Ex 3.4.9 Rational numbers at the boundary #84

Hi.

First of all, just wanted to say thanks for documenting these solutions. These are quite helpful when trying to get through the book.

I have a question about the proof for the Exercise 3.4.9 (apologies if was raised before). Specifically this fragment,

To construct an $(r_n)$ which satisfies this condition, we start with an arbitrary $(q_n)$. Define $R_n = \bigcup^n_{i=1} r_i$. For each element of $(q_n)$, we add $q_i$ to $(r_n)$ only if either $V_{\epsilon_n}(q_i) \subseteq R_n$ or $V_{\epsilon_n}(q_i) \cap R_n = \emptyset$. Otherwise, we procrastinate on adding $q_i$ by appending any rational number $s > \max(R_n) + 2 \epsilon_n$. Clearly, for any $q_i$ there will eventually be $\epsilon_n$ small enough that $q_i$ can be added to $(r_n)$ without violating our restrictions, and we don't need to worry about $s$'s being added since they're far enough away from everything that they can't affect the restrictions.

This line of reasoning does not describe rational numbers that are exactly on the boundary of the set we are currently defining. If we have an $\epsilon$ neighbourhood around $x$, we can get an $x + \epsilon$ as our next rational number $r_n$ (both $x$ and $\epsilon$ are rational numbers, so their sum is as well). For $x + \epsilon$, there is no $\epsilon$ small enough such that either of these criteria are met. In fact, the only hope for that number would be if something later swallows it, but since this situation isn't mentioned in here, I suspect it might also be a hole in the argument (and I am not even sure if it's possible due to disjointness condition we imposed earlier).