If the space $X$ is completely regular, we know that the collection
{${\rm int}\,Z(f)$:$f$ 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 $f\colon X\to \mathbb{R}$ such that $x\in {\rm int}\, Z(f)\subseteq Z(f)\subseteq U_x)$.

I have two questions about converse of this theorem. these questions are almost the same, but I think these are different.

If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f\colon X\to \mathbb{R}$ such that $x\in {\rm int}\, Z(f)\subseteq U_x$, then $X$ is completely regular.

If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f\colon X\to \mathbb{R}$ such that $x\in {\rm int}\,Z(f)\subseteq Z(f) \subseteq U_x$, then $X$ is completely regular.

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

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