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, Am. J. Math. 86, 602-607 (1964). ZBL0129.38101.

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.


1 Answer 1


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 $F\subseteq V\subseteq G\subseteq U$".


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