why does $u$ lies in an interval $I_k=[x_{k-1},x_k]$ of $\dot P_1$ in Example 7.1.4, introduction to real analysis by Bartle & Sherbert?

Question

I want to refer that similar question was asked here before (with no answers)

Let $g:[0,3]\to \mathbb{R}$ be defined by $g(x)=2$ for $0\le x\le 1,$ and $g(x)=3$ for $1<x \le3.$

Let $\dot P$ be a tagged partition of $[0,3]$ with norm$ < \delta$; we will show how to determine $\delta$ in order to ensure that $|S(g;\dot P)-8|<\epsilon.$ Let $\dot P_1$ be the subset of $\dot P$ having its tags in $[0,1]$ where $g(x)=2,$ and let $\dot P_2$ be the subset of $\dot P$ with its tags in $(1,3]$ where $g(x)=3$. It is obvious that we have $S(g;\dot P)=S(g;\dot P_1)+S(g;\dot P_2).$

the proof on the book If we let $U_1$ denote the union of the subintervals in $\dot P_1,$ then it is readily shown that $$[0,1-\delta]\subset U_1 \subset [0,1+\delta]\tag1.$$ (i) $[0,1-\delta]\subset U_1$
let $u\in[0,1-\delta].$ Then $\color{blue}{u \ lies \ in \ an \ interval}$ $ \color{blue}{I_k=[x_{k-1},x_k]}$ $ \color{blue}{of \dot P_1 }$ and since $||\dot P||<\delta,$ we have $x_k-x_{k-1}<\delta.$ Then $x_{k-1}\le u\le 1-\delta$ implies that $x_k< x_{k-1}+\delta \le(1-\delta)+\delta \le 1.$ Thus the tag $t_k$ in $I_k$ satisfies $t_k \le 1$ and therefore $u$ belongs to a subinterval whose tag is in $[0,1,]$ that is, $u\in U_1.$

I don't understand how do the author reach the conclusion in blue and also if that was true if for any $u \in [0,1-\delta] \ then $ u lies on $I_k=[x_{k-1},x_k]$ of $ \dot P_1 $then what else to prove ? I mean $U_1$ is defined as the union of all sub interval and since $u \in I_k$ then $u \in U_1$ in my first reed to that section for some reason I read $ \dot P_1 $ as $ \dot P$ and I didn't have any problem with that since $I_k $ is a part of $ \dot P $ anyway

Answer 1

The intervals belonging to $\dot{P}$ actually fall into three groups:

1. Intervals $I_k$ that are subsets of $[0, 1]$. All of these intervals will fall in $\dot{P}_1$, since the tag $t_k$ will always fall within this interval.

2. Intervals $I_k$ that are subsets of $(1, 3]$. All of these intervals will fall in $\dot{P}_2$ for similar reasons.

3. Intervals $I_k$ that have non-empty intersections with both $[0, 1]$ and $(1, 3]$. Any such intervals could belong to either $\dot{P}_1$ or $\dot{P}_2$, since it's possible to choose a tag $t_k$ on either side of $x = 1$.

If you look closely, group 3 in fact contains one interval, and that's whichever $I_k$ happens to contain 1. Now, since the norm of $\dot{P}$ is less than $\delta$, that means that the width of this $I_k$ must also be less than $\delta$, meaning that the furthest left it can be is $[1 - \delta, 1]$ and the furthest right is $[1, 1 + \delta]$.

That means that if you take any point in $[0, 1 - \delta]$, it cannot be in that group 3 interval, it must be in one of the intervals from group 1, which we have said is a subset of $\dot{P}_1$, which is why we have $[0, 1 - \delta] \subset U_1$.

Then if we consider an arbitrary point in an interval from $\dot{P}_1$, we know that it must be within $[0, 1 + \delta]$ because there are two possibilities:

1. The point is in one of the intervals from group 1, which are all contained within $[0, 1]$; or

2. The interval from group 3 happens to be in $\dot{P}_1$, and we know that the largest value a point from that interval can take is $1 + \delta$.

That's what lets us say that $U_1 \subset [0, 1 + \delta]$.