[proofplan] Fix a well-ordering $X$ of order type $\alpha$. The ordinals below $\alpha$ correspond, as isomorphism classes, to the order types of proper initial segments of $X$. By the characterisation of initial segments, these are exactly the $I_x = \{y \in X : y <_X x\}$ for $x \in X$. The assignment $x \mapsto \operatorname{otp}(I_x)$ is therefore a surjection from $X$ onto the ordinals below $\alpha$, and we show it is order-preserving and injective. This exhibits an order-preserving bijection between $X$ and $I_\alpha := \{\beta : \beta < \alpha\}$, which is exactly the statement that $I_\alpha$ is a well-ordering of order type $\alpha$. [/proofplan] [step:Fix a well-ordering $X$ representing $\alpha$ and set up the candidate map] By definition, an ordinal $\alpha$ is the order type of some well-ordering. Fix a well-ordering $(X, <_X)$ with $\operatorname{otp}(X) = \alpha$. Recall that the comparison order on ordinals is defined by \begin{align*} \beta < \alpha \iff \text{ there is a well-ordering of type } \beta \text{ that is isomorphic to a proper initial segment of } X, \end{align*} independently of the representative of $\alpha$ chosen, and that this is well-defined by the [Antisymmetry](/theorems/1468) of the comparison order on well-orderings. Define \begin{align*} \Phi: X &\to \{\beta : \beta < \alpha\} \\ x &\mapsto \operatorname{otp}(I_x), \quad \text{where } I_x = \{y \in X : y <_X x\}. \end{align*} Note that $I_x$ is a proper initial segment of $X$ (proper because $x \notin I_x$), hence $\operatorname{otp}(I_x) < \alpha$ by the definition of $<$ on ordinals. So $\Phi$ is well-typed. We will prove $\Phi$ is an order-preserving bijection from $X$ onto $I_\alpha := \{\beta : \beta < \alpha\}$. Under the inherited order $<$ on ordinals, this makes $I_\alpha$ a well-ordering of order type $\alpha$. [guided] We need to produce a well-ordering of order type $\alpha$ from the set $I_\alpha = \{\beta : \beta < \alpha\}$. The strategy is to start from *any* concrete representative $X$ of $\alpha$ — the order types of its proper initial segments are the ordinals $<\alpha$, by definition — and then exhibit $I_\alpha$ as a copy of $X$ via the "take the order type of the initial segment below $x$" map. Pick any well-ordering $(X, <_X)$ with $\operatorname{otp}(X) = \alpha$. The definition of $<$ on ordinals, combined with the [Antisymmetry](/theorems/1468) theorem (which certifies that the order type of a well-ordering is a well-defined invariant up to isomorphism), says that $\beta < \alpha$ iff some well-ordering of type $\beta$ embeds isomorphically onto a proper initial segment of $X$. Set \begin{align*} \Phi: X &\to \{\beta : \beta < \alpha\} \\ x &\mapsto \operatorname{otp}(I_x), \quad I_x = \{y : y <_X x\}. \end{align*} Each $I_x$ is a proper initial segment of $X$ (it misses $x$), so $\operatorname{otp}(I_x) < \alpha$, and $\Phi$ lands in $I_\alpha$ as required. The rest of the proof is: $\Phi$ is order-preserving, injective, and surjective onto $I_\alpha$. [/guided] [/step] [step:Prove $\Phi$ is order-preserving] Let $x_1 <_X x_2$. Then $I_{x_1} = \{y : y <_X x_1\}$ is a proper initial segment of $I_{x_2} = \{y : y <_X x_2\}$: it is contained in $I_{x_2}$ (since $y <_X x_1 <_X x_2$ gives $y <_X x_2$), and $x_1 \in I_{x_2} \setminus I_{x_1}$ witnesses strict containment. Moreover $I_{x_1}$ is downward closed in $I_{x_2}$ (the trace of a downward-closed set is downward closed in any superset that is itself downward closed in $X$). By the [Initial Segments of Well-orderings](/theorems/1462) characterisation applied to $I_{x_2}$: $I_{x_1}$ is a proper initial segment