Proposition

Let $X$ be a topological space and let $A \subset X$ be any subset.

A point is in $\Int{A}$ if and only if it has a neighborhood contained in $A$.
A point is in $\Ext{A}$ if and only if it has a neighborhood contained in $X \setminus A$.
A point is in $\partial A$ if and only if every neighborhood of it contains both a point of $A$ and a point of $X \setminus A$.
A point is in $\overline{A}$ if and only if every neighborhood of it contains a point of $A$.
$\overline{A} = A \cup \partial A = \Int{A} \cup \partial A$.
$\Int{A}$ and $\Ext{A}$ are open in $X$, while $\overline{A}$ and $\partial{A}$ are closed in $X$.
The following are equivalent:
- $A$ is open in $X$.
- $A = \Int{A}$.
- $A$ contains none of its boundary points.
- Every point of $A$ has a neighborhood contained in $A$.

Solution

1

Let $x \in X$.

Suppose $x \in \Int{A}$. Since $\Int{A}$ is the union of all open subsets of $X$ that are contained in $A$, there must exist an open $U \subset X$ such that $x \in U \subset A$. Then $U$ is a neighborhood of $x$ contained in $A$.

On the other hand, suppose that $x$ has a neighborhood contained in $A$. Let $U$ denote such a neighborhood. Then $U$ is an open subset of $X$ contained in $A$, so $U \subset \Int{A}$. Therefore, $x \in \Int{A}$.

2

Let $x \in X$.

Suppose $x \in \Ext{A}$. Then $x \notin \overline{A}$. Since $\overline{A}$ is the intersection of all closed sets containing $A$, there exists a closed set $C \subset X$ such that $x \notin C$ and $A \subset C$. Then $C^c$ is a neighborhood of $x$ that is disjoint from $A$. In other words, $x \in C^c \subset X \setminus A$.

On the other hand, suppose that $x$ has a neighborhood contained in $X \setminus A$. Let $U$ denote such a neighborhood. Then $U^c$ is a closed subset of $X$ containing $A$. Therefore, $\overline{A} \subset U^c$. This implies that $x \notin \overline{A}$, so $x \in \Ext{A}$.

3

Let $x \in X$.

$x \notin \Int{A}$ if and only if every neighborhood of $x$ is not contained in $A$ by (1). In other words, $x \notin \Int{A}$ if and only if every neighborhood of $x$ contains a point in $X \setminus A$.
Since $x \notin \Ext{A}$, every neighborhood of $x$ is not contained in $X \setminus A$. In other words, $x \notin \Ext{A}$ if and only if every neighborhood of $x$ contains a point in $A$.

Let $x \in \partial A$. Then $x \notin \Int{A}$ and $x \notin \Ext{A}$. Therefore, every neighborhood of $x$ contains a point in $A$ and a point in $X \setminus A$.

On the other hand, if every neighborhood of $x$ contains both a point of $A$ and a point of $X \setminus A$. Then $x \notin \Int{A}$ and $x \notin \Ext{A}$, so $x \notin \partial A$.

4

A point is in $\overline{A}$ if and only if it is not in $\Ext{A}$. A point is not in $\Ext{A}$ if and only if it does not have a neighborhood contained in $X \setminus A$ by (2). A point does not have a neighborhood contained in $X \setminus A$ if and only if every neighborhood of it contains a point in $A$.

5

We will first show that $\overline{A} = \Int{A} \cup \partial A$.

$\overline{A} \subset \Int{A} \cup \partial A$.

Let $x \in \overline{A}$. Suppose $x \notin \Int{A}$. Then $x$ has no neighborhood contained in $A$. In other words, every neighborhood of $x$ contains a point of $X \setminus A$.

Suppose that $x$ has a neighborhood $U$ that does not contain a point of $A$. Then $U^c$ is a closed subset of $X$ that contains $A$. Thus $\overline{A} \subset U^c$. However, $x \in \overline{A}$ and $x \notin U^c$, so this is impossible. Thus every neighborhood of $x$ contains a point of $A$. By (3), $x \in \partial A$. Therefore, $\overline{A} \subset \Int{A} \cup \partial A$.

$\Int{A} \cup \partial A \subset \overline{A}$.

$\Int{A} \subset A \subset \overline{A}$. Let $x \in (\partial A) \setminus \Int{A}$. If there exists no such element, we are done.

Suppose $x \notin \overline{A}$. Then there exists a closed set $C$ such that $A \subset C$ and $x \notin C$. Then $C^c$ is a neighborhood of $x$ that contains no points of $A$. However, this is impossible by (3) since $x$ is in $\partial A$. Therefore, $x \in \overline{A}$.

Since $\Int{A} \subset A \subset \overline{A}$, $\Int{A} \cup \partial A \subset A \cup \partial A \subset \overline{A}$. Since $\Int{A} \cup \partial A = \overline{A}$, the three sets are equal to each other.

6

$\Int{A}$ is the union of open sets, so it is open. $\overline{A}$ is the intersection of closed sets, so it is closed.

$\Ext{A} = X \setminus \overline{A}$, so $\Ext{A}$ is open. This implies $\Int{A} \cup \Ext{A}$ is open, so $\partial A = X \setminus (\Int{A} \cup \Ext{A})$ is open.

7

If $A$ is open in $X$, every point has a neighborhood contained in $A$. Thus every point of $A$ is in $\Int{A}$, so $A = \Int{A}$.

Suppose $A = \Int{A}$. Let $x$ be a boundary point of $A$. By (3), every neighborhood of $x$ contains both a point of $A$ and a point of $X \setminus A$. By (1), $x$ is not a point of $\Int{A}$. Thus $A$ contains none of its boundary point.

Suppose $A$ contains none of its boundary points. Let $x \in A$. Then every neighborhood of $x$ contains $x$, which is a point of $A$. Since $x$ is not a boundary point, there exists a neighborhood of $x$ that doesn’t contain a point of $X \setminus A$. In other words, there exists a neighborhood of $x$ that is completely contained in $A$. Therefore, $A$ is open.