# 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 in$$X$$).

**Example:**

Let $$X = \left\{ {1,2,3,4,5} \right\}$$ be a non-empty set and $$\tau = \left\{ {\phi ,X,\left\{ 3 \right\},\left\{ {3,4} \right\},\left\{ {2,3} \right\},\left\{ {2,3,4} \right\}} \right\}$$ 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.

Aeriana Narbonne

September 23@ 12:49 amexample with {1,2,3,4,5} is wrong