a]$ is an initial segment of $W[b]$. This contradicts Lemma 4.
If $f(b)\in f(a)$, then $\alpha_b\in\alpha_a$ and thus $\alpha_b\subsetneq\alpha_a$. It follows that $W[b]$ is isomorphic to an initial segment of $\alpha_a$ and in turn is isomorphic to an initial segment of $W[a]$, whereas $W[a]$ is an initial segment of $W[b]$. This contradicts Lemma 4.
As a result, $a<b\implies f(a)\in f(b)$.
Hence $f$ is an isomorphism between $A$ and $\alpha$.
We next prove that $A=W$. If not, $A\neq W$ and thus $A=W[c]$ for some $c\in W$. Then $f:W[c]\to \alpha$ is an isomorphism between $W[c]$ and $\alpha$. It follows that $c\in A$ and thus $c<c$. This is a contradiction.
Hence $A=W$ and $f:A\to \alpha$ is an isomorphism between $A$ and $\alpha$.
Uniqueness
By Lemma 1, two different ordinals are not isomorphic. Then $\alpha$ is uniquely determined by $A$.