Use of choice function in Urysohn's lemma? Let $A = \mathbb Q \cap [0,1]$. In the proof of Urysohn's lemma, we construct a family of open set such that: $\{V_q\}_{q \in A}$, where if $r_1 < r_2$, then $V_{r_1} \subset \subset V_{r_2}$.
The construction was made possible by some choice $\alpha: A \rightarrow \{\text{open set}\}$. Is there a particular reason why we chose $A = \mathbb Q \cap [0,1]$, i.e. a countable, dense set of $[0,1]$? Why can't we choose $A = [0,1]$? We still have a choice function $\alpha: [0,1] \rightarrow \{\text{open sets}\}$ such that if $r_1 < r_2$, then $\alpha(r_1) \subset\subset \alpha(r_2)$.
 A: The proof does not assume there is a function $\alpha$ as you describe. The construction of the $V_q$ for $q\in Q=\mathbb{Q}\cap(0,1)$ uses a choice function $C$ defined on the set of pairs $(F,U)$, where $F$ is closed, $U$ is open, and $F\subseteq U$; the codomain of $C$ is the family of open sets and $C(F,U)$ is such that $F\subseteq C(F,U)\subseteq\overline{C(F,U)}\subseteq U$.
Given the disjoint closed sets $A$ and $B$ the construction is by recursion on $\mathbb{N}$ after you enumerate $Q$ as $\{q_n:n\in\mathbb{N}\}$ with $q_0=0$ and $q_1=1$.  First one takes $V_{q_0}=C(a,X\setminus B)$ and $V_1=X\setminus B$.At stage $n\ge2$ look where $q_n$ is relative to $\{q_i:i<n\}$ take the largest $q_i$ and smallest $q_j$ with $q_i<q_n<q_j$ and let $V_{q_n}=C(\overline{V_{q_i}}, V_{q_j})$; as $q_0<q_n<q_1$ there are always such $i$ and $j$.
The reason we use this $Q$ is that it is countable and hance that we can do the recursion along $\mathbb{N}$.
You can then define $V_x$ for all $x\in(0,1)$ by $V_x=\bigcup\{V_q:q\in Q, q<x\}$.
