Strictly Increasing Sequence induces Partition

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

Suppose that:


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

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

Since:


 * $\left \langle {r_k} \right \rangle_{0 \le k \le n}$ is strictly increasing, and
 * $0 \le j - 1 < j \le n$

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

Also, as $\left \langle {r_k} \right \rangle_{0 \le k \le n}$ 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.

Now we need to establish 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 $k \ne 0$ since $r_0 < m$.

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 $m \in A_k$.