TITLE: Uncertainity regarding the proof from Rosenlicht's "Introduction to Analysis"
QUESTION [0 upvotes]: I would be very grateful if somebody could explain me the part in the red rectangle in the proof below. I don't quite get it why the author claims that if $\delta$ is sufficiently small then $p+1\le q-1$. I don't get it why it is the consequence of the inequality $\beta-\alpha \le x_q-x_{p-1}<(q-p+1)\delta$. Thank you for the explanation.

REPLY [1 votes]: If $\delta \leq (\beta - \alpha)/2$, then 
$$\beta - \alpha < (q-p+1)\delta \leq (q-p+1)(\beta-\alpha)/2,$$
hence 
$$2 < q - p + 1.$$
The desired inequality follows from this.
Put simply, as the step $\delta$ of the partition becomes small, some $x_i$'s must begin to appear between $\alpha$ and $\beta$. When that happens, the first $x_i$ to the right of $\alpha$ is less than or equal to the last $x_i$ to the left of $\beta$. The calculation given determines how small this $\delta$ has to be.
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