Separable Space

A topological space \left( {X,\tau } \right) is said to be a 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 a separable space if there is a subset A of X such that (1) A is countable (2) \overline A = X (A is dense inX).

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\} be a topology defined on X. Suppose a subset A = \left\{ {1,3,5} \right\} \subseteq X. The closed sets 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, then X is a separable space.

Example:

Consider that the set of rational numbers \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}, then \mathbb{R} is also separable in \mathbb{R}. However, the 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 a second countable space.
• Every separable metric space is a second countable space.
• The continuous image of a separable space is separable.