Prove that these four definition of $T_{3{\frac 1 2}}$ space are equivalent.

Question

Definition

A subset $Y$ of a topological space $X$ is a zero-set if there exist a continuous real valued function $f:X\rightarrow\Bbb R$ such that $$ Y=f^{-1}[0] $$ So we say that a subset $Z$ of $X$ is cozero-set if $X\setminus Z$ is zero-set.

So we observe that a zero set is closed whereas a cozero-set is open and moreover the finite intersection or finite union of zero-set or cozero-set is a zero-set or a cozero-set respectively. Finally, we observe that the sets $$ \{x\in X:f(x)>a\}\quad\text{and}\quad\{x\in X:a<f(x)<b\}\quad\text{and}\quad\{x\in X:f(x)<b\} $$ are cozero-sets.

So we let give this definition.

Definition

A topological space $X$ is said completely regular if for any closed $C$ and for any $x\notin C$ there exists a continuous function $f$ from $X$ to $[0,1]$ such that $$ f(x)=0\quad\text{and}\quad f[C]\subseteq{1} $$

However it is a well know result that $[0,1]$ is homeomorphic to any other intervals $[a,b]$ between a function $\varphi$ such that $$ \varphi(0)=b\quad\text{and}\quad \varphi(1)=a $$ so that we observe that $X$ is completely regular if and only if for any for any closed set $C$ and for any $x\notin C$ there exists a continuous function $f$ from $X$ to $[0,1]$ such that $$ f(x)=b\quad\text{and}\quad f(x)=a $$

Now we give the following definition

Definition

A space $X$ is said $T_{3_{\frac 1 2}}$ if it is $T_1$ and completely regular.

So I tried to prove the following relevant result.

Theorem

For any $T_1$ space $X$ the following four statement are equivalent.

1. $X$ is $T_{3_{\frac 1 2}}$
2. the collection $C([0,1]^X)$ of continuous functions from $X$ to $[0,1]$ generates the topology on $X$, that is the collection $$ \mathcal S:=\big\{f^{-1}[V]: f\in C([0,1]^X)\quad\text{and}\quad V\,\text{open in }[0,1]\big\} $$ is a subbase for $X$.
3. the collection of cozero-sets is a base for $X$
4. there exists a base $\mathcal B$ such such that

4.1. for any open basic neighborhood $B_x$ of $x\in X$ there exists a basic open $B$ disjoint from $x$ and such that $$ B_x\cup B=X $$

4.2. for any pair $B_1$ and $B_2$ of basic open set there exists another pair $A_1$ and $A_2$ of basic open and disjoint set such that $$ X\setminus B_1\subseteq A_1\quad\text{and}\quad X\setminus B_2\subseteq A_2 $$

Proof.

1. So if $A$ is an open set of $X$ then $X\setminus A$ is closed and so for any $x\in A$ there exists $f\in C([0,1]^X)$ such that $$ f(x)=1\quad\text{and}\quad f[X\setminus A]\subseteq\{0\} $$ that is such that $$ f(x)=1\quad\text{and}\quad X\setminus A\subseteq f^{-1}[0] $$ so that we conclude that $$ x\in f^{-1}\big[(0,1]\big]=f^{-1}\big[[0,1]\setminus\{0\}\big]=f^{-1}\big[[0,1]\big]\setminus f^{-1}[0]=X\setminus f^{-1}[0]\subseteq A $$ which proves that the collection $$ \big\{f^{-1}\big[(0,1]\big]:f\in C([0,1]^X)\big\} $$ is a base for $X$ and so the statement follows observing that $$ \big\{f^{-1}\big[(0,1]\big]:f\in C([0,1]^X)\big\}\subseteq\cal S $$
2. First of all we observe that the inclusion $$ C([0,1]^X)\subseteq C(\Bbb R^X) $$ holds and thus by above observed for any $f\in C([0,1]^X)$ the set $$ f^{-1}\big[(\epsilon,1]\big] $$ with $\epsilon\in(0,1)$ is a cozero-set: thus clearly the statement follows proving that if $A$ is open then for any $x\in A$ there exists $f\in C([0,1]^X)$ such that $$ x\in f^{-1}\big[(\epsilon,1]\big]\subseteq A $$ So if $C([0,1]^X)$ generates the topology on $X$ then for any $x_0\in A$ there exists $f\in C([0,1]^X)$ and open basic $B$ of $[0,1]$ such that $$ x_0\in f^{-1}[B]\subseteq A $$ and thus we distinguish the case where $$ y_0:=f(x_0) $$ is zero from the case where it is not. $$ \bf{ CASE\,I}\\y_0=0 $$ So if $y_0$ is zero then without loss of generality we suppose that $$ B=[0,\epsilon) $$ for any $\epsilon\in(0,1)$ so that we observe that the position $$ \varphi(y):=1-y $$ for any $y\in [0,1]$ defines a homeomorphism $\varphi$ from $[0,1]$ to $[0,1]$ such that $$ \varphi^{-1}\big[(1-\epsilon,1]\big]=[0,\epsilon) $$ and thus we conclude that $\varphi\circ f$ is a continuous function from $X$ to $[0,1]$ such that $$ x_0\in(\varphi\circ f)^{-1}\big[(1-\epsilon,1]\big]\subseteq A $$ $$ \bf CASE\, II\\ y_0\neq 0 $$ If $y_0$ is not zero then we distinguish the following two cases

