# Closure of a set (Definition+4 Theorems proof with examples)

#### Definition

 \begin{array}{l}\text{Let }\left( {\text{X,}\tau } \right)\text{ be a topological space, and A}\subseteq \text{X }\left( {\text{A subset of set X or equal to X}} \right)\\\text{ then the closure of A is the intersection of all closed supersets of A}\text{.}\\\text{it is denoted by }\!\!~\!\!\text{ }\overline{\text{A}}\text{ orCl(A)}\text{.}\end{array}

##### Example
 \begin{array}{l}Let\text{ }X=\left\{ {1,2,3,4} \right\}\\\tau =\{\varphi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {1,2} \right\},\left\{ {2,3} \right\},\left\{ {1,3} \right\},\left\{ {1,2,3} \right\},X\}\\A=\left\{ {1,3} \right\}~~~~~~~\overline{A}=?\\B=\left\{ 4 \right\}~~~~\overline{B}=?\end{array}
##### Solution
\begin{array}{l}\text{closed subsets of x}\\\{X,\left\{ {2,3,4} \right\},\left\{ {1,3,4} \right\},\left\{ {1,2,4} \right\},\left\{ {3,4} \right\},\left\{ {1,4} \right\},\left\{ {2,4} \right\},\left\{ 4 \right\},\theta \}\\\\\text{closed supersets of A}\\X,\left\{ {1,3,4} \right\}\\\overline{\text{A}}\text{(closure of A) = intersection of all closed supersets of A}\\\overline{A}=X\bigcap \left\{ {1,3,4} \right\}\\\overline{A}=\left\{ {1,3,4} \right\}\\\\\text{closed supersets of B}\\X,\left\{ {2,3,4} \right\},\left\{ {1,3,4} \right\},\left\{ {1,2,4} \right\},\left\{ {3,4} \right\},\left\{ {1,4} \right\},\left\{ {2,4} \right\},\left\{ 4 \right\}\\\overline{\text{B}}\text{(closure of B) = intersection of all closed supersets of B}\\\overline{B}=X\bigcap \left\{ {2,3,4} \right\}\bigcap \left\{ {1,3,4} \right\}\bigcap \left\{ {1,2,4} \right\}\bigcap \left\{ {3,4} \right\}\bigcap \left\{ {1,4} \right\}\bigcap \left\{ {2,4} \right\}\bigcap \left\{ 4 \right\}\\\overline{B}=\left\{ 4 \right\}\end{array}

#### Theorem1: The closure of a subset A of a topological space is the smallest closed superset of A.

##### Proof

Since the intersection of any number of closed sets is closed so the closure of A, being the intersection of its closed supersets, is also closed and is a superset of A. Since the intersection of any number of sets is always a subset of each of the sets so, the closure of A is the smallest closed superset of A.

#### Theorem2:

 \begin{array}{l}In\text{ }a\text{ }topological\text{ }spaceX\\i.~\overline{\varphi }=\varphi ~~~ii.~\overline{X}=X\end{array}
##### Proof
\begin{array}{l}i.~\text{ }Since\text{ }\varphi \text{ }isclosed,\text{ }so\text{ }it\text{ }is\text{ }the\text{ }smallest\text{ }closed\text{ }superset\text{ }of\text{ }itself~~~~~~~~~~~~~~~~~~~\\~\Rightarrow \overline{\varphi }=\varphi \\~ii.~\text{ }Since\text{ }X\text{ }isclosed\text{ }so\text{ }it\text{ }is\text{ }the\text{ }smallest\text{ }closed\text{ }superset\text{ }of\text{ }itself~\\\Rightarrow \overline{X}=X\end{array}
##### Example
\begin{array}{l}let\text{ }X=\{a,b\}\\\tau =\{\varphi ,\{a\},\{b\},X\}\\A=X\text{  }\\\text{  }\overline{A}=?\\B=\varphi \text{    }\\\text{ }\overline{B}=?\end{array}
##### Solution
 \begin{array}{l}\text{closed subsets of X}\\X,\{a\},\{b\},\varphi \\\\\text{closed supersets of X}\\X\text{ itsef only the closed superset of x}\\so\\\overline{X}=X\\\\\text{closed supersets of }\varphi \\\varphi \text{ itsef only the closed superset of }\varphi \\so\\\overline{\varphi }=\varphi \end{array}

#### Theorem3:

 A\text{ }subset\text{  }\!\!'\!\!\text{ }A'of\text{ }a\text{ }topological\text{ }space\text{ }X\text{ }is\text{ }closed\text{ }iff\text{ }\overline{A}=A.
##### Proof

Let A be a closed subset of topological space X, then A itself be the smallest closed superset of A.

