# 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 belong 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 open sets $$U$$ and $$V$$ such that $$x \in U,{\text{ A}} \subseteq U$$ and $$U \cap V = \phi $$.

**Example:**

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 is $$X$$, so for any $$x \in X$$, there does not exist a 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.

**T3-Space**

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

**Theorems**

• 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 exists an open set $$V$$ such that $$a \in V \subseteq \overline V \subseteq U$$; **(3)** every point of $$X$$ has a local neighborhood basis consisting of closed sets.