i.$B=B(y_0,\epsilon):=(y_0-\epsilon,y_0+\epsilon)$ with $\epsilon\in\Bbb R^+$

First of all we observe that the positions $$ \text{a. }f_1(y):=\frac y{y_0} \\ \text{b. }f_2(y):=\frac{1-y}{1-y_0} $$ for any $y\in\Bbb R$ define two continuous functions $f_1$ and $f_2$ so that we observe that the position $$ \varphi(y):=\begin{cases}f_1(y),\quad\text{if }y\le y_0\\ f_2(y),\quad\text{otherwise}\end{cases} $$ for any $y\in[0,1]$ define a continuous function from $[0,1]$ to $[0,1]$ - with respect the subspace topology. Now put $$ \varepsilon:=\max\big\{\varphi(y_0-\epsilon),\varphi(y_0+\epsilon)\big\} $$ and thus we let to prove that $$ \varphi^{-1}\big[(\varepsilon,1]\big]\subseteq B(y_0,\epsilon) $$ First of all we observe that $$ \varphi^{-1}\big[(\varepsilon,1]\big]=f^{-1}_1\big[(\varepsilon,1]\big]\cup f^{-1}\big[(\varepsilon,1]\big] $$ where $f^{-1}_i$ for $i=1,2$ is restriced to $[0,1]$, the domain of $\varphi$: so the statement follows proving that $$ f^{-1}_i\big[(\varepsilon,1]\big]\subseteq B(y_0,\epsilon) $$ for $i=1,2$. So we observe that $f_1$ is strictly increasing (and thus injective too) so that the inequality $$ y_0-\epsilon\le f^{-1}_1(\varepsilon)\le y_0 $$ holds and thus we conclude that $$ f^{-1}\big[(\varepsilon,1]\big]\subseteq (y_0-\epsilon,y_0]\subseteq B(y_0,\epsilon) $$ However we observe that $f_2$ is strictly decreasing (and thus injective too) so that the inequality $$ y_0\le f^{-1}_2(\varepsilon)\le y_0+\epsilon $$ holds and thus we conclude that $$ f^{-1}_2\big[(\varepsilon,1]\big]\subseteq [y_0,y_0+\epsilon)\subseteq B(y_0,\epsilon) $$

ii.$B=(\epsilon, 1]$ with $\epsilon\in(0,1)$

In this case $f^{-1}[B]$ is trivially a cozero neighborhood of $x_0$.

3. Now let be $Z_f$ the set of zero points for any $f\in C(\Bbb R^X)$ whereas let be $Z_f^*$ its complement and thus finally let be $\cal Z^*$ the collection of this such sets which now we assume is a base so that we let to prove that it satisfies $4.1$ and $4.2$.

4.1 Well, if $x_0$ is an element of any $Z^*_f$ then putting $$ y_0:=f(x_0) $$ we observe that the position $$ \varphi(y):=y_0-y $$ for any $y\in\Bbb R$ defines a continuous function so that also $\varphi\circ f$ is continuous. So observing that $$ (\varphi\circ f)(x)=y_0\neq 0 $$
for any $x\in Z_f$ we conclude that $$ Z_f\subseteq Z^*_{\varphi\circ f} $$ that is $$ Z^*_f\cup Z^*_{\varphi\circ f}=X $$ Moreover we observce that $$ (\varphi\circ f)(x_0)=0 $$ that is $$ x_0\notin Z^*_{\varphi\circ f} $$ So $\mathcal Z^*$ satisfies $4.1$.

