# Separable Space

A topological space $\left( {X,\tau } \right)$, is said to be separable space, if it has a countable dense subset in $X$. i.e., $A \subseteq X$, $\overline A = X$, or $A \cup U \ne \phi$, where $U$ is an open set.

In other words, a space $X$ is said to be separable space if there is a subset $A$ of $X$ such that (1) $A$ is countable (2) $\overline A = X$ ($A$ is dense in$X$).

Example:

Let $X = \left\{ {1,2,3,4,5} \right\}$ be a non-empty set and $\tau = \left\{ {\phi ,X,\left\{ 3 \right\},\left\{ {3,4} \right\},\left\{ {2,3} \right\},\left\{ {2,3,4} \right\}} \right\}$ is a topology defined on $X$. Suppose a subset $A = \left\{ {1,3,5} \right\} \subseteq X$. The closed set are $X,\phi ,\left\{ {1,2,4,5} \right\},\left\{ {1,2,5} \right\},\left\{ {1,4,5} \right\},\left\{ {1,5} \right\}$. Now we have $\overline A = X$. Since $A$ is finite and dense in $X$. So, $X$ is a separable space.

Example:

Consider the set of rational number $\mathbb{Q}$ a subset of $\mathbb{R}$ (with usual topology), then the only closed set containing $\mathbb{Q}$ is $\mathbb{R}$ which shows that $\overline {\Bbb Q} = \mathbb{R}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, so $\mathbb{R}$ is also separable in $\mathbb{R}$. But set of irrational numbers is dense in $\mathbb{R}$ but not countable.

Theorems:
• Every second countable space is a separable space.
• Every separable space is not second countable space.
• Every separable metric space is second countable.
• The continuous image of a separable space is separable.