For example the norm of a partition is the widest subinterval
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$\begingroup$Here is a formal definition of partitions of intervals:
In mathematics, a partition of an interval $[a, b]$ on the real line is a finite sequence $x_0, x_1, x_2, ..., x_n$ of real numbers such that$$ a = x_0 < x_1 < x_2 < ... < x_n = b. $$
Such partition gives you $n$ "subintervals" of $[a,b]$,
$$
[x_{k-1},x_{k}],\quad k=1,\cdots,n\tag{1}
$$
The norm (or mesh) of the partition$$ x_0 < x_1 < x_2 < ... < x_n $$is the length of the longest of the subintervals in (1):$$ \max\{|x_i − x_{i−1}| : i = 1, ... , n \}. $$
To see a simple example for the interval $[0,10]$, take the partition:$$ 0<1<6<8.5<10 $$This partition gives you four subintervels:$$ [0,1],\quad [1,6],\quad [6,8.5],\quad [8.5,10] $$and the norm of this partition is $5$.
$\endgroup$ $\begingroup$A partition of an interval is a division of the interval into subintervals; the subintervals are the ‘pieces’ of the partition, so to speak, the subintervals of which the partition is composed.
The norm of a partition is not the sidest subinterval of the partition: it is the width of the widest subinterval. Consider, for instance, the partition of $[0,1]$ into subintervals $\left[0,\frac13\right],\left[\frac13,\frac12\right]$, and $\left[\frac12,1\right]$. The partition is this collection of subintervals. The lengths of the subintervals are $\frac13,\frac16$, and $\frac12$, so the norm of the partition is $\frac12$; it is *not the interval $\left[\frac12,1\right]$.
$\endgroup$ 4 $\begingroup$A partition of a set $S$ is a collection of subsets $S_i\subset S$ such that the union $\bigcup S_i=S,$ and each distinct $S_i$ have empty intersection. In the case of $S=[a,b]\subset \Bbb R^1$, this gives you non-overlapping subintervals of $[a,b]$ which together forms the entirety of $[a,b]$.
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