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 Precedes Next: $$r_0 \le r_{j-1} < r_j \le r_n \Longrightarrow \left({r_0 + 1}\right) \le \left({r_{j-1} + 1}\right) \le r_j \le r_n$$.

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:

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

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 minimal 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 Precedes Next, $$r_{k-1} + 1 \le m \le r_k$$.

Therefore $$m \in A_k$$.