Restriction of Total Ordering is Total Ordering

Theorem
Let $\left({S, \preceq}\right)$ be a total ordering.

Let $T \subseteq S$.

Let $\preceq'$ be the restriction of $\preceq$ to $T$.

Then $\preceq'$ is a total ordering of $T$.

Proof
By Restriction of Ordering is Ordering, $\preceq'$ is an ordering.

Let $x, y \in T$.

As $T \subseteq S$ it follows by definition of subset that:
 * $x, y \in S$

As $\preceq$ is a total ordering:
 * $\left({x, y}\right) \in {\preceq}$

or:
 * $\left({y, x}\right) \in {\preceq}$

Suppose $\left({x, y}\right) \in {\preceq}$.

As $x, y \in T$, it follows by definition of cartesian product that:
 * $\left({x, y}\right) \in T \times T$

Thus:
 * $\left({x, y}\right) \in \left({T \times T}\right) \cap {\preceq}$

By definition of the restriction of $\preceq$ to $T$:
 * $\left({T \times T}\right) \cap {\preceq} = {\preceq'}$

That is:
 * $\left({x, y}\right) \in {\preceq'}$

A similar argument shows that:
 * $\left({y, x}\right) \in {\preceq} \implies \left({y, x}\right) \in {\preceq'}$

Thus $\preceq'$ is a total ordering of $T$.