The Interior and Closure of a Set of Points

# The Interior and Closure of a Set of Points

## 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 $U$ of $x$, $U \cap A \neq \emptyset$. The set of all points of closure of $A$ is called the Closure of $A$ and is denoted by $\overline{A}$ or $\mathrm{cl}(A)$.
 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.