4.2 So we observe that if any $Z^*_{f_1}$ and $Z^*_{f_2}$ are disjoint when their union is $X$ then $\cal Z^*$ satisfies $4.2$ trivially so that we suppose there exists a not disjoint pair $Z^*_{f_1}$ and $Z^*_{f_2}$ whose union is $X$. So in this case we pick $x_0$ in $Z^*_{f_1}\cap Z^*_{f_2}$ and thus we observe that the positions $$ \text{a. }\varphi_1(x):=\min\big\{f_1^2(x)-f^2_2(x),0\big\} \\ \text{b. }\varphi_2(x):=\max\big\{f_1^2(x)-f_2^2(x),0\big\} $$ for any $x\in X$ define two continuous functions $\varphi_1$ and $\varphi_2$ so that we let to prove that $Z^*_{\varphi_1}$ and $Z^*_{\varphi_2}$ are two disjoint open set containg $Z_{f_1}$ and $Z_{f_2}$ respectively. So we observe that if $\varphi_1(x)$ is not zero for any $x\in X$ then $\varphi_2(x)$ is zero and vice versa so that $X$ is union of $Z_{\varphi_1}$ and of $Z_{\varphi_2}$ and thus finally $$ Z_{\varphi_1}^*\cap Z_{\varphi_2}^*=(X\setminus Z_{\varphi_1})\cap(X\setminus Z_{\varphi_2})=X\setminus(Z_{\varphi_1}\cup Z_{\varphi_2})=X\setminus X=\emptyset $$ After all, $X$ is union of $Z_{f_1}^*$ and of $Z_{f_2}^*$ so that $Z_{f_1}$ and $Z_{f_2}$ are disjoint an thus $\varphi_i$ is not zero in $Z_{f_1}$ for $i=1,2$, that is $$ Z_{f_1}\subseteq Z^*_{\varphi_i} $$

4. So if $C$ is a closed set then $X\setminus C$ is open and thus for any $x_0\in X\setminus C$ there exists a basic open set $U$ such that $$ x\in U\subseteq X\setminus C $$ Nowe by assumption there exists a basic open set $V$ disjoint form $x$ and such that $$ U\cup V=X $$ Moreover by assumption there exists two basic disjoint open set $A$ and $V$ such that $$ X\setminus U\subseteq B\quad\text{and}\quad X\setminus V\subseteq A $$ So we observe that $$ x_0\in X\setminus V\subseteq A\quad\text{and}\quad F\subseteq X\setminus U\subseteq B $$ and moreover if $A$ and $B$ are disjoint then $X$ is union of $X\setminus A$ and $X\setminus B$. So with these achievements we put $$ f(x):=\begin{cases}0,\quad\text{if }x\in A\\ \frac 1 2,\quad\text{if }x\in (X\setminus A)\cap (X\setminus B)\\ 1,\quad\text{if }x\in B\end{cases} $$ for any $x$ and thus we let to that the last position defines a continuous function.

So as you can see I was not able to prove that the statement $4$ implies the statement $1$ so that I thought to put a specific question where additionally I asked if the proof of the first three statement are correct: in particular I tried to use the pasting lemma to prove the continuity of $f$ but unfortunately it seem not work now. So could someone help me, please?

Answer 1

First we prove:

If $(X, \mathcal{T})$ is a topological space an $M_k := \lbrace \frac{l}{2^k} : l = 0, \ldots, 2^k\rbrace$,$k \in \mathbb{N}_0$ and $M := \bigcup^{\infty}_{k=0}M_k$. Furthermore if there exists a function, which assigns every $r \in M$ a set $O_r \in \mathcal{T}$ such that if $r,s \in M$ and $r < s$ then $\overline{O_r} \subseteq O_s$ and $O_0 = \emptyset,O_1 = X$ follows. Under these conditions \begin{equation} f : \begin{cases} X \to [0,1] \\ x \mapsto \inf \lbrace r \in M : x \in O_r \rbrace \end{cases} \end{equation} is continuous.

