Norm of Refinement is no Greater than Norm of Subdivision

Theorem

Let $P$ be a finite subdivision of $\closedint a b$.

Let $P'$ be a refinement of $P$.

Then:

$\norm {P'} \le \norm P$

where $\norm P$ denotes the norm of $P$.

Proof

Let $P = \set {x_0, x_1, \dotsc, x_{n - 1}, x_n}$.

Let $P' = \set {y_0, y_1, \dotsc, y_{m - 1}, y_m}$.

By definition of refinement:

$P \subseteq P'$

Therefore, every $x_i \in P'$.

Thus, every $i \in \set {0, 1, \dotsc, n - 1, n}$, define $k_i \in \set {0, 1, \dotsc, m - 1, m}$ such that:

$y_{k_i} = x_i$

Now, for each $j \in \set {1, 2, \dotsc, m - 1, m}$, let $i_j$ be the smallest $i$ such that $k_i \ge j$.

Since $k_{i_j} \ge j$, it follows that $y_j \le y_{k_{i_j}}$.

Therefore, by definition of $k_i$:

$y_j \le x_{i_j}$

Aiming for a contradiction, suppose $k_{i_j - 1} > j - 1$.

Then, $k_{i_j - 1} \ge j$, contradicting the fact that $i_j$ is the smallest $i$ satisfying the property.

Therefore, by Proof by Contradiction, $k_{i_j - 1} \le j - 1$.

Hence, $x_{i_j - 1} = y_{k_{i_j - 1}} \le y_{j - 1}$.

Let $Z = \norm {P'}$ be the norm of $P'$.

By definition, there is some $j \in \set {1, 2, \dotsc, m - 1, m}$ such that:

$y_j - y_{j - 1} = Z$

But from:

$x_{i_j - 1} \le y_{j - 1} \le y_j \le x_{i_j}$

it follows that:

$x_{i_j} - x_{i_j - 1} \ge y_j - y_{j - 1} = Z$

Therefore, by definition of the norm of $P$:

$\norm P \ge Z = \norm {P'}$

$\blacksquare$