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.

For a basic example, 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:

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

Furthermore, the interior of $A$ is:

(2)
\begin{align} \quad \mathrm{int} (A) = (0, 1) \end{align}

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

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

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

(4)
\begin{align} \quad \bar{B} = [0, 1] \cup [2, 3] \end{align}

The interior of $B$ is:

(5)
\begin{align} \quad \mathrm{int} (B) = (0, 1) \cup (2, 3) \end{align}

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

(6)
\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}

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: