# Closure of a Set

Let $\left( {X,\tau } \right)$ be a topological space and $A$ is a subset of $X$, then the closure of $A$ is denoted by $\overline A$ or ${\text{cl}}\left( A \right)$ is the intersection of all closed sets containing $A$ or all closed super set of $A$. i.e. the smallest closed set containing $A$.

On the other hand it can also be as let $\left( {X,\tau } \right)$ be a topological space and let $A$ be any subset of $X$. A point $x \in X$ is said to be adherent to $A$ if each neighborhood of $x$ contains a point of $A$ (which may be $x$ itself). The set of all points of $X$ adherent to $A$ is called closure (or adherence) of $A$ and is denoted by $\overline A$. In symbols

Remarks:
• Every set is always contained in its closure. i.e. $A \subseteq \overline A$
• Closure of a set by definition (being intersection of closed set is always closed set).

Example:

Let $X = \left\{ {a,b,c,d} \right\}$ with topology $\tau = \left\{ {\phi ,\left\{ a \right\},\left\{ {b,c} \right\},\left\{ {a,b,c} \right\},X} \right\}$ and $A = \left\{ {b,d} \right\}$ be a subset of $X$.

Open sets are $\phi ,\left\{ a \right\},\left\{ {b,c} \right\},\left\{ {a,b,c} \right\},X$
Closed sets are $X,\left\{ {b,c,d} \right\},\left\{ {a,d} \right\},\left\{ d \right\},\phi$
Closed sets containing A are $X,\left\{ {b,c,d} \right\}$
Now $\overline A = \left\{ {b,c,d} \right\} \cap X = \left\{ {b,c,d} \right\}$

Theorem: Let $\left( {X,\tau } \right)$ be a topological space, and $A$ and $B$ be subsets of $X$, then
$A$ is closed if and only if $A = \overline A$
$\overline A = \overline {\overline A }$
$A \subseteq B \Rightarrow \overline A \subseteq \overline B$
$\overline {A \cup B} = \overline A \cup \overline B$
$\overline {A \cap B} \subseteq \overline A \cap \overline B$

Dense Subset of a Topological Space:

Let $\left( {X,\tau } \right)$ be a topological space and $A$ be a subset of $X$, then $A$ is said to be dense subset of $X$. (i.e. dense in $X$), if $\overline A = X$

Example:

Consider the set of rational number $\mathbb{Q} \subseteq \mathbb{R}$ (with usual topology), then the only closed set containing $\mathbb{Q}$ in $\mathbb{R}$. Which shows that $\mathbb{Q} = \mathbb{R}$. Hence $\mathbb{Q}$ is dense in $\mathbb{R}$.

Remarks:
• It may be noted that the set of irrational numbers is also dense in $\mathbb{R}$. i.e. $\overline {\left( {\mathbb{R} - \mathbb{Q}} \right)} = \mathbb{R}$.
• Rational are dense in $\mathbb{R}$ and countable but irrational numbers are also dense in $\mathbb{R}$ but not countable.