Strictly Increasing Sequence on Ordered Set

Lemma
Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $\left \langle {r_k} \right \rangle_{p \le k \le q}$ be a sequence of elements of $\left({S, \preceq}\right)$.

Then $\left \langle {r_k} \right \rangle_{p \le k \le q}$ is strictly increasing iff:


 * $\forall k \in \left[{p + 1 .. q}\right]: r_{k - 1} \prec r_k$

Proof

 * Let $\left \langle {r_k} \right \rangle_{p \le k \le q}$ be strictly increasing.

Because $\forall k \in \N^*: k - 1 < k$, it follows directly that $\forall k \in \left[{p + 1 .. q}\right]: r_{k - 1} \prec r_k$.


 * Now suppose $\left \langle {r_k} \right \rangle_{p \le k \le q}$ is not strictly increasing.

Let $K$ be the set of all $k \in \left[{p .. q}\right]$ such that $\exists j \in \left[{p .. q}\right]$ such that $j < k$ and $r_k \preceq r_j$.

The set $K$ is not empty because $\left \langle {r_k} \right \rangle_{p \le k \le q}$ is not strictly increasing.

As $K \subset \N$ and the latter is well-ordered, then so is $K$. Thus it has a minimal element $m$.

Thus there exists $j \in \left[{p .. q}\right]$ such that $j < m$ and $r_m \preceq r_j$.

Then $j \le m - 1$, and $m - 1 \notin K$ as $m - 1 < m$.

So $r_j \preceq r_{m-1} \prec r_m \preceq r_j$, which is a contradiction.

Therefore there can be no such element $m$, and therefore $K$ is empty.

The result follows.