Strictly Increasing Sequence on Ordered Set

Theorem
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 $$\forall k \in \mathbb{N}^*: k - 1 < k$$, 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 $$K \subset \mathbb{N}$$ 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.