# Isolated Point of a Set

Let $$A$$ be a subset of a topological space $$X$$, then a point $$x \in A$$ is said to be an isolated point of $$A$$ if there exists an open set containing $$x$$ which does not contain any point of $$A$$ different from $$x$$. In other words, a point $$x \in A$$ is said to be an isolated point of $$A$$ if there exists an open set $$U$$ containing $$x$$ such that $$A \cap U = \left\{ x \right\}$$. It is obvious from the definition of an isolated point of a set that an isolated point of $$A$$ can never be the limit point of $$A$$. The set of all isolated points of $$A$$ is usually denoted by $$I\left( A \right)$$.

**Theorem**

Any closed subset of a topological space $$X$$ is the disjoint union of its set of isolated points and its set of limit points in the sense that it contains these sets, they are disjoint, and it is their union.

**Perfect Set**

A subset of a topological space is said to be a perfect set if it is equal to its derived set. Thus, a subset $$A$$ of a topological space $$X$$ is said to be a perfect set if $$A = {A^d}$$.

**Theorem**

A subset of a topological space is perfect if and only if it is closed and has no isolated points.

**Proof**

Let $$A$$ be a perfect subset of a topological space $$X$$, then $$A = {A^d}$$. Since $${A^d}$$ is the set of all limit points of $$A$$ and a limit point is not an isolated point, so $$A$$ has no isolated points.

Conversely, let $$A$$ be closed and have no isolated points. Then $$A$$ is equal to its derived set, i.e., $$A = {A^d}$$, so $$A$$ is perfect.