# Strictly Increasing Sequence induces Partition

## Theorem

Let $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ be a strictly increasing finite sequence of natural numbers.

Let:

- $\forall k \in \left[{1 \,.\,.\, n}\right]: A_k := \left[{r_{k-1} + 1 \,.\,.\, r_k}\right]$

Then:

- $\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ is a partition of $\left[{r_0 + 1 \,.\,.\, r_n}\right]$.

## Proof

First we show that the elements of $\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ are disjoint.

Let $j \in \left[{1 \,.\,.\, n}\right]$.

We have that:

- $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ is strictly increasing

and:

- $0 \le j - 1 < j \le n$

Thus by Sum with One is Immediate Successor in Naturally Ordered Semigroup:

- $r_0 \le r_{j-1} < r_j \le r_n \implies \left({r_0 + 1}\right) \le \left({r_{j-1} + 1}\right) \le r_j \le r_n$

So:

- $\varnothing \subset A_j \subseteq \left[{r_0 + 1 \,.\,.\, r_n}\right]$

Because $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ is strictly increasing:

- $1 \le j < k \le n \implies A_j \cap A_k = \varnothing$

So the elements of $\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ are disjoint, as we were to show.

It remains to be established that:

- if $m \in \left[{r_0 + 1 \,.\,.\, r_n}\right]$

then:

- $m \in A_k$ for some $k \in \left[{1\,.\,.\, n}\right]$.

Let $m \in \left[{r_0 + 1 \,.\,.\, r_n}\right]$.

Consider the set:

- $J = \left\{{j \in \left[{0 \,.\,.\, n}\right]: m \le r_j}\right\}$

Clearly $j \ne \varnothing$ as $n \in J$.

Let $k$ be the smallest element of $J$.

Then because $r_0 < m$:

- $k \ne 0$

Thus:

- $k \in \left[{1 \,.\,.\, n}\right]$

By its definition:

- $r_{k-1} < m \le r_k$

Thus, by Sum with One is Immediate Successor in Naturally Ordered Semigroup:

- $r_{k-1} + 1 \le m \le r_k$

Therefore $m \in A_k$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 18$: Theorem $18.3$