Why does it follow?

Question

In this proof, I am trying to understand why "It follows that $w\in\mathbb{R}$ and that $w$ can be expressed as $\sup\left\{a_{r}\mid r\in\left(0,\frac{\epsilon}{2}\right)\right\}$." So here are my questions:

Here, $\operatorname{dist}\left(a,S\right)=\inf\left\{d\left(a,b\right)\mid b\in S\right\}$.

$1)$ Why/how does it follow?

$2)$ Why did we not introduce $w$ as $\sup\left\{a_{r}\mid r\in\left(0,\frac{\epsilon}{2}\right)\right\}$ first, rather than $\sup\left\{a_{r}\mid r\in\mathbb{R^{+}}\right\}$?

Answer 1

The proof is arguing that $w$ is real as opposed to infinity. This follows from the fact that the $a_r$'s are bounded above, since $a_r\le b_1$ for every $r$ (by B.7.4), and $b_1$ is a real number.

The assertion that we can consider $r\in(0,\frac\epsilon2)$ instead of $r\in{\mathbb R}^+$ follows from the fact that $a_r$ increases as $r$ decreases (B.7.4 again). In other words, wlog we can restrict attention to small positive $r$. We need $r$ to be small because the $a_p$ chosen later has to have a small value of $p$.

I agree the author could have introduced $(0,\frac\epsilon2)$ from the start. Even better, stick with $\mathbb R^+$ and omit the fragment "and that $w$ can be expressed as $\sup\{a_r\mid r\in(0,\frac\epsilon2)\}$". The argument still goes through; just appeal to (B.7.4) again for the reason we can take $p<\frac\epsilon2$.