I am studying a proof that every well-ordered set is isomorphic to a unique ordinal. However, I don't understand why $A = pred(\omega)$ (see yellow).
One direction is clear:
Let $x \in pred(\omega)$, then $x < \omega$. Now since $\omega$ is the least element of $W \setminus A$, $x \in A$.
Hint: $pred(x)$ is the set of predecessors of $x$.
In the recording of the lecture (at 31:58) the professor argues that the other direction also follows from the fact that $\omega$ is the least element of $W \setminus A$, but I don't see how. Does the converse really follow from this fact alone?
The only thing I could come up with is the following:
Assume there exists $v \in A$ such that $v \not \in pred(\omega)$. Then $v \not \in pred(\omega)$ implies that $\omega<v$, and so $pred(\omega) \subsetneq pred(v)$. Since $v \in A$, there exists an ordinal $\alpha_v$ s.t. $(pred(v), <)$ is isomorphic to $(\alpha_v, \in_{\alpha_v})$ by the definition of $A$. Let $f$ be the order-isomorphism between those sets and consider its restriction to the set $pred(\omega)$:
$f \restriction_{pred(\omega)} \to f[pred(\omega)]$
Then $f \restriction_{pred(\omega)}$ is also an order isomorphism and $f[pred(\omega)]$ is a proper subset of an ordinal, and thus also an ordinal. But then $\omega \in A$ by the definition of $A$, which contradicts the definition of $\omega$.
This is basically a proof by contradiction within a proof by contradiction (to show that $A = W$), but don't need the fact that $\omega$ is the least element of $W \setminus A$ for this.
Could anyone please clarify the step in the proof? Also is my proof above correct?
Thank you very much!
Edit: My proof actually does not work since a proper subset of an ordinal is not necessarily an ordinal.
Best Answer
Let's recall some basic facts: For a partially ordered set $P$ and $x\in P$, you can more generally define $pred(x)=\{y\in P,\, y<x\}$. These subsets (I've seen them called initial segments) are preserved by order isomorphisms. In other words, if $f:P\to Q$ is some order isomorphism between posets, then $f(pred(x))=pred(f(x))$. Moreover, for ordinals $\alpha\in\beta$ ordered by membership, $pred(\alpha)=\alpha$.
Bearing all that in mind, let's go back to your problem. Using the given notation, notice now how if $v\in A$, then $pred(v)\subset A$. Indeed let $f:pred(v)\to\alpha$ be an isomorphism to an ordinal number, then for any $u\prec v$, $f(pred(u))=pred(f(u))=f(u)$ an ordinal. So the appropriate restriction of $f$ is indeed an isomorphism between $pred(u)$ and the ordinal $f(u)$. Therefore $u\in A$ and $pred(v)\subset A$.
This allows to fix your argument.