Let $X$ be a topological space. (All of the spaces I'm considering are $T\_0$, but in general they are *not* $T\_1$. To be even more concrete, one can even consider $X={\rm Spec}(R)$ to be the space of prime ideals of a [not necessarily commutative] ring $R$, under the Zariski topology.)


Let us say that $X$ has *clopen separation* if for any two closed, disjoint sets $A,B\subseteq X$, then there is a clopen set $U$ containing $A$, whose complement contains $B$.


I'm reading a paper which states that this is the definition of a *strongly zero-dimensional* space; but other sources give a (seemingly) different definition. Is this equivalence true? If not, how close to true is it?



Zero-dimensional spaces and clopen separations
$\textbf{All the spaces mentioned in this answer are completely regular}$


I usually call a space $T\_{1}$-space ultranormal if whenever there are two disjoint closed sets $A,B\subseteq X$ there is a clopen set with $A\subseteq C,B\subseteq C^{c}$. A strongly zero-dimensional space is usually defined to be a completely regular space $X$ such that the Stone-Cech compactification $\beta X$ is zero-dimensional. Equivalently, a completely regular space $X$ is strongly zero-dimensional if and only if whenever $f:X\rightarrow[0,1]$ is continuous, then there is a clopen set $C$ with $f^{-1}[\{0\}]\subseteq C,f^{-1}[\{1\}]\subseteq C^{c}$. The ultranormal spaces are precisely the normal strongly zero-dimensional spaces, but not every strongly zero-dimensional space is ultranormal. For example, the Tychonoff plank
$(\omega+1)\times(\omega\_{1}+1)\setminus\{(\omega,\omega\_{1})\}$ is strongly zero-dimensional since $\beta((\omega+1)\times(\omega\_{1}+1)\setminus\{(\omega,\omega\_{1})\})=(\omega+1)\times(\omega\_{1}+1)$ but not ultranormal. I gave a long answer [here](https://mathoverflow.net/a/134184/22277) giving (probably too much) information about the distinction between various notions related to zero-dimensionality.


Let me now prove that every normal strongly zero-dimensional space is ultranormal. Suppose that $X$ is strongly zero-dimensional and normal. Let $C,D\subseteq X$ be disjoint closed sets. Then by normality, there is some continuous $f:X\rightarrow[0,1]$ with $C\subseteq f^{-1}[\{0\}],D\subseteq f^{-1}[\{1\}]$. However, since $X$ is strongly zero-dimensional, there is a clopen set $R$ with $f^{-1}[\{0\}]\subseteq R,f^{-1}[\{1\}]\subseteq R^{c}$. Therefore $C\subseteq f^{-1}[\{0\}],D\subseteq f^{-1}[\{1\}]$.


