Closure of a Set

Let $$\left( {X,\tau } \right)$$ be a topological space and $$A$$ be 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 sets of $$A$$; i.e. the smallest closed set containing $$A$$.

On the other hand it can also be written 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 the 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. $$A \subseteq \overline A $$
• The closure of a set by definition (the intersection of a closed set is always a 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 a dense subset of $$X$$ (i.e. dense in $$X$$), if $$\overline A = X$$



Consider the set of rational numbers $$\mathbb{Q} \subseteq \mathbb{R}$$ (with usual topology), then the only closed set containing $$\mathbb{Q}$$ in $$\mathbb{R}$$. This 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 numbers are dense in $$\mathbb{R}$$ and countable but irrational numbers are also dense in $$\mathbb{R}$$ but not countable.