The Boundary of Clopen Sets in a Topological Space

# The Boundary of Clopen Sets in a Topological Space

Recall from The Boundary of a Set in a Topological Space page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then 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 $A$, that is, $x \in \bar{A} \setminus \mathrm{int} (A)$.

Furthermore, the set of all boundary points of $A$ was called the boundary of $A$ and was denoted $\partial A$.

We will now look at a very simple yet significant theorem which will tell us that the boundary of any clopen set is always the empty set.

 Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is a clopen set if and only if $\partial A = \emptyset$.
• Proof: $\Rightarrow$ Suppose that $A$ is a clopen set. Then $A$ is both open and closed. Now note that:
(1)
\begin{align} \quad \partial A = \bar{A} \setminus \mathrm{int} (A) \end{align}
• Since $A$ is closed, we have that the smallest closed set containing $A$ is $A$ itself. Therefore $\bar{A} = A$.
• Similarly, since $A$ is open, we have that the largest open set contained in $A$ is $A$ itself. Therefore $\mathrm{int} (A) = A$.
• Hence we conclude that $\partial A = A \setminus A = \emptyset$.
• $\Leftarrow$ Suppose that $\partial A = \emptyset$. Since $\partial A = \bar{A} \setminus \mathrm{int} (A) = \emptyset$ we must have that:
(2)
\begin{align} \quad \bar{A} \subseteq \mathrm{int} (A) \end{align}
• By definition $\bar{A}$ is the is the smallest closed set containing $A$ so $A \subseteq \bar{A}$ and $\mathrm{int} (A)$ is the largest open set contained in $A$, so $\mathrm{int} (A) \subseteq A$. From this we see that $\bar{A} = A = \mathrm{int} (A)$. Since $\bar{A}$ is a closed set this implies that $A$ is closed, and since $\mathrm{int} (A)$ is an open set this implies that $A$ is open. Therefore $A$ is a clopen set. $\blacksquare$
 Corollary 1: Let $(X, \tau)$ be a topological space. Then $\partial \emptyset = \emptyset$ and $\partial X = \emptyset$.
• Proof: We know that $\emptyset$ and $X$ are both clopen sets so by Theorem 1 we have that $\partial \emptyset = \emptyset$ and $\partial X = \emptyset$. $\blacksquare$