## Table of Contents

#### Dense set definition

Let (X,τ) be a topological space and ‘A’ is a subset of X. then ‘A’ is said to be dense in X

\text{if }\overline{A}=X

**Example**

\begin{array}{l}Let\text{ }X=\left\{ {a,b,c,d,e} \right\}\\\tau =\{\varphi ,\left\{ a \right\},\left\{ {a,b} \right\},\left\{ {a,c,d} \right\},\left\{ {a,b,c,d} \right\},\left\{ {a,b,e} \right\},X\}\\\text{check which sets are dence in X}\\A=\left\{ a \right\}~~~~~~~\\B=\left\{ {c,e} \right\}~~\\C=\{b\}\end{array}

**Solution**

\begin{array}{l}\text{closed subsets of x}\\\{X,\left\{ {b,c,d,e} \right\},\left\{ {c,d,e} \right\},\left\{ {b,e} \right\},\left\{ e \right\},\left\{ {c,d} \right\},\varphi \}\\\\\text{closed supersets of A }\\X\\\overline{\text{A}}\text{(closure of A) = intersection of all closed supersets of A}\\\text{Sine X is the only superset of A }\\\text{so,}\\\overline{A}=X\\\text{A is dense in X because }\overline{A}\text{ is equal to X }\\\\\text{closed supersets of B}\\X,\left\{ {b,c,d,e} \right\},\left\{ {c,d,e} \right\}\\\overline{\text{B}}\text{(closure of B) = intersection of all closed supersets of B}\\\overline{B}=\left\{ {a,b,c,d,e} \right\}\bigcap \left\{ {b,c,d,e} \right\}\bigcap \left\{ {c,d,e} \right\}\\\overline{B}=\left\{ {c,d,e} \right\}\\\text{Here B is not dencse in X because }\overline{B}\text{ is not equal to X}\\\\\text{closed supersets of C}\\X,\left\{ {b,c,d,e} \right\},\left\{ {b,e} \right\}\\\overline{C}\text{(closure of C) = intersection of all closed supersets of C}\\\overline{C}=\left\{ {a,b,c,d,e} \right\}\bigcap \left\{ {b,c,d,e} \right\}\bigcap \{b,e\}\\\overline{C}=\left\{ {b,e} \right\}\\\text{Here C is not dencse in X because }\overline{C}\text{ is not equal to X}\end{array}

## Add a Comment