Regular Space

Let \left( {X,\tau } \right) be a topological space, then for every non-empty closed set F and a point x which does not belongs to F, there exist open sets U and V, such that x \in U,{\text{ }}F \subseteq U and U \cap V = \phi .

In other words, a topological space X is said to be a regular space if for any x \in X and any closed set A of X, there exist an open sets U and V such that x \in U,{\text{ A}} \subseteq U and U \cap V = \phi .


Show that a regular space need not be a Hausdorff space.

For this, let X be an indiscrete topological space, then the only non-empty closed set X, so for any x \in X, there does not exist and closed set A which does not contain x. so X is trivially a regular space. Since for any x,y \in X,\;x \ne y, there is only one open set X itself containing these points, so X is not a Hausdorff space.


A regular {T_1} space is called a {T_3} space.

• Every subspace of a regular space is a regular space.
• Every {T_3} space is a Hausdorff space.
• Let X be a topological space, then the following statements are equivalent. (1) X is a regular space. (2) For every open set U in X and a point a \in U there exist an open set V such that a \in V \subseteq \overline V  \subseteq U. (3) Every point of X has a local neighbourhood basis consisting of closed sets.