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There is a particular characterization of Tychonoff spaces (also known as completely regular spaces) which uses cozero sets which I'm trying to prove. As a quick reminder,

For a space $X$, a set $A \subseteq X$ is said to be a zero set of $X$ if there exists a continuous function $f: X \rightarrow [0,1]$ such that $A = f^{-1}(\{0\})$. We say that a set $B \subseteq X$ is a cozero set of $X$ if it is the complement of a zero set; i.e., there exists a continuous function $g: X \rightarrow [0,1]$ such that $B = g^{-1}((0,1])$.

With this in mind, I'm trying to prove that a $T_1$ topological space $X$ is Tychonoff if and only if it has a base of cozero sets. This seems to be a pretty well known result, but I'm struggling to prove it (or for that matter find a proof). I've shown that if a space is Tychonoff then it has a base of cozero sets, but after a couple of weeks of failing, I can't prove the fact that if a space has a base of cozero sets, then it must be Tychonoff.

Any pointers / proofs / articles which have this proof on them?

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Suppose $X$ has a base of cozero sets.

Let $p \in O$ where $O$ is open in $X$.

As the cozero sets form a base, there is a cozero set $U$ such that

$$p \in U \subseteq O$$ This $U$ is of the form $g^{-1}[[(0,1]]$ for some $g: X \to [0,1]$ that is continuous. (This is your own definition of a cozero set).

Note that this means $g(p) \in (0,1]$ and $g(x) = 0$ whenever $x \notin O$ (or else $x \in g^{-1}[[(0,1]] = U \subseteq O$, contradiction). Now take any continuous function $h: [0,1] \to [0,1]$ such that $h(g(p)) = 0$ and $h(0) = 1$ (a simple linear descending function from $(0,1)$ to $(f(p),0))$ continuing with $0$ (as the graph) will do). Then $f:= h \circ g$ is also continuous, maps $X$ to $[0,1]$ also and obeys $f(p) = 0$ and $f[X\setminus O] = \{1\}$

As we have such $f$ for every such $x \in O$, $X$ is completely regular.

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  • $\begingroup$ This was it! Thanks a bunch! I'm assuming you meant $h(f(p))$ on the choice of $h$, correct? Thanks again man! $\endgroup$ – Ando Escalando Sep 1 '18 at 0:16
  • $\begingroup$ @AndoEscalando Yes, I edited the post to correct the $g$ to $h$. Glad I could help. $\endgroup$ – Henno Brandsma Sep 1 '18 at 5:00
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If we prove that $\mathcal{Z}[X]$ (the zero sets of $X$) are a basis for the closed sets we are done (only take complements).

Let $X$ be a Tychonoff space. We want to prove that $\mathcal{Z}[X]$ is a basis for closed sets, .i.e., for all $F$ closed there exist $\mathscr{F}\subseteq\mathcal{Z}[X]$ such that $\displaystyle\bigcap\mathscr{F}=F$. Let $F\subseteq X$ be a closed set. If $F=X$ then $\mathscr{F}=\{X\}$. If $F\subseteq X$ and $F\neq X$ then for all $p\in X\setminus F$ there exist $f_p:X\to[0,1]$ (because $X$ is Tychonoff) a continuous function such that $f_p(p)=1$ and $f_p[F]=\{0\}$. Let $\mathscr{F}=\{Z(f_p):p\in X\setminus F\}$ (here, $Z(f_p)$ is the zero set of $f_p$). Clearly, because for all $p\in X\setminus F$ we have that $F\subseteq Z(f_p)$ then $F\subseteq\displaystyle\bigcap \mathscr{F}$. If we take $p\in\displaystyle\bigcap\mathscr{F}$ then, if we suposse that $p\notin F$ we conclude that $f_p(p)=1$ but $p\in Z(f_p)$, i.e., $f_p(p)=0$. This is a contradiction. Therefore $p\in F$ and thus $\displaystyle\bigcap\mathscr{F}=F$. In this way, $\mathcal{Z}[X]$ is a basis for a closed sets, therefore, $\text{co}\mathcal{Z}[X]$ (the cozero sets) is a basis for $X$.

Now, if $\mathcal{Z}[X]$ is a basis for the closed sets ($\text{co}\mathcal{Z}[X]$ (the cozero sets) is a basis for $X$), take $F\subseteq X$ a closed set and $p\in X\setminus F$. By hypothesis there exist $\mathscr{F}\subseteq\mathcal{Z}[X]$ such that $F=\displaystyle\bigcap\mathscr{F}$. Because $p\notin F$ then there exist $Z\in\mathscr{F}$ such that $p\notin Z$. Let $f$ be the function that gives the zero set $Z$. Then, $f(p)\neq 0$. If we define $$g(x)=\dfrac{f(x)}{f(p)}$$then $Z(g)=Z(f)$, i.e., $f$ and $g$ vanishes in the same points. Therefore $F\subseteq Z(g)$ and $g(p)=1$.

Note that your general hypothesis should be $X$ a $T_1$ space.

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  • $\begingroup$ I thought about something similar when defining $g$, but couldn't figure out how to guarantee that $Im(g) \subseteq [0,1]$, since $f(p) \in (0,1]$. How do you work around this? $\endgroup$ – Ando Escalando Aug 31 '18 at 19:05
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This is Theorem 3.12 in Chapter 7 of Freiwald's book An Introduction to Set Theory and Topology, which also gives two other characterizations.

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  • $\begingroup$ Would the downvoter dare to explain the dowvote? The question was "Any pointers / proofs / articles which have this proof on them?". I think I answered the question and I actually spent quite a long time to find a reachable link on the web. $\endgroup$ – J.-E. Pin Aug 30 '18 at 8:12

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