User:Dfeuer/Lexicographic Ordering is Ordering

Theorem
Let $(I,\preceq)$ be a well-ordered set.

Let $(X_i,\le_i)$ be an Ordered Set for each $i \in I$.

Let $X = \prod_{i \mathop\in I} X_i$.

Let $\le$ be the lexicographic ordering on $X$.

Then $\left({X,\le}\right)$ is an ordered set.

Proof
Reflexivity: Follows immediately from the definition of lexicographic ordering.

Transitivity: Suppose that $x \le y$ and $y \le z$. If $x = y$ or $y = z$, the result holds trivially. Assume, then, that neither of these is true.

Let $M = \left\{{i \in I \mid x_i \ne y_i}\right\}$ and

let $N = \left\{{i \in I \mid y_i \ne z_i}\right\}$.

Since $x \ne y$ and $y \ne z$, by the definition of product, $M$ and $N$ are non-empty.

Since $I$ is well-ordered and $M,\,N \subseteq I$, $M$ has a least element $m$ and $N$ has a least element $n$.

Let $w = \min\left\{{m,n}\right\}$.

Then if $i < w$, $x_i = y_i$ and $y_i = z_i$, so $x_i = z_i$.

Furthermore, we know that $x_w \le_w y_w$, $y_w \le_w z_w$, and either $x_w <_w y_w$, $y_w <_w z_w$, or both. By transitivity of $\le_w$, $x_w < z_w$. Thus by the definition of lexicographic ordering, $x < z$, so $x \le z$.