Vtyjkyuk

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?

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 $f[Xsetminus O] = 1$

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

If we prove that $mathcalZ[X]$ (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 $mathcalZ[X]$ is a basis for closed sets, .i.e., for all $F$ closed there exist $mathscrFsubseteqmathcalZ[X]$ such that $displaystylebigcapmathscrF=F$. Let $Fsubseteq X$ be a closed set. If $F=X$ then $mathscrF=X$. If $Fsubseteq X$ and $Fneq X$ then for all $pin Xsetminus F$ there exist $f_p:Xto[0,1]$ (because $X$ is Tychonoff) a continuous function such that $f_p(p)=1$ and $f_p[F]=0$. Let $mathscrF=Z(f_p):pin Xsetminus F$ (here, $Z(f_p)$ is the zero set of $f_p$). Clearly, because for all $pin Xsetminus F$ we have that $Fsubseteq Z(f_p)$ then $Fsubseteqdisplaystylebigcap mathscrF$. If we take $pindisplaystylebigcapmathscrF$ then, if we suposse that $pnotin F$ we conclude that $f_p(p)=1$ but $pin Z(f_p)$, i.e., $f_p(p)=0$. This is a contradiction. Therefore $pin F$ and thus $displaystylebigcapmathscrF=F$. In this way, $mathcalZ[X]$ is a basis for a closed sets, therefore, $textcomathcalZ[X]$ (the cozero sets) is a basis for $X$.

Now, if $mathcalZ[X]$ is a basis for the closed sets ($textcomathcalZ[X]$ (the cozero sets) is a basis for $X$), take $Fsubseteq X$ a closed set and $pin Xsetminus F$. By hypothesis there exist $mathscrFsubseteqmathcalZ[X]$ such that $F=displaystylebigcapmathscrF$. Because $pnotin F$ then there exist $ZinmathscrF$ such that $pnotin Z$. Let $f$ be the function that gives the zero set $Z$. Then, $f(p)neq 0$. If we define $$g(x)=dfracf(x)f(p)$$then $Z(g)=Z(f)$, i.e., $f$ and $g$ vanishes in the same points. Therefore $Fsubseteq Z(g)$ and $g(p)=1$.

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

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.