Question

If the space $X$ is completely regular, we Know that The collection {$intZ(f)$:$f$ is a continuous function from $X$ to the real numbers} is an open base for open subsets of the space $X$ (i.e. If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆Z(f)⊆U_x)$. I have two questions about converse of this theorem. these questions are almost the same, but I think these are different.

1.If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆U_x$, then $X$ is completely regular.

2.If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆Z(f)⊆U_x$, then $X$ is completely regular.

I think these two claims have counterexamples and these conditions don't emply the complete regularity of $X$.

Answer 1

Here's a counterexample to 1.

Let $T$ be the Tychonoff plank, i.e., the product $(\omega_1+1)\times(\omega+1)$ with the point $\langle\omega_1,\omega\rangle$ removed.

Consider the set $\omega\times T\cup\lbrace \infty\rbrace $ topologized so that $\omega\times T$ has the product topology and is an open subset itself, and the basic neighbourhoods of $\infty$ are of the form $U_n(\infty) = (\omega\setminus n)\times T\cup\lbrace \infty\rbrace $.

We construct a quotient space by identifying $\langle n,\alpha,\omega\rangle$ and $\langle n+1,\alpha,\omega\rangle$ whenever $n$ is odd and $\alpha\in\omega_1$, and identifying $\langle n,\omega_1,i\rangle$ and $\langle n+1,\omega_1,i\rangle$ whenever $n$ is even and $i\in\omega_1$.

The resulting space $C$, the Tychonoff corkscreww, is regular but not completely regular (the copy of $\lbrace 0\rbrace \times T$ and $\infty$ cannot be separated by continuous functions).

For each odd $n$ define $f_n:C\to[0,1]$ by $f_n(\infty)=0$ and $$ f(m,\alpha,i)= \begin{cases} 2^{-i} &\text{ if } m\le n \text{ and }i<\omega\cr
0 &\text{ if } m> n \text{ and }i<\omega
\end{cases} $$ and $f_n(m,\alpha,\omega)=0$ for all $m$ and $\alpha$.

Then the interiors $\operatorname{int}Z(f_n)$ form a local base at $\infty$.