# Lexicographic Order on Products of Well-Ordered Sets

## Theorem

Let $S$ be a set which is well-ordered by $\preccurlyeq$.

Let $\preccurlyeq_l$ be the lexicographic order on the set $T$ of all ordered tuples of $S$.

Then:

- $(1): \quad$ For a given $n \in \N: n > 0$, $\preccurlyeq_l$ is a well-ordering on the set $T_n$ of all ordered $n$-tuples of $S$;
- $(2): \quad \preccurlyeq_l$ is
**not**a well-ordering on the set $T$ of*all*ordered tuples of $S$.

## Proof

It is straightforward to show that $\preccurlyeq_l$ is a total ordering on both $T_n$ and $T$.

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It remains to investigate the nature of $\preccurlyeq_l$ as to whether or not it is a well-ordering.

### Proof for Finite Product

Consider $T_n$ where $n \in \N_{>0}$.

It is clear that $\struct {T_1, \preccurlyeq_l}$ is order isomorphic to $\struct {S, \preccurlyeq}$.

Thus as $\preccurlyeq$ is a well-ordering on $S$, $\preccurlyeq_l$ is a well-ordering on $T_1$.

Now, let us assume that $\preccurlyeq_l$ is a well-ordering on $T_k$ for some $k \in \N: k \ge 1$.

Let $A$ be a non-empty subset of $T_{k + 1}$.

Let $A_1$ be the set of all of the first components of the ordered $n$-tuples that comprise $A$.

Since $A_1$ is a non-empty subset of $S$, and $S$ is itself well-ordered by $\preccurlyeq$, it follows that $A_1$ contains a minimal element $x$ by $\preccurlyeq_l$.

Let $A_x$ be the subset of $A$ in which the first component equals $x$.

We may consider $A_x$ to be a subset of $T_k$ where this first component $x$ has been suppressed.

But we assumed that $T_k$ is well-ordered by $\preccurlyeq_l$.

So $A_x$ contains a minimal element $\tuple {x, x_2, x_3, \ldots, x_{k + 1} }$ by $\preccurlyeq_l$.

This element $\tuple {x, x_2, x_3, \ldots, x_{k + 1} }$ is the minimal element of $A$ by $\preccurlyeq_l$.

Hence, by definition, $T_{k + 1}$ is well-ordered by $\preccurlyeq_l$.

The result follows by induction.

$\blacksquare$

### Proof for Infinite Product

Consider a set $S = \set {a, b}$ such that $a \prec b$.

Then the set $\set {\tuple b, \tuple {a, b}, \tuple {a, a, b}, \tuple {a, a, a, b}, \ldots}$ has no minimal element by $\preccurlyeq_l$.

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Thus $T$ is not well-ordered by $\preccurlyeq_l$.

$\blacksquare$