Unique Isomorphism between Equivalent Finite Totally Ordered Sets

Theorem
Let $S$ and $T$ be finite sets such that:
 * $\left|{S}\right| = \left|{T}\right|$

Let $\left({S, \preceq}\right)$ and $\left({T, \preccurlyeq}\right)$ be totally ordered sets.

Then there is exactly one order isomorphism from $\left({S, \preceq}\right)$ to $\left({T, \preccurlyeq}\right)$.

Proof
It is sufficient to consider the case where $\left({T, \preccurlyeq}\right)$ is $\left({\N_n, \le}\right)$ for some $n \in \N$.

Let $A$ be the set of all $n \in \N$ such that if:


 * $(1): \quad S$ is any set such that $\left|{S}\right| = n$, and
 * $(2): \quad \preceq$ is any total ordering on $S$

then there is exactly one isomorphism from $\left({S, \preceq}\right)$ to $\left({\N_n, \le}\right)$.

$\varnothing \in A$ from Empty Mapping is Mapping and Same Cardinality Bijective Injective Surjective.

Now let $n \in A$.

Let $\left({S, \preceq}\right)$ be a totally ordered set with $n + 1$ elements.

By Finite Subset of Totally Ordered Set, $S$ has a greatest element $b$.

Then $\left|{S \setminus \left\{{b}\right\}}\right| = n$ by Cardinality Less One.

The total ordering on $S \setminus \left\{{b}\right\}$ is the one induced from that on $S$.

So as $n \in A$, there exists a unique order isomorphism $f: S \setminus \left\{{b}\right\} \to \left({\N_n, \le}\right)$.

Let us define the mapping $g: S \to \N_{n+1}$ as follows:


 * $\forall x \in S: g \left({x}\right) = \begin{cases}

f \left({x}\right): & x \in S \setminus \left\{{b}\right\} \\ n: & x = b \end{cases}$

This is the desired order isomorphism from $\left({S, \preceq}\right)$ to $\left({\N_{n+1}, \le}\right)$.

Now let $h: \left({S, \preceq}\right)$ to $\left({\N_{n+1}, \le}\right)$ be an order isomorphism.

Then $h \left({b}\right) = n$ (it has to be).

So the restriction of $h$ to $S \setminus \left\{{b}\right\}$ is an isomorphism from $S \setminus \left\{{b}\right\}$ to $\left({\N_n, \le}\right)$, and hence $h = f$ (as $n \in A$, any such order isomorphism is unique).

Thus:
 * $\forall x \in S \setminus \left\{{b}\right\}: h \left({x}\right) = f \left({x}\right) = g \left({x}\right)$

and:
 * $h \left({b}\right) = n = g \left({b}\right)$

thus $h = g$.

Therefore $n + 1 \in A$.

The result follows from the Principle of Finite Induction.