of the well-ordering $I_{x_2}$, so \begin{align*} \operatorname{otp}(I_{x_1}) < \operatorname{otp}(I_{x_2}), \end{align*} by the definition of $<$ on ordinals (a proper initial segment has strictly smaller order type). That is, $\Phi(x_1) < \Phi(x_2)$ in $I_\alpha$. [/step] [step:Prove $\Phi$ is injective] Strict monotonicity proved in the previous step immediately gives injectivity: if $x_1 \neq x_2$, the trichotomy of $<_X$ yields $x_1 <_X x_2$ or $x_2 <_X x_1$, and by monotonicity $\Phi(x_1) \neq \Phi(x_2)$. [/step] [step:Prove $\Phi$ is surjective] Let $\beta \in I_\alpha$, i.e., $\beta < \alpha$. By the definition of $<$ on ordinals, there is a well-ordering of order type $\beta$ that is order-isomorphic to a proper initial segment $J \subsetneq X$ of our fixed representative $X$. By [Initial Segments of Well-orderings](/theorems/1462), the proper initial segment $J$ has the form \begin{align*} J = I_x = \{y \in X : y <_X x\} \end{align*} for a unique $x \in X$ (namely $x = \min_X(X \setminus J)$, which exists as a minimum because $X \setminus J$ is a non-empty subset of the well-ordering $X$). Therefore $\operatorname{otp}(I_x) = \operatorname{otp}(J) = \beta$, so $\Phi(x) = \beta$. Hence $\Phi$ is surjective. [guided] Surjectivity is where the structural fact "proper initial segments of $X$ are exactly the $I_x$" does its work. Let $\beta \in I_\alpha$. The meaning of $\beta < \alpha$ is that $\beta$ is the order type of some proper initial segment of $X$ (any representative of $\alpha$ works because the comparison order on ordinals is well-defined, which in turn rests on the [Antisymmetry theorem](/theorems/1468)). So pick a proper initial segment $J \subsetneq X$ with $\operatorname{otp}(J) = \beta$. Now we want to pull $J$ into the form $I_x$ for some $x$, because $\Phi$ is defined on elements of $X$, not on subsets. The [Initial Segments of Well-orderings](/theorems/1462) characterisation says exactly this: every proper initial segment of a well-ordering is of the form $I_x$ for a unique $x$. Explicitly, $x = \min_X(X \setminus J)$, which is well-defined because $X \setminus J$ is a non-empty subset of a well-ordering. With $J = I_x$, we have $\Phi(x) = \operatorname{otp}(I_x) = \operatorname{otp}(J) = \beta$, so $\beta$ is in the range of $\Phi$. [/guided] [/step] [step:Conclude that $I_\alpha$ is a well-ordering of order type $\alpha$] Combining the previous three steps, $\Phi: X \to I_\alpha$ is an order-preserving bijection, where $I_\alpha$ carries the inherited order $<$ from the ordinals. *$I_\alpha$ is totally ordered.* The restriction of a total order to a subset is a total order; since $<$ on the ordinals is a total order (this is the comparison-order totality, cf. [Totality of the Comparison Order](/theorems/1467)), the restriction to $I_\alpha$ is total. *$I_\alpha$ is a well-ordering.* Let $\varnothing \neq S \subseteq I_\alpha$. Then $\Phi^{-1}(S) \subseteq X$ is a non-empty subset of the well-ordering $X$. Let $x_0 := \min_X(\Phi^{-1}(S))$. We claim $\Phi(x_0) = \min_< S$. Indeed, $\Phi(x_0) \in S$ since $x_0 \in \Phi^{-1}(S)$. For any other $\beta \in S$, $\beta = \Phi(x)$ for some $x \in \Phi^{-1}(S)$, so $x_0 \leq_X x$ by minimality; by order-preservation of $\Phi$, $\Phi(x_0) \leq \Phi(x) = \beta$. Thus every non-empty subset of $I_\alpha$ has a minimum, so $I_\alpha$ is a well-ordering. *Order type is $\alpha$.* Since $\Phi$ is an order-preserving bijection from $X$ (of order type $\alpha$) to $I_\alpha$, the two are isomorphic as well-orderings, so $\operatorname{otp}(I_\alpha) = \operatorname{otp}(X) = \alpha$. This completes the proof. [/step]