The Closure of Closed Sets in a Topological Space

# The Closure of Closed Sets in a Topological Space

Recall from The Closure of a Set in a Topological Space page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then the closure of $A$ is the smallest closed subset containing $A$ denoted $\bar{A}$.

On the The Closure of a Set Equals the Union of the Set and Its Accumulation Points page, we found a nice relationship between the closure $\bar{A}$, the set $A$, and the set of accumulation points $A'$ in an equivalent definition of the closure of a set:

(1)\begin{align} \quad \bar{A} = A \cup A' \end{align}

We will now look at a rather simple but nice theorem which says that the $A \subseteq X$ is closed if and only if $\bar{A} = A$.

Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is closed if and only if $\bar{A} = A$. |

**Proof:**$\Rightarrow$ Suppose that $A$ is closed. Then the smallest closed subset containing $A$ is $A$ and hence:

\begin{align} \quad \bar{A} = A \end{align}

- $\Leftarrow$ Suppose that $\bar{A} = A$. By definition $\bar{A}$ is a closed set so $A$ is a closed set. $\blacksquare$

Corollary 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. If $A$ is clopen then $\bar{A} = A$. |

**Proof:**Suppose that $A$ is clopen. Then $A$ is closed and by Theorem 1 we have that $\bar{A} = A$. $\blacksquare$