The Closure of a Set in a Topological Space

# The Closure of a Set in a Topological Space

Recall from the Accumulation Points of a Set in a Topological Space that if $(X, \tau)$ is a topological space and $A \subseteq X$ then a point $x \in X$ is called an accumulation point of $A$ if every open neighbourhood $U$ ($U \in \tau$) contains elements of $A$ different from $x$.

We denoted the set of all accumulation points of $A$ as:

(1)
\begin{align} \quad A' = \{ x \in X : x \: \mathrm{is \: an \: accumulation \: point \: of \:} A \} \end{align}

We will now look at another important term known as the closure of $A$.

 Definition: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then Closure of $A$ denoted by $\bar{A}$ or $\mathrm{cl} (A)$ is the smallest closed set such that $A \subseteq \bar{A}$.

By $U$ being the "smallest" closed set containing $A$ we mean that if $U$ and $V$ are both closed sets containing $A$ then $A \subseteq U \subseteq V$. Equivalently, the closure of $A$ can be defined to be the the intersection of all closed sets which contain $A$ as a subset.

 Proposition 1: Let $(X, \tau)$ be a topological space. a) The closure of the whole set $X$ is $X$, that is, $\overline{X} = X$. b) The closure of the empty set is the empty set, that is, $\overline{\emptyset} = \emptyset$.
 Proposition 2 (Idempotency of the Closure of a Set): Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then the closure of the closure of $A$ is equal to the closure of $A$, that is, $\bar{\bar{A}} = \bar{A}$.
• Proof: By definition, $\bar{\bar{A}}$ is the smallest closed set containing $\bar{A}$. But $\bar{A}$ is closed, and so $\bar{\bar{A}} = \bar{A}$. $\blacksquare$
 Proposition 3: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then $a \in \bar{A}$ if and only if $U \cap A \neq \emptyset$ for every open neighbourhood $U$ of $a$.