Doubt involving set inclusion step in Rudin's proof of Urysohn's lemma

Question

In the proof of the proposition, $X$ is normal $\implies$ $\forall E,F \subset X$ disjoint and closed $\exists f : X \rightarrow [0,1]$ continuous such that $\forall x \in E, f(x) = 0$ and $\forall x \in F, f(x) = 1$. In the proof they construct $\{ U_{q} : q \in D\}$, where $D$ denotes the dyadic rationals, which has the property that $p,q \in D, p < q \implies \overline{U_{p}} \subset U_{q}$.

In particular, the claim I am having trouble with is: given $U(\frac{k}{2^{n}})$, we can find $U(\frac{2k+1}{2^{n+1}})$ such that $\overline{U(\frac{k}{2^{n}})} \subseteq U(\frac{2k+1}{2^{n+1}}) \subseteq \overline{U(\frac{2k+1}{2^{n+1}})} \subseteq U(\frac{k+1}{2^{n}})$, which is guaranteed by the fact that $X$ is a normal space. In the case when, $n = 1$, I understand that $E = \overline{U(0)} \subseteq U(1/2) \subseteq \overline{U(1/2)} \subseteq U(1) = X \setminus F$, because $N_{E} \cap N_{F} = \emptyset \implies N_{E} \subset X \setminus F$, where $N_{E}$ is the open neighborhood of $E$, and so $U(1/2) = N_{E}$, but I am unable to interpolate for higher $n$. Please refer to this post: here .

Answer 1

It is well-known that a space $X$ is normal if and only for each closed $C \subset X$ and each open $V \subset X$ such that $C \subset V$ there exists an open $W \subset X$ such that $C \subset W \subset \overline W \subset V$.

We inductively construct the $U(\frac{k}{2^n})$ such that $\overline{U(\frac{k}{2^n}) }\subset U(\frac{k+1}{2^n})$ for $k = 0,\ldots,2^n-1$. If this has been done for some $n$, we get the $U(\frac{k}{2^{n+1}})$ by defining $U(\frac{2k}{2^{n+1}}) = U(\frac{k}{2^{n}})$ and choosing the $U(\frac{2k+1}{2^{n+1}})$ such that $$\overline{U(\frac{2k}{2^{n+1}})} = \overline{U(\frac{k}{2^n})} \subset U(\frac{2k+1}{2^{n+1}}) \subset \overline{U(\frac{2k+1}{2^{n+1}})} \subset U(\frac{k+1}{2^n}) = U(\frac{2k+2}{2^{n+1}})$$ for $k = 0,\ldots,2^n-1$.

Answer 2

This is an add-on to the answer given by @Paul Frost. I will prove why $X$ is normal $\iff \forall C \subset X$ closed, $V \subset X : C \subset V$, $V$ open, $\exists W \subset X$, $W$ open, such that $C \subset W \subset \overline{W} \subset V$ $(2)$. Let the standard definition of normality be $\forall E,F \subset X : E \cap F = \emptyset$, $E,F$ closed sets, $\exists N_{E}, N_{F} \subset X$ open such that $E \subset N_{E} , F \subset N_{F}$ with $N_{E} \cap N_{F} = \emptyset$ $(1)$ .
$(\implies)$ Let $X$ be a normal space as in $(1)$. Let $C \subset X$ be a closed set and $V \subset X$ open such that $C \subset V$. Observe that $T = X \setminus C$ is a closed set such that $C \cap T = \emptyset$. By definition $(1)$, $\exists N_{C} , N_{T} \subset X : N_{C} \cap N_{T} = \emptyset$, $N_{C}, N_{T}$ are open and $N_{C} \supset C , N_{T} \supset T$. This implies that $\overline{N_{C}} \subset X \setminus T = V$. Hence, we have $C \subset N_{C} \subset \overline{N_{C}} \subset V$ as desired.
$(\impliedby)$ Suppose that $X$ is normal as in definition $(2)$. Let $C,F \subset X$ be disjoint closed sets. Since $C \subset X \setminus F$, where $X \setminus F$ is an open set, by $(2)$, $\exists N_{C} \subset X$ open, such that $C \subset N_{C} \subset \overline{N_{C}} \subset X \setminus F$. Similarly, $F \subset X \setminus \overline{N_{C}}$, where $X \setminus \overline{N_{C}}$ is an open set. Once again by $(2)$, there exists $N_{F} \subset X$ open such that $F \subset N_{F} \subset \overline{N_{F}} \subset X \setminus \overline{N_{C}}$. Under this construction, $N_{C} \cap N_{F} = \emptyset$ and are open neighborhoods of $C$ and $F$ respectively.