The problem was taken from “Introduction to Real Analysis” by R. Bartle and D. Sherbert, Section 7.1, Exercise 4.
Let $\dot{\mathcal{P}}$ be a tagged partition of $[0,3]$.
(a) Show that the union $U_1$ of all subintervals in $\dot{\mathcal{P}}$ with tags in $[0,1]$ satisfies $[0,1-||\dot{\mathcal{P}}||]\subset U_1\subset [0,1+||\dot{\mathcal{P}}||]$.
(b) Show that the union $U_2$ of all subintervals in $\dot{\mathcal{P}}$ with tags in $[1,2]$ satisfies $[1+||\dot{\mathcal{P}}||,2-||\dot{\mathcal{P}}||]\subset U_2\subset [1-||\dot{\mathcal{P}}||,2+||\dot{\mathcal{P}}||]$.
For part (a), I mainly have problem on proving $[0,1-||\dot{\mathcal{P}}||]\subset U_1$:
Suppose $x\in [0,1-||\dot{\mathcal{P}}||]$.
I was thinking that if there exists an interval $[x_{i-1},x_i]$ such that $x_{i-1}\le x\le x_i$, then I would have to argue that the tag of $[x_{i-1},x_i]\in [0,1]$. But I have no idea to prove either of them.
For part (b), I found a solution online but I have no idea how the proof of $[1+||\dot{\mathcal{P}}||,2-||\dot{\mathcal{P}}||]\subset U_2$ works. The proof and my question are as follows:
Suppose $1+||\dot{\mathcal{P}}||\le v\le 2-||\dot{\mathcal{P}}||$. Further suppose $v\in [x_{i-1},x_i]$ (Why can we do so?). Then $1+ ||\dot{\mathcal{P}}||\le x_i$ which implies $1\le x_i-||\dot{\mathcal{P}}||\le x_{i-1}\le t_i$ and $x_{i-1}\le 2-||\dot{\mathcal{P}}||$ which implies $t_i\le x_i\le x_{i-1}+ ||\dot{\mathcal{P}}||\le 2$. Therefore we get $t_i\in [1,2]$.
In this part, I only don’t understand why we can let $v\in [x_{i-1},x_i]$.