The Interior and Closure of a Set of Points

Table of Contents

The Interior of a Set of Points

Definition: Let

(E, \tau)

be a topological space and let

A \subseteq E

. A point

x \in A

is an Interior Point of

$A$

if there exists an open set

$U$

with

x \in U \subseteq A

. The set of all interior points of

$A$

is called the Interior of $$A$$ and is denoted by

\mathrm{int}(A)

Proposition 1: Let

(E, \tau)

be a topological space and let

A \subseteq E

. Then

$A$

is an open set if and only if

A = \mathrm{int}(A)

, and consequently,

\mathrm{int}(A)

is the largest open set contained in

$A$

Proof: $\Rightarrow$ Suppose that $$A$$ is an open set. Then for each $x \in A$ , take the open set $$U := A$$ . It satisfies $x \in U = A$ , and so $x \in \mathrm{int}(A)$ . So $A \subseteq \mathrm{int}(A)$ . But trivially, we also have that $A \supseteq \mathrm{int}(A)$ by definition. So $A = \mathrm{int}(A)$ .

$\Leftarrow$ Suppose that $A = \mathrm{int}(A)$ . Then, for each $x \in A$ , there exists an open set $$U_x$$ with $x \in U_x \subseteq A$ . Therefore:

(1)

\begin{align} \quad A = \bigcup_{x \in X} U_x \end{align}

This shows that $$A$$ is an arbitrary union of open sets, which is an open set. $\blacksquare$

The Closure of a Set of Points

Definition: Let

(E, \tau)

be a topological space and let

A \subseteq E

. A point

x \in E

is a Point of Closure of

$A$

if for every neighbourhood

$U$

$x$

U \cap A \neq \emptyset

. The set of all points of closure of

$A$

is called the Closure of $$A$$ and is denoted by

\overline{A}

\mathrm{cl}(A)

Proposition 2: Let

(E, \tau)

be a topological space and let

A \subseteq E

. Then

$A$

is a closed set if and only if

A = \overline{A}

, and consequently,

\overline{A}

is the smallest closed set containing

$A$

Proof: $\Rightarrow$ Suppose that $$A$$ is a closed set. Certainly $A \subseteq \overline{A}$ , since if $x \in A$ , then every neighbourhood $$U$$ of $$x$$ is such that $a \in U \cap A$ . So suppose that there exists a point $x \in \overline{A} \setminus A$ . Then $x \in A^c$ , which is open. So there exists an open set $$V$$ with $x \in V \subseteq A^c$ . But then $$V$$ is an (open) neighbourhood of $$x$$ with $V \cap A = \emptyset$ , contradicting $x \in \overline{A}$ . Thus $\overline{A} \setminus A = \emptyset$ and so $A = \overline{A}$ .

$\Leftarrow$ Suppose that $A = \overline{A}$ . Let $x \in A^c$ . Since $A = \overline{A}$ , this implies that $$x$$ is not a point of closure of $$A$$ . So there exists an neighbourhood $$U$$ of $$x$$ for which $U \cap A = \emptyset$ . But since $$U$$ is a neighbourhood of $$x$$ , there exists an open set $$V$$ with $x \in V \subseteq U$ , and so $V \cap A = \emptyset$ . Thus $x \in V \subseteq A^c$ . Since this holds true for all $x \in A^c$ , we have that $$A^c$$ is open, and consequently, $$A = (A^c)^c$$ is closed.