So\text{ }\overline{A}=A

Conversely,

\begin{array}{l}Let\text{ }\overline{A}=A~\text{ }\\\overline{{A\text{ }}}is\text{ }the\text{ }intersection\text{ }of\text{ }all\text{ }closed\text{ }supersets\text{ }of\text{ }A.\\Since\text{ }\overline{{A\text{ }}}is\text{ }closed\text{ }so\text{ }A\text{ }is\text{ }closed.\end{array}
##### Example
 \begin{array}{l}Let\text{ }X=\left\{ {1,2,3,4} \right\}\\\tau =\{\varphi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {1,2} \right\},\left\{ {2,3} \right\},\left\{ {1,3} \right\},\left\{ {1,2,3} \right\},X\}\\A=\left\{ {2,3} \right\}~~~~~~~\overline{A}=?\\B=\left\{ {1,4} \right\}~~~~\overline{B}=?\end{array}
##### Solution
 \begin{array}{l}\text{closed subsets of x}\\\{X,\left\{ {2,3,4} \right\},\left\{ {1,3,4} \right\},\left\{ {1,2,4} \right\},\left\{ {3,4} \right\},\left\{ {1,4} \right\},\left\{ {2,4} \right\},\left\{ 4 \right\},\varphi \}\\\\\text{closed supersets of A}\\X,\left\{ {2,3,4} \right\}\\\overline{\text{A}}\text{(closure of A) = intersection of all closed supersets of A}\\\overline{A}=X\bigcap \left\{ {2,3,4} \right\}\\\overline{A}=\left\{ {2,3,4} \right\}\\\text{Here }\overline{A}\text{ is not equal to A because A is subset of topological space X }\\\text{but A is not a closed subset of topological space X}\text{.}\\\text{so, A is not equal to }\overline{A}.\\\\\text{closed supersets of B}\\X,\left\{ {2,3,4} \right\},\left\{ {1,3,4} \right\},\left\{ {1,2,4} \right\},\left\{ {3,4} \right\},\left\{ {1,4} \right\},\left\{ {2,4} \right\},\left\{ 4 \right\}\\\overline{\text{B}}\text{(closure of B) = intersection of all closed supersets of B}\\\overline{B}=X\bigcap \left\{ {2,3,4} \right\}\bigcap \left\{ {1,3,4} \right\}\bigcap \left\{ {1,2,4} \right\}\bigcap \left\{ {3,4} \right\}\bigcap \left\{ {1,4} \right\}\bigcap \left\{ {2,4} \right\}\bigcap \left\{ 4 \right\}\\\overline{B}=\left\{ 4 \right\}\\\text{Here B is equal to }\overline{B}\text{ because B is closed subset of topological space X}\end{array}

#### Theorem4:

 A\text{ }subset\text{  }\!\!'\!\!\text{ }A'of\text{ }a\text{ }topological\text{ }space\text{ }X\text{ }is\text{ }closed\text{ }iff\text{ }\overline{\overline{A}}=A.
##### Proof
 \begin{array}{l}As\text{ }the\text{ }closure\text{ }of\text{ }a\text{ }closed\text{ }set\text{ }is\text{ }equal\text{ }to\text{ }itself\text{ }(\overline{A}=A).\\Now\text{ }because~\text{ }\overline{A}\text{ }is\text{ }closed\text{ }set.\\\Rightarrow cl\left( {\overline{A}} \right)=~\overline{A}~~~~~~~\\\overline{\overline{A}}=A\end{array}
##### Example
 \begin{array}{l}Let\text{ }X=\left\{ {1,2,3,4} \right\}\\\tau =\{\varphi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {1,2} \right\},\left\{ {2,3} \right\},\left\{ {1,3} \right\},\left\{ {1,2,3} \right\},X\}\\A=\left\{ {3,4} \right\}~~~~~~~\overline{A}=?\end{array}
##### Solution
 \begin{array}{l}\text{closed subsets of x}\\\{X,\left\{ {2,3,4} \right\},\left\{ {1,3,4} \right\},\left\{ {1,2,4} \right\},\left\{ {3,4} \right\},\left\{ {1,4} \right\},\left\{ {2,4} \right\},\left\{ 4 \right\},\varphi \}\\\text{closed supersets of A}\\X,\left\{ {2,3,4} \right\},\left\{ {1,3,4} \right\},\left\{ {3,4} \right\}\\\overline{A}\text{(closure of A) = intersection of all closed supersets of A}\\\overline{A}=X\bigcap \left\{ {2,3,4} \right\}\bigcap \left\{ {1,3,4} \right\}\bigcap \left\{ {3,4} \right\}\\\overline{A}=\left\{ {3,4} \right\}\\\text{Here A is equal to }\overline{A}\text{ because A is closed subset of topological space X}\text{.}\\\text{As the closure of a closed set is equal to itself (}\overline{\text{A}}\text{=A)}\text{.}\\\\\text{we want to show that }\overline{\overline{A}}=A\\\overline{A}=\left\{ {3,4} \right\}\\\text{closed superset of }\overline{\text{A}}\\X,\left\{ {2,3,4} \right\},\left\{ {1,3,4} \right\},\left\{ {3,4} \right\}\\\overline{\overline{\text{A}}}\text{(closure of }\overline{\text{A}}\text{) = intersection of all closed supersets of }\overline{\text{A}}\\\overline{\overline{A}}=X\bigcap \left\{ {2,3,4} \right\}\bigcap \left\{ {1,3,4} \right\}\bigcap \left\{ {3,4} \right\}\\\overline{\overline{A}}=\left\{ {3,4} \right\}\\so\overline{\overline{A}}=A\\\left\{ {3,4} \right\}\text{=}\left\{ {3,4} \right\}\end{array}