The Boundary of a Set in a Topological Space

# The Boundary of a Set in a Topological Space

 Definition: Let $(X, \tau)$ be a topological space and $A \subseteq X$. A point $x \in X$ is said to be a Boundary Point of $A$ if $x$ is in the closure of $A$ but not in the interior of $A$, i.e., $x \in \bar{A} \setminus \mathrm{int} (A)$. The set of all boundary points of $A$ is called the Boundary of $A$ and is denoted $\partial A = \bar{A} \setminus \mathrm{int} (A)$.

Equivalently, $x \in \partial A$ if every $U \in \tau$ with $x \in U$ intersects $A$ and $A^c = X \setminus A$ nontrivially.

Another equivalent definition for the boundary of $A$ is the set of all points $x \in X$ such that every open neighbourhood of $x$ contains at least one point of $A$ and at least one point of $X \setminus A$. I.e., $x \in \partial A$ if and only if for every open neighbourhood $U$ of $x$ we have that $A \cap U \neq \emptyset$ and $(X \setminus A) \cap U \neq \emptyset$.

 Lemma 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\overline{X \setminus A} = X \setminus \mathrm{int}(A)$.
• Proof: Let $x \in \overline{X \setminus A}$. Then $x$ is in the closure of $X \setminus A$. Then for every $U \in \tau$ with $x \in U$ we have that:
(1)
\begin{align} \quad U \cap (X \setminus A) \neq \emptyset \end{align}
• So there does NOT exist an open neighbourhood of $x$ that is fully contained in $A$. Thus, $x \not \in \mathrm{int}(A)$, i.e., $x \in X \setminus \mathrm{int}(A)$. So $\overline{X \setminus A} \subseteq X \setminus \mathrm{int}(A)$.
• Now let $x \in X \setminus \mathrm{int}(A)$. Then $x \not \in \mathrm{int}(A)$. So for every open neighbourhood $U$ of $x$ we have that $U \not \subseteq A$. So $U \cap (X \setminus A) \neq \emptyset$ for every open neighbourhood $U$ of $x$. Thus $x \in \overline{X \setminus A}$. So $\overline{X \setminus A} \supseteq X \setminus \mathrm{int}(A)$.
• So we conclude that:
(2)
\begin{align} \overline{X \setminus A} = X \setminus \mathrm{int}(A) \quad \blacksquare \end{align}
 Proposition 2: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\partial A = \overline{A} \cap \overline{X \setminus A}$. • Proof: By definition, $\partial A = \overline{A} \setminus \mathrm{int}(A)$, or equivalently:
(3)
\begin{align} \quad \partial A = \overline{A} \cap (X \setminus \mathrm{int}(A)) \end{align}
• By lemma 1 we have that:
(4)
\begin{align} \quad \partial A = \overline{A} \cap \overline{X \setminus A} \quad \blacksquare \end{align}
 Corollary 3: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\partial A$ is closed.
• Proof: By proposition 2, $\partial A$ can be written as an intersection of two closed sets and so $\partial A$ is closed. $\blacksquare$.
 Corollary 4: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $\partial A = \partial (X \setminus A)$.
• Proof: By proposition 2 we have that:
(5)
\begin{align} \quad \partial A = \overline{A} \cap \overline{X \setminus A} \end{align}
• And since $X \setminus (X \setminus A) = A$, we also have by proposition that:
(6)
\begin{align} \quad \partial (X \setminus A) = \overline{X \setminus A} \cap \overline{X \setminus (X \setminus A)} = \overline{X \setminus A} \cap \overline{A} \end{align}
• Comparing the two above expressions yields $\partial A = \partial (X \setminus A)$. $\blacksquare$.

## Example 1

Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}$ of open intervals and consider the set $A = [0, 1) \subset \mathbb{R}$. The closure of $A$ is:

(7)
\begin{align} \quad \bar{A} = [0, 1] \end{align}

Furthermore, the interior of $A$ is:

(8)
\begin{align} \quad \mathrm{int} (A) = (0, 1) \end{align}

Hence we see that the boundary of $A$ is as expected:

(9)
\begin{align} \quad \partial A = \bar{A} \setminus \mathrm{int} (A) = [0, 1] \setminus (0, 1) = \{0, 1 \} \end{align}

## Example 2

For another example, consider the set $B = [0, 1) \cup (2, 3) \subset \mathbb{R}$. The closure of $A$ is:

(10)
\begin{align} \quad \bar{B} = [0, 1] \cup [2, 3] \end{align}

The interior of $B$ is:

(11)
\begin{align} \quad \mathrm{int} (B) = (0, 1) \cup (2, 3) \end{align}

Hence we see that the boundary of $B$ is:

(12)
\begin{align} \quad \partial B = \bar{B} \setminus \mathrm{int} (B) = [[0, 1] \cup [2, 3]] \setminus [(0, 1) \cup (2, 3)] = \{ 0, 1, 2, 3 \} \end{align}

## Example 3

For a third example, consider the set $X = \mathbb{R}^2$ with the the usual topology $\tau$ containing open disks with positive radii. Let $A = [0, 1) \times [0, 1) \subseteq \mathbb{R}^2$. Then $A$ can be depicted as illustrated: Then the boundary of $A$, $\partial A$ is therefore the set of points illustrated in the image below: 