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) ?

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

Further considerations:

(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.

  • $\begingroup$ Sierpinski space is useful for many examples and counter-examples. $\endgroup$ – DanielWainfleet Jan 5 at 0:58

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.


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