Woset is Isomorphic to Set of its Initial Segments

Theorem
Let $\left({S, \preceq}\right)$ be a well-ordered set.

Let:
 * $A = \left\{{a^\prec: a \in S}\right\}$

where $a^\prec$ is the strict lower closure of $S$ determined by $a$.

Then:
 * $\left({S, \preceq}\right) \cong \left({A, \subseteq}\right)$

where $\cong$ denotes order isomorphism.

Proof
Define $f: S \to A$ as:
 * $\forall a \in S: f \left({a}\right) = a^\prec$

where $a^\prec$ is the initial segment determined by $a$.

$f$ is surjective
$f$ is trivially surjective by the definition of $A$.

$f$ is strictly increasing
Let $x, y \in S$ with $x \prec y$.

Let $z \in f \left({x}\right)$.

Then $z \prec x$ by the definition of initial segment.

By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is also transitive.

Thus $z \prec y$.

Thus by the definition of initial segment, $z \in y^\prec = f \left({y}\right)$.

As this holds for all such $z$, $f \left({x}\right) \subseteq f \left({y}\right)$.

Since $x \prec y$, $x \in y^\prec = f \left({y}\right)$.

But since $\prec$ is antireflexive, $x \nprec x$, so $x \notin f \left({x}\right)$.

Thus $f \left({x}\right) \subsetneqq f \left({y}\right)$.

As this holds for all such $x$ and $y$, $f$ is strictly increasing.

Since a well-ordering is a total ordering, $f$ is an order embedding by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing.

Thus $f$ is a surjective order embedding and therefore an order isomorphism.