Proof:
Obviously from $O_1 = X$ it follows that $\lbrace r \in M : x \in O_r \rbrace$ for $x \in X$ is never empty and therefore from $M \subseteq [0,1]$ we get that $f$ is well defined. Similarly from $O_0 = \emptyset$ follows that $\lbrace r \in M : x \in \overline{O_r}^C \rbrace \neq \emptyset$ hence \begin{equation} g : \begin{cases} X \to [0,1] \\ x \mapsto \sup \lbrace r \in M : x \in \overline{O_r}^C \rbrace \end{cases} \end{equation} is well defined. For $s \in \lbrace r \in M : x \in \overline{O_r}^C \rbrace$ and $t \in \lbrace r \in M : x \in O_r \rbrace$ where $x \in X$ is fixed we get $s < t$ otherwise $x \in \overline{O_s}^C \cap O_t \subseteq \overline{O_s}^C \cap O_s = \emptyset$ which is a contradiction. Therefore $g(x) \leq f(x)$ for $x \in X$. If there exists $x \in X$ such that $g(x) \leq f(x)$ then from the density of $M$ in $[0,1]$ one gets $r,s \in M$ such that $g(x) < r < s < f(x)$. Since $\overline{O_r} \subseteq O_s$ which implies $x \in O_s$ or $x \in O_s^C \subseteq \overline{O_r}^C$ hence $f(x) \leq s$ or $g(x) \geq r$, contradiction. Therefore $f(x) = g(x)$ for all $x \in X$. The continuity of $f : X \to [0,1]$ is equivalent to the continuity of $f : X \to \mathbb{R}$ since the subspace topology on $[0,1]$ is the initial topology and $f$ is continous iff $\iota \circ f$ is continuous, where $\iota : [0,1] \to \mathbb{R}$ is the canonical embedding. Also it is enough to show that the preimages of a subbasis of the euclidian topology on $\mathbb{R}$ are in $\mathcal{T}$. For this we choose the sets $(-\infty, t),(t, +\infty)$,$t \in \mathbb{R}$. Then \begin{align} f(x) &= \inf \lbrace r \in M : x \in O_r \rbrace < t \iff \exists r \in M, r < t : x \in O_r \iff x \in \bigcup_{\substack{r \in M \\ r < t}}O_r, \\ g(x) &= \sup \lbrace r \in M : x \in \overline{O_r}^C\rbrace > t \iff \exists r \in M, r > t : x \in \overline{O_r}^C \iff x \in \bigcup_{\substack{r \in M \\ r > t}}\overline{O_r}^C. \end{align} From this $f^{-1}(-\infty,t) = \bigcup_{\substack{r \in M \\ r < t}}O_r \in \mathcal{T}$ and $f^{-1}(t,+\infty) = \bigcup_{\substack{r \in M \\ r > t}}\overline{O_r}^C \in \mathcal{T}$ follows hence $f$ is continuous.

Now we use this to proof:

If $(X,\mathcal{T})$ is a topological space and $\mathcal{B} \subseteq \mathcal{P}(X)$ is a normal basis, then two disjoint sets $A,B \in \mathcal{B}^C = \lbrace B^C : B \in \mathcal{B} \rbrace$ can be separated by a continuous function $f : X \to [0,1]$, i.e. $f(A) \subseteq \lbrace 0 \rbrace$,$f(B) \subseteq \lbrace 1 \rbrace$.

