$\Bbb{R}_\ell$ is Lindelöf...Why is this Map Well-defined?

Question

Here is Munkres' proof that $\Bbb{R}_\ell$, the Sorgenfrey line ($\Bbb{R}$ with the lower limit topology), is Lindelöf (example 3 in section 30):

Let $C = \bigcup_{i \in I}(a_i,b_i)$. . .Let $x$ be a point of $\Bbb{R}-C$. We know that $x$ belongs to no open interval $(a_i,b_i)$; therefore $x = a_i$ for some index $i$. Choose such a $i$ and then choose $q_x$ to be a rational number belonging to the interval $(a_i,b_i)$. It follows that if $x$ and $y$ are two points of $\Bbb{R}-C$ with $x < y$, then $q_x < q_y$. . .Therefore the map $x \mapsto q_x$ of $\Bbb{R}-C$ into $\Bbb{Q}$ is injective, so that $\Bbb{R}-C$ is countable.

My question is pretty simple: why is this map--well--a map? It doesn't appear to be well-defined, since there are may rationals in $(a_i,b_i)$.

Answer 1

Since only one point $q_i $ is choosen in each interval $(a_i, b_i) $, we do indeed have a function $f $...

The axiom of choice guarantees we can do this...

Answer 2

Being very explicit:

Enumerate the rationals as $\mathbb{Q} = \{q_1, q_2 ,q_3, \ldots\}$.

Now, if $X$ is is in $\mathbb{R} - C$, we know $I_x = \{i \in I: x_i \in [a_i, b_i)\}$ is a non-emty set, because the $[a_i, b_i), i \in I$ form a cover of $\mathbb{R}$. So we pick one such $i(x) \in I_x$ for each $x$, by the Axiom of Choice. Indeed this means that $x= a_{i(x)}$, because it must be in $[a_{i(x)}, b_{i(x)})$ but cannot be in $(a_{i(x)}, b_{i(x)}) \subseteq C$, as $x \notin C$.

The open interval $(a_{i(x)}, b_{i(x)})$ must contain some rational number so let $n(x)$ be the smallest $n$ such that $q_{n(x)} \in (a_{i(x)}, b_{i(x)})$. Note that by the enumeration of the rationals we can avoid choice here. This $q_{n(x)}$ is then denoted by $q_x$. It's clear such a $q_x$ exists (if we assume choice for the picking of the $i(x)$). It's not a "map" as some formula, but a "choice function" for some simultaneous choices; this happens a lot in topology.

Now, if $x \neq y$ are both in $\mathbb{R} - C$ we must have $x < y$ or $y < x$, so assume WLOG that $x < y$. We know $x =a_{i(x)}$ and again $y \notin (a_{i(x)}, b_{i(x)}) \subseteq C$, so that in fact $y \ge b_{i(x)}$.

Hence we have the situation $$x = a_{i(x)} < q_x = q_{n(x)} < b_{i(x)} \le y = a_{i(y)} < q_{n(y)} = q_y$$ so indeed $x < y$ implies $q_x < q_y$, as claimed.