# Why do we need singletons to be closed in the definition of Normal/Regular spaces?

In the book of Topology by Munkres, at page 193, it is given that

Definition. Suppose that one-point sets are closed in X. Then X is said to be regular if for each pair consisting of a point x and a closed set B disjoint from x, there exist disjoint open sets containing x and B, respectively. The space X is said to be normal if for each pair A, B of disjoint closed sets of X, there exist disjoint open sets containing A and B, respectively.

It is clear that a regular space is Hausdorff, and that a normal space is regular. (We need to include the condition that one-point sets be closed as part of the definition of regularity and normality in order for this to be the case. A two-point space in the indiscrete topology satisfies the other part of the definitions of regularity and normality, even though it is not Hausdorff.)

But, how does assuming one-point sets are closed allow regular (normal) space to be Hausdorff(regular) ?

• Because you can take B={y}, a different point, in your definition of regular. – T.J. Gaffney Jan 3 at 6:51
• @Gaffney In case when $B= \{y\}$ is open ? I didn't full get it . – onurcanbektas Jan 3 at 6:53
• If it's open then you can't conclude that the space is Hausdorff. You don't conclude anything. – T.J. Gaffney Jan 3 at 7:04
• @Gaffney Yes, I know that, but I still don't understand what are you trying to say in your first comment. – onurcanbektas Jan 3 at 7:05
• @Gaffney Why did you deleted your answer ? – onurcanbektas Jan 3 at 7:20

(i). Consider Sierpinski space $$S=\{x,y\}$$ with $$x\ne y,$$ where $$S,\emptyset,$$ and $$\{x\}$$ are open but $$\{y\}$$ is not open. If $$A, B$$ are disjoint closed subsets of $$S$$ then at least one of $$A, B$$ is empty so $$A,B$$ are covered by disjoint open sets. But $$S$$ does not satisfy the condition for regularity: $$x$$ does not belong to the closed subset $$\{y\}$$ but the only open set covering $$\{y\}$$ is the whole space $$S$$.
(ii). Let $$X$$ be a normal space. Let $$p\in X$$ and let $$Y$$ be a closed subset of $$X$$ with $$p\not \in Y.$$ Since $$X$$ is a $$T_1$$ space (one-point subsets are closed), the sets $$\{p\}, Y$$ are closed and disjoint. So, since $$X$$ is normal, there are disjoint open sets with $$\{p\}\subset U$$ and $$Y\subset V.$$ That is, $$p\in U$$ and $$Y\subset V.$$ So $$X$$ is regular.
If $$x \neq y$$ then $$\{y\}$$ is a closed set adn $$x \notin \{y\}$$ so there exist disjoint open sets $$U,V$$ such that $$x \in U$$ and $$\{y\} \subset V$$. This shows that the space is Hausdorff.