If $X$ is perfectly normal then for any closed set there exists a continuous function $φ:X⟶[0,1]$ such that $φ^{-1}[0]=C$ and $φ^{-1}[1]=∅$.

Question

What shown below is a reference from Engelking Topology text

So I would be sure that the statement iv implies that if $C$ is closed then there exists a continuous function $\varphi$ form $X$ to $[0,1]$ such that $$ \varphi^{-1}[0]=C\quad\text{and}\quad \varphi^{-1}[1]=\emptyset $$ I know that this could seem a trivial question but the issue is that my professor defined perfect normality using the statement iv so that by it I have to infer the others and in particular the second: indeed if iv holds for any closed and disjoint sets then in particular holds for a closed set $C$ and for empty so that there exist a the above function $\varphi$. Anyway I am quite sure that for a metrica space the function $\varphi$ exists: indeed, if $d$ is a metric on $X$ then I know that the position $$ d^*(x,y):=\min\{d(x,y)1\} $$ defines an equivalent bounded metric so that the position $$ f(x):=\inf_{c\in C}d(x,c) $$ defines a continuous (see here for details) function from $X$ to $[0,1]$ such that $$ f^{-1}[0]=C $$ but the function $$ \psi:[0,1]\ni x\longrightarrow \frac x 2\in\Big[0,\frac 1 2\Big] $$ is a bijection such that $$ \psi(0)=0 $$ so that finally we can put $$ \varphi\equiv\psi\circ f $$

So could someone help me, please?

Answer 1

I assume $I=[0,1]$ here.

First a simple observation: If the case $B=\emptyset$ is excluded in (iv) then we can put $B=\{b\}$ for some $b\notin A$ (if $A\neq X$) and find a function $f\colon X\to I$ separating $A$ and $\{b\}$. Then the function $\psi\circ f$ (that is $x\mapsto\frac 12f(x)$) satisfies (iv) for $A$ and $\emptyset$.

Definition The set $U\in X$ is functionally open iff $U=f^{-1}(V)$, where $f\colon X\to I$ is continuous and $V\subset I$ is open. The set $C\in X$ is functionally closed iff $C=f^{-1}(K)$, where $f\colon X\to I$ is continuous and $K\subset I$ is closed.

Proof (of the theorem). (ii)$\implies$(iii): If $C$ is closed then the set $U=X\setminus C$ is open and therefore it's functionally open, so $U=f^{-1}(V)$, where $f\colon X\to I$ is continuous and $V\subset I$ is open. Then $C=X\setminus f^{-1}(V)=f^{-1}(I\setminus V)$, so $C$ is functionally closed.

(iii)$\implies$(ii): as above.

(iv)$\implies$(ii): Take any closed $C\subset X$. From (iv) we know that there exists $f\colon X\to I$ such that $C=f^{-1}(\{0\})$ and $\emptyset=f^{-1}(\{1\})$. Since $\{0\}$ is closed in $I$ we have that $C$ is functionally closed.

(iv)$\implies$(i) Let $C$ be closed. Then from (iv) we have a continuous $f\colon X\to I$ such that $C=f^{-1}(\{0\})$. Then $$C=\bigcap_{n=1}^\infty f^{-1}([0,1/n)).$$ Since $[0,1/n)$ are open in $I$ we see that $C$ is $G_\delta$.

(i)$\implies$(iii) Let $C\subset X$ be closed. Then $C=\bigcap_{n=1}^\infty U_n$. Since $X$ is normal, from Urysohn Theorem we have a continuous $f_n\colon X\to I$ such that $f_n|C\equiv 0$ and $f_n|X\setminus U_n\equiv 1$. Then the function $f$ defined by $$f(x)=\sum_{n=1}^\infty \frac 1{2^n}f_n(x)$$ safisfies $C=f^{-1}(\{0\})$.

(iii)$\implies$(iv) Let $C,D$ be disjoint and closed. Let $f,g,h\colon X\to I$ be continuous and such that $$C=f^{-1}(\{0\}),\ D=g^{-1}(\{1\}),\ h|C\equiv 0,\ h|D\equiv 1.$$ Let $$f'=\max(f,h),\ g'=\min(g,h).$$ Then $$C=f'^{-1}(\{0\}),\ D\subset f'^{-1}(\{1\}) \ D=g'^{-1}(\{1\}),\ C\subset g'^{-1}(\{0\})$$ and $s:=\frac 12(f'+g')$ satisfies (iv).

Answer 2

Engelking assumes Hausdorff for all separation properties above $T_2$, in particular normal spaces (and perfectly normal spaces) are assumed to be $T_1$, which implies Hausdorff in that case. That's why Vedenissoff's theorem is stated with the $T_1$ assumption. But the theorem holds with the same proof even if the space is not Hausdorff/$T_1$.

With the more modern definition of normal/perfectly normal that does not assume $T_1$, the various equivalent characterizations of a "perfectly normal" space $X$ are:

• $X$ is normal and every closed set is a $G_\delta$.
• Every closed set is a zero-set (= your (iii))
• Disjoint closed sets $A$ and $B$ can be precisely separated by a function (= your (iv))

As you noticed, it is essential to allow $B$ empty in that last condition, which then shows that any closed set $A$ is a zero-set and hence a $G_\delta$.

It would not be equivalent to use the following condition:

(*) Nonempty disjoint closed sets can be precisely separated by a function.

Ultraconnected spaces have no nonempty disjoint closed sets, so they vacuously satisfy (*). But ultraconnected spaces are not perfectly normal in general. For example, in the Sierpinski space the unique closed singleton is not a $G_\delta$.