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?
Best Answer
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.