> [!definition] > > Let $-\infty < a \le b < \infty$, then a **partition** of the interval $[a, b]$ is a sequence of the form > $ > P = \bracs{x_j}_{0}^n = [a = x_0 < x_1 < \cdots < x_n = b] > $ > where the **size/mesh** of $P$ is > $ > \sigma(P) = \max_{1 \le j \le n}(x_{j} - x_{j - 1}) > $ > A **tagged partition** is a pair $(P = \bracs{x_j}_0^n, c = \bracs{c_{j}}_1^n)$ such that $c_j \in [x_{j - 1}, x_j]$ for each $1 \le j \le n$. > [!definition] > > Let $\scp([a, b])$ be the set of all partitions of $[a, b]$. For each $P = \seqfz{x_j}, Q = \seqfz[m]{y_j} \in \scp([a, b])$, $Q$ is **finer** than $P$ if for every $0 \le j \le n$, there exists $0 \le k \le m$ such that $x_j = y_k$. Denote $P \le Q$ if $Q$ is finer than $P$, then > 1. $\scp([a, b])$ equipped with $\le$ is a [[Directed Set|directed set]]. > 2. If $P \le Q$, then $\sigma(P) \ge \sigma(Q)$. Moreover, the [[Net|net]] $\anglesn{\sigma(P)}_{P \in \scp([a, b])}$ converges to $0$. > > For any $(P, c), (Q, d) \in \scp_t([a, b])$, let $(P, c) \le (Q, d)$ if $P \le Q$, then $\scp_t([a, b])$ inherits the above properties from $\scp([a, b])$. > > *Proof*. $(1)$: $P \le Q$ if and only if $\bracs{x_j: 0 \le j \le n} \subset \bracs{y_j: 0 \le j \le n}$. Thus $\le$ is symmetric and transitive. Let $S = \bracs{x_j: 0 \le j \le n} \cup \bracs{y_j: 0 \le j \le n}$ and enumerate $S = \seqfz[\ell]{z_j}$, then $S \ge P, Q$. > > $(3)$: Let $1 \le j \le m$ and $1 \le k \le n$ be the smallest $k$ such that $y_j \le x_{k}$. As the elements of $P$ and $Q$ are placed in ascending order and $k$ is minimal, $x_{k - 1} \le y_j$. Thus $x_{k - 1} \le y_{j - 1} \le y_j \le x_k$, and $y_{j} - y_{j- 1} \le x_k - x_{k - 1}$. As this holds for each $1 \le j \le m$, $\sigma(P) \ge \sigma(Q)$. > > Lastly, let $\eps > 0$, then there exists $N \in \nat$ such that $(b - a)/N < \eps$, so if > $ > R = \braks{a, a + \frac{b - a}{N}, \cdots, b} > $ > then $\sigma(R) = (b - a)/N < \eps$, and for all $S \in \scp([a, b])$ with $S \ge R$, $\sigma(S) < \eps$. Therefore $\sigma(P) \to 0$.