Let $X$ be a normal space, let $A$ and $B$ be disjoint closed subsets of $X$. Let $[a,b]$ be a closed interval in the real line. Then there exists a continuous map $f:X\to [a,b]$ such that $f(x)=a$ for every $x$ in $A$, and $f(x)=b$ for every $x$ in $B$.
I have read complete proof of this theorem. Everything in the proof depends on step $1$. let $P= \Bbb{Q} \cap [0,1]$. In step $1$, our goal is the following: $\forall p\in P$, $\exists U_p \in \mathcal{T}_X$ such that if $q\in P$ with $p \lt q$, then $\overline{U_p} \subseteq U_q$. Since $P$ is countable, $P=\{ x_n |n\in \Bbb{N}\}$. Let $x_1=1$ and $x_2=0$. Define $U_1=X-B$. $S_n:$ $\forall n\in \Bbb{N}$, $\exists U_{x_n}\in \mathcal{T}_X$ such that if $p,q\in P_n=\{x_1,..,x_n\}$ and $p\lt q$, then $\overline{U_p}\subseteq U_q$. I think statement $S_n$ assume existence of $U_{x_i}$, $\forall i\in \{1,..,n-1\}$. Is my $S_n$ statement correct? Base case: $n=2$. Since $X$ is $T_4$, $\exists U_0 \in \mathcal{N}_A$ such that $\overline{U_0}\subseteq U_1$. $P_2=\{x_1, x_2\}=\{1,0\}$. Thus $S_2$ holds. Inductive step: suppose $S_n$ is true. Then I don’t satisfactory understand following paragraph of Munkres’
In a finite simply order set, every element (other than the smallest and the largest) has an immediate predecessor and an immediate successor. (see theorem 10.1) the number $0$ is the smallest element, and $1$ is the largest element, of the simply ordered set $P_{n+1}$, and $r=x_{n+1}$ is neither $0$ nor $1$. So $r$ has an immediate predecessor $p$ in $P_{n+1}$ and an immediate successor $q$ in $P_{n+1}$. The sets $U_p$ and $U_q$ are already defined, and $\overline{U_p}\subseteq U_q$ by the inductive hypothesis.
Rest of the proof is easy to follow. In step $4$, apparently Munkres’ shows $f:X\to \Bbb{R}$ is continuous. By theorem 18.2(e), $f:X\to [0,1]$ is continuous. Let say, if were doing it (proving continuity of $f$) in conventional way. $f(x_0)=1$ and $(c,1] \in \mathcal{N}_{f(x_0)}$. Then what $U\in \mathcal{N}_{x_0}$ should we take such that $f(U)\subseteq (c,1]$? Can you use different definition of continuity to make proof easier?
Best Answer
Your statement $S_n$ is misleading. The aim is to construct a function $\phi : P \to \mathcal T_X$ (written as $\phi(p) = U_p$) such that
Your statement $S_n$ only requires that for each $n$ there exist suitable $U_{x_i} = U^n_{x_i}$ for $i = 1,\ldots,n$, but it does not explicitly require that $U^m_{x_i} = U^n_{x_i}$ for $i \le \min(m,n)$. Only then we get the desired $\phi$.
What Munkres does is this: He considers pairs $(P',\phi')$, where $P' \subset P$ is a finite subset such that $0,1 \in P'$ and $\phi' : P' \to \mathcal T_X$ is a function such that 1. and 2. are satisfied. The set $\mathcal P$ of these pairs can be ordered by $(P',\phi') \le (P'',\phi'')$ provided $P' \subset P''$ and $\phi'' \mid_{P'} = \phi'$.
The set $\mathcal P$ is nonempty: The smallest $P'$ is $\{0,1\}$, and a function $\phi'$ can be chosen such that $A \subset \phi'(0) \subset \overline{\phi'(0)} \subset X \setminus B = \phi'(1)$.
Given $(P',\phi') \in \mathcal P$, he picks $r \in P \setminus P'$ and shows that $\phi'$ extends to a function $\phi'' : P'' = P' \cup \{r\} \to \mathcal T_X$ satisfying 1. How does this work? Since $P''$ is finite and $r \ne 0,1$, there exist unique elements $p,q \in P''$ such that $p < r < q$. Clearly $p,q \in P'$, i.e. we have $\overline{\phi'(p)} \subset \phi'(q)$. Now normality allows us to find $\phi''(r)$ such that $\overline{\phi'(p)} \subset \phi''(r) \subset \overline{\phi''(r)} \subset \phi(q)$. This proves that the extension $\phi''$ satisfies 1.
Although Munkres claims that he uses induction and the above step resembles the inductive step of mathematical induction, it is not a proof by induction. Actually we have to make infinitely many choices in extending pairs $(P',\phi')$ to a bigger pairs $(P'',\phi'')$ and we therefore need the axiom of choice to get the desired $\phi : P \to \mathcal T_X$.