Need help understanding proof that $\text{gap} P_{1} \leq \text{gap} P_{2}$ under certain conditions. I have the following question:

Let $P_{1}$ and $P_{2}$ be partitions of $[a, b]$. Show that if $P_{1}$ is a refinement of $P_{2}$ then $\text{gap } P_{1} \leq \text{ gap } P_{2}$.

Here's a proof I have but don't understand:
Suppose $P_{1}$ is a refinement of $P_{2}$. So, we can let $P_{2} = \{x_{0}, \ldots x_{n}\}$ and $P_{1} = \{x_{0}', \ldots, x_{n}', \ldots x_{n + r}'$}.  So
$$ \text{gap } P_{1} = \text{max}\{x_{i}' - x_{i - 1}'\} \leq \text{max}\{x_{1} - x_{i - 1}\} = \text{gap } P_{2}.$$
Thus,
$$\text{gap} P_{1} \leq \text{gap} P_{2} $$
So, I'm not sure if this proof is even correct. Why are they defining $P_{1}$ to be a larger set than $P_{2}$? Any explanation would be nice. Thanks
 A: I don't think the proof is correct. It is best to rename $P_1,P_2$ as $P', P$ respectively. If $P'$ is a refinement of $P$ it means that $P\subseteq P'$ ie $P'$ is obtained by adding some more points in $P$. Without loss of generality it is sufficient to prove the result when $P'$ has exactly one more point extra compared to $P$.
Let $$P=\{x_0,x_1,\dots,x_{k-1},x_{k},\dots,x_n\} $$ and $$P'=\{x_0,x_1,\dots, x_{k-1},x',x_{k},\dots,x_n\}$$ Note that $$x_k-x_{k-1}= (x_k-x') +(x'-x_{k-1})$$ The LHS as well as each expression in parentheses on RHS is positive and hence $$x'-x_{k-1}<x_k-x_{k-1},x_k-x'<x_k-x_{k-1}$$ and therefore by applying definition of definition of gap of a partition we can see that $\operatorname {gap} P'\leq \operatorname {gap} P$ as desired. 
A: Here is another method to prove this
Let  $P'=\{y_0,y_1,…,y_m \}$ be a refinement of $P=\{x_0,x_1,…,x_n \}$ for some $n,m\in\mathbb{N}$.
Suppose $\text{gap }P'>\text{gap }P$. Then $\text{gap }P'=y_j-y_{j-1}$ for some $1≤j≤m$ and $y_j-y_{j-1}>x_i-x_{i-1}$ for all $1≤i≤n$. Also $x_i-x_{i-1}\geq x_k-x_l$ for all $1\leq l<k\leq n$.
But since $y_j,y_{j-1}∈P$, we have $y_j-y_{j-1}>y_j-y_{j-1}$ which is a contradiction.