Proof: Since $\mathcal{B}$ is normal there exists $G_0,H_0 \in \mathcal{B} : G_0 \supseteq A$,$H_0 \supseteq B$ and $G_0 \cap H_0 = \emptyset$. Set $H_1 := \emptyset$ and $G_1 := B^C \supseteq H^C_0 \supseteq \overline{G_0}$ then $G_1 \in \mathcal{B}$. Then from this definitions for $k = 0$ we have \begin{equation}\tag{1}\label{eq:simple} \forall \zeta,\xi \in M_k : \zeta < \xi \implies \overline{G_{\zeta}} \subseteq H^C_{\zeta} \subseteq G_{\xi}. \end{equation} For $n \in \mathbb{N}_0$ and all $\zeta \in M_n$ let the sets $G_{\zeta},H_{\zeta} \in \mathcal{B}$ be defined such that the statement above holds for $k = n$. Now for the induction step let $\zeta \in M_{n+1}$. Then $\zeta$ is of the form $\frac{l}{2^{n+1}}$ for some $l \in \lbrace 0 , \ldots,2^{n+1} \rbrace$. If $l$ is even then $G_{\zeta},H_{\zeta}$ are already defined, since $\zeta \in M_n$. If $l$ is odd then $G_{\zeta},H_{\zeta}$ are not defined yet and \begin{equation} M_n \supseteq \zeta' := \frac{l-1}{2^{n+1}} < \zeta < \frac{l+1}{2^{n+1}} =: \zeta'' \in M_n. \end{equation} From \eqref{eq:simple} we get $G_{\zeta'} \subseteq H^C_{\zeta'} \subseteq G_{\zeta''}$ hence $H^C_{\zeta'} \cap G^C_{\zeta''} = \emptyset$. Since $\zeta' \neq 1$ we have $H_{\zeta'}$,$G_{\zeta''} \in \mathcal{B}$. Therefore there exists disjoint sets $G_{\zeta},H_{\zeta} \in \mathcal{B}$ with $G_{\zeta} \supseteq H^C_{\zeta'}$ and $H_{\zeta} \supseteq G^C_{\zeta''}$. From this follows $\overline{G_{\zeta'}} \subseteq H^C_{\zeta'} \subseteq G_{\zeta}$ and $\overline{G_{\zeta}}\subseteq H^C_{\zeta} \subseteq G_{\zeta''}$ hence \eqref{eq:simple} holds for $M_{n+1}$. Let $t \in M \setminus \lbrace 1, 0 \rbrace$ where \begin{equation} M := \bigcup^{\infty}_{k=0}M_k. \end{equation} Then define $O_t := G_t$ and $O_0 := \emptyset$,$O_1 := X$. Now there exists (why?) a continuous function
\begin{equation} f : \begin{cases} X \to [0,1] \\ x \mapsto \inf \lbrace r \in M : x \in O_r \rbrace \end{cases}. \end{equation} For $x \in A$ we have $x \in G_0$ thus $x \in O_t$ if $M \supseteq t > 0$. Therefore $f(x) = 0$. Let $\zeta \in M$ then there exists $k \in \mathbb{N}_0$ such that $\zeta \in M_k$ meaning $\zeta = \frac{l}{2^k}$ for some $l \in \lbrace 0,\ldots,2^k \rbrace$. Then since \begin{equation} \overline{G_{\zeta}} \subseteq H_{\xi}^C \subseteq G_{\xi} \end{equation} holds for all $\xi > \zeta$ with $\xi \in M_k$. Since $1 \in M_k$ and $1 > \zeta$ we have \begin{equation} G_{\zeta} \subseteq \overline{G_{\zeta}} \subseteq G_{1} = B^C. \end{equation} The set $G_1$ is already defined at the start of the induction and is always $B^C$. The only sets that have to be additionaly construct are $H_{\zeta},G_{\zeta}$ for $\zeta \neq 0$ and $\zeta \neq 1$ such that these satisfy the condition \eqref{eq:simple} since those are already defined. So we squeeze those sets $H_{\zeta},G_{\zeta}$ in between. Hence by construction $G_{\zeta} \subseteq B^C$ for all $\zeta \in M$. For $M \supseteq t < 1$ one has $B \subseteq O_t^C$ and $f(x) = 1$ for $x \in B$ follows.

Finally:

$(iv) \implies (i)$: Let $\mathcal{B} \subseteq \mathcal{P}(X)$ be a normal (4.2) and weakly regular (4.1) (equivalent to $\forall B \in \mathcal{B} \forall x \in B \exists C \in \mathcal{B} : x \in C^C \subseteq B$) Basis of $\mathcal{T}$ as stated in your $(iv)$ point and $A \subseteq X$ closed with $x \in A^C \in \mathcal{T}$. Since $\mathcal{B}$ is a basis, there exists a $B \in \mathcal{B}$ with $x \in B \subseteq A^C$. Also since $\mathcal{B}$ is weakly regular there exists a $C \in \mathcal{B}$ with $x \in C^C \subseteq B \subseteq A^C$. Since $B$ and $C$ are in $\mathcal{B}$, $C^C$ and $B^C$ disjoint and since $\mathcal{B}$ is a normal basis there exists (why?) a continuous function $f : X \to [0,1]$ such that $f(C^C) \subseteq \lbrace 0 \rbrace$ and $f(B^C) \subseteq \lbrace 1 \rbrace$. From $x \in C^C$ and $A \subseteq B^C$ one gets $f(x) = 0$ and $f(A) \subseteq \lbrace 1 \rbrace$ hence $(X, \mathcal{T})$ is completely regular and $(T_{3.5})$ holds.