Def: Suppose X is topological space and B is a base for it. We say, that B is normal base, if following properties hold:


a. For any x∈X and A∈B, with x∈A, there exist A′∈B, such that x∉A′ and A∪A′=X.


b. If U and V are open sets from B, such that U∪V=X, than there exist U′ and V′, disjoint sets from B, such that X∖U⊆V′ and X∖V⊆U′.


It is well known that the space is completely regular iff it has normal base. This characterizations first published in:
*Frink, Orrin*, [**Compactifications and semi-normal spaces**](http://dx.doi.org/10.2307/2373025), Am. J. Math. 86, 602-607 (1964). [ZBL0129.38101](https://zbmath.org/?q=an:0129.38101).


It's easy part to prove that every completely regular space has normal base, because the set of co-zero sets is such a base, which is well-known to be a base for a completely regular space.


Inverse part, that every space which has normal base is completely regular, maybe is also easy, because there is a hint that this is actually the rewriting of proof of Urysohn lemma. But, however I still have a problem with this proof and can't construct the scheme like from Urysohn's lemma. Please, help me if you can.



O. Frink's characterization of completely regular spaces
You can copy any standard proof of Urysohn's Lemma and substitute "member of $\mathcal{B}$" for "open set" and "complement of member of $\mathcal{B}$" for closed set.


Let $\mathcal{C}$ denote $\{X\setminus B:B\in\mathcal{B}\}$. Then point a says: if $x\in X$ and $C\in\mathcal{C}$ are such that $x\notin C$ then there is a $C'\in\mathcal{C}$ such that $x\in C'$ and $C'\cap C=\emptyset$. And point b says: if $C$ and $D$ are disjoint members of $\mathcal{C}$ then there are disjoint members $U$ and $V$ of $\mathcal{B}$ such that $C\subseteq U$ and $D\subseteq V$.


Now any standard proof will yield, given disjoint members of $\mathcal{C}$, a continuous function that separates them.


The lemma that given disjoint closed sets $F$ and $G$ one can find disjoint open sets $U$ and $V$ with disjount closures, and such that $F\subseteq U$ and $G\subseteq V$, can be replaced by the following: if $F$ and $G$ are disjoint members of $\mathcal{C}$ then there are $C$ and $D$ in $\mathcal{C}$ and $U$ and $V$ in $\mathcal{B}$ such that $F\subseteq U\subseteq C$, $G\subseteq V\subseteq D$ and $C\cap D=\emptyset$.


Likewise "if $F$ is closed and $U$ is open with $F\subseteq U$ then there is an open $V$ with $F\subseteq V\subseteq\overline{V}\subseteq U$" becomes: "if $F\in\mathcal{C}$ and $U\in\mathcal{B}$ with $F\subseteq U$ then there are $V\in\mathcal{B}$ and $G\in\mathcal{C}$ with $F\subseteq V\subseteq G\subseteq U$".


