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

 \overline A  = \left\{ {x \in X:{\text{ for all }}N\left( x \right),N\left( x \right) \cap A \ne \phi } \right\}

• 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).


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


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}.

• 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.