Consider the interval $[1, 5]$ and the partition $P = \{1, 3, 4, 5\}$, what is the mesh of this partition?
So i'd say the mesh of this is $\text{mesh}(P) = \max{(|P_{i} - P_{i - 1}|)}$ for $i = 1, ..., 4$.
Now what do I do here? If we choose $P_{1} = 1$ there is no $P_{0} \in P$.
When an interval $[a,b]$ is partitioned into $n$ parts, the standard notation is to use the subscript $0$ for $a$ and subscript $n$ for $b$, and if as you do the partition is $P$ and you just put your subscripts on the letter $P,$ then the $3$ part partition you have of $[1,5]$ would have interval endpoints (in the standard notation) $P_0=1,P_1=3,P_2=4,P_3=5.$
In this notation, the $k$-th subinterval is $[P_{k-1},P_k]$ and has length $P_k-P_{k-1}.$ Because of the convention, there is always a $P_0$ namely the left end of the partitioned interval.
Note that the number of dividing points is always one more than the number of intervals of the partition, a fact sometimes found to be confusing on first doing partitions.
As you say, mesh(P) $=\max\{P_i-P_{i-1}:1\le i\le 4\}$ where $ P_0=1, P_1=3, P_2=4, P_3=5$,
so mesh(P)=2.
