Thimoteus
commented
9 years ago I just answered my own question: the construction doesn't work for half-open intervals, because the union of half-open intervals with rational endpoints (as specified in the construction) will never include the supremum/infimum.
Or in other words, the limit of the construction I did would yield an open interval and not a half-open interval.
So originally I wrote up a convoluted answer to question 8a before rereading it and realizing how it's "supposed" to be done. The problem I have is the convoluted answer I had originally conflicts with what 8b is asking. I'll try to explain the construction I had that I thought meant the negation of 8b:
By lemma 13.1, a topology is just the closure of its basis under unions. So for any open set $U$, we can write it as a union of basis elements: $U = \bigcup B_i$, for an arbitrary index set $I$.
Take an arbitrary $B_i$. It has the form $(a, b)$ for reals $a, b$. Now the question is: can we write $(a, b)$ itself as a union of intervals $(p, q)$ for rationals $p, q$? Supposing that we can do this construction for every interval with real-valued endpoints, we can just unite all the constructions and get the $U$ back as a union of unions of intervals with rational-valued endpoints.
So, we take a decreasing sequence of rationals $p_i$ with $p_0 < b$ and $\lim p_i = a$, an increasing sequence of rationals $q_i$ with $q_0 > a$ and $\lim q_i = b$. Then each interval $(p_i, q_i)$ is a subset of $(a, b)$ and taking the union of all of them gives $(a, b)$ back.