on a connected normal space, applying Urysohn's lemma to show that $f^{-1}(r)$ has nonempty interior for each $r \in \mathbb{Q}\cap I$

Question

I know that it is already asked (Strong form of Urysohn Lemma)

Typing the question here again:

Let $A$ and $B$ be two disjoint closed subsets of a connected normal space $X$. Prove there exists a continuous function $f:X\rightarrow [0,1]$ such that $f(A)=\{0\}, \space f(B)=\{1\}.$ Also for all $r\in \mathbb{Q}\cap[0,1]$ , the interior of $f^{-1}(r)$ is not empty.

Though I read two answers to the question, I couldn't proceed.

and I couldn't see where the connectedness was used.

To clear what I want to know, I want to know how to show that $f^{-1}(r)$ has nonempty interior for each $r \in \mathbb{Q}\cap I$.

Answer 1

"Where is the connectedness of $X$ used?"

Let $A, B$ be disjoint non-empty subsets of $X$ and let $g:X\to [0,1]$ be continuous with $g[A]=\{0\}$ and $g[B]=\{1\}.$ Let $h:[0,1]\to [0,1]$ be a continuous with $h(0)=0$ and $h(1)=1,$ such that $int_{[0,1]}h^{-1}\{r\}\ne \emptyset $ for each $r\in \Bbb Q\cap [0,1].$ Let $f=h\circ g.$

Then for $r\in \Bbb Q\cap [0,1] $ we have $f^{-1}\{r\}=g^{-1}h^{-1}\{r\}\supset g^{-1} int_{[0,1]} (h^{-1}\{r\}).$

Now the set $S=int_{[0,1]}(h^{-1}\{r\})$ is a non-empty open subset of $[0,1],$ and $g$ is continuous, so $g^{-1}S$ is open in $X,$ so $$g^{-1}S \subset int_X(f^{-1}\{r\}).$$

BUT how do we know that $g^{-1}S$ is not empty?

BY THIS: $X$ is connected so its continuous image $g[X]$ is connected, with $\{0,1\}\subset g[X]\subset [0,1]$, so $g[X]=[0,1].$ And $\emptyset \ne S\subset [0,1].$ So $g^{-1}S\ne \emptyset.$

To see how this fails if $X$ is not connected, suppose $X=A\cup B$ where $A, B$ are disjoint non-empty open-and-closed subsets of $X.$ Then the $only$ continuous $f:X\to [0,1]$ with $f[A]=\{0\}$ and $f[B]=\{1\}$ is $f=(A\times \{0\})\cup (B\times \{1\}),$ and we have $f^{-1}\{r\}=\emptyset$ if $0\ne r\ne 1.$

$Addendum.$ At the proposer's request, here is how to obtain the function $h$. Let $C$ be the Cantor set. Let $[0,1]\setminus C=\cup S$ where $S$ is a family of non-empty open intervals. For $s,s'\in S$ let $s<^*s'$ iff $\sup s<\inf s'.$

Now $<^*$ is a linear order on the countably infinite set $S,$ and $<^*$ is order-dense (That is, if $s<^*s'$ then there exists $s''$ with $s<^*s''<^*s'$), and there is no $<^*$-max or $<^*$-min member of $S$.... Theorem. (Cantor): Such a linear order is order-isomorphic to $\Bbb Q$ (with the usual order on $\Bbb Q$).

And $\Bbb Q$ is order-isomorphic to $\Bbb Q\cap (0,1).$ So let $\psi: S\to \Bbb Q\cap (0,1)$ be an order-isomorphism.

Now for $x\in s\in S$ let $\phi(x)=\psi (s).$ Extend the domain of $\phi$ from $\cup S$ to $(\cup S)\cup C =[0,1]$ by letting $\phi(0)=0$ and letting $\phi(x)=\sup \{\phi (y): x>y\in \cup S\}$ when $ 0<x\in C.$ I assert without proof that $\phi:[0,1]\to [0,1]$ is continuous.

Finally for $x \in (1/4,3/4)$ let $h(x)=\phi (2x-1/2).$ For $x\in [0,1/4]$ let $h(x)=0.$ For $x\in [3/4,1]$ let $h(x)=1.$

Answer 2

i'm still wondering around, I've checked it is really uniformly convergent.

Here is my argument, when @Wlod AA construct $f_n$ inductively, he set out four regions where the first one is a disjoint union of two intervals, $\ [0;1]\setminus(a_n;b_n)\ $ (the other three are intervals).

Consider only $|f_{n+1}(x)-f_n(x)| $ on these intervals,

On the first disconnected region, it is $0$.

On the second region the difference $\ |f_{n+1}(x)-f_n(x)|\ $ is $\ \le\ \frac{b_n-a_n}6$ when $\ p\ $ is the center of the given interval; and

$$ |f_{n+1}(x)-f_n(x)|\ \le\ \frac 13\cdot\max(p\!-\!a_n,\ \ b_n\!-\!p) \ \le\ \frac 13\cdot(b_n-a_n) $$

for the whole second region, especially in the non-central case.

On the 3rd and 4th, $p-a_n, b_n -p$, respectively.

Since $\frac13 (a_n +b_n -2p) \leq max(p-a_n, b_n -p)$,

for convenience, assume $p-a_n \leq b_n -p$ for every $ n \in \mathbb{Z}_+$.

Then let $d_n = b_n - p_n$, and then $d_{2k+1} \leq d_{2k}$ and $d_{2k+2} \leq \frac12 d_{2k}$.

(Here for each determined p at n-th stage, denote it by $p_n$)

So if we choose $N=2^i$ so large enough that $d_n \lt \frac12 \epsilon$ for a given $\epsilon>0$,

Then for $n,m \geq N$ implies that (assume m is largen than n)

$$\begin{align} |f_m(x) - f_n (x)|& \leq d_n + d_{n+1}+...+d_{m-1}\\ &\leq 2d_n +2d_{n+2}+.... \\ &\lt 2(\epsilon +\frac12 \epsilon+ \frac14 \epsilon+....)\\ &=\epsilon \end{align}$$

This shows that the sequence of functions converges uniformly to a continuous function $f$.

And also by construnction of $Q(n) \subseteq \mathbb{Q}\cap [0,1]$,

For a given $r \in \mathbb{Q}\cap [0,1]$,

There exists $n \in \mathbb{Z}_+$ such that $r \in Q(n) \setminus Q (n-1)$.

Then ${f_n}^{-1}(r) \supseteq {f_{n-1}}^{-1}([\frac13 a_{n-1} + \frac23 r, \frac23 r+ \frac13 b_{n-1}])$,

The latter set contains clearly nonempty interior because of connectedness of $X$.

So for all $k \geq n$, ${ f_k}^{-1}(r)$ has nonempty interior.

Hence the uniform limit function $f^{-1} (r)$ has to have nonempty interior.

Is this right?