Strictly Increasing Sequence induces Partition

Theorem

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

Let:

$\forall k \in \closedint 1 n: A_k := \closedint {r_{k - 1} + 1} {r_k}$

Then:

$\set {A_k: k \in \closedint 1 n}$ is a partition of $\closedint {r_0 + 1} {r_n}$.

Proof

First we show that the elements of $\set {A_k: k \in \closedint 1 n}$ are disjoint.

Let $j \in \closedint 1 n$.

We have that:

$\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is strictly increasing

and:

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

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

$r_0 \le r_{j - 1} < r_j \le r_n \implies \paren {r_0 + 1} \le \paren {r_{j - 1} + 1} \le r_j \le r_n$

So:

$\O \subset A_j \subseteq \closedint {r_0 + 1} {r_n}$

Because $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is strictly increasing:

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

So the elements of $\set {A_k: k \in \closedint 1 n}$ are disjoint, as we were to show.

It remains to be established that:

if $m \in \closedint {r_0 + 1} {r_n}$

then:

$m \in A_k$ for some $k \in \closedint 1 n$.

Let $m \in \closedint {r_0 + 1} {r_n}$.

Consider the set:

$J = \set {j \in \closedint 0 n: m \le r_j}$

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

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

Then because $r_0 < m$:

$k \ne 0$

Thus:

$k \in \closedint 1 n$

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): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.3$