The Boundary of Any Set is Closed in a Topological Space

# The Boundary of Any Set is Closed 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 contained in the closure of $A$ and not in the interior of $A$, i.e., $x \in \bar{A} \setminus \mathrm{int} (A)$.

We also noted that the set of all boundary points of $A$ is called the boundary of $A$ and is denoted:

(1)\begin{align} \quad \partial A = \bar{A} \setminus \mathrm{int} (A) \end{align}

We will now look at a nice theorem that says the boundary of any set in a topological space is always a closed set.

Theorem 1: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then $\partial A$ is closed. |

**Proof:**To show that $\partial A$ is closed we only need to show that $(\partial A)^c = X \setminus \partial A$ is open. Notice that $X \setminus \partial A$ can be written as:

\begin{align} \quad X \setminus \partial A = (A \setminus \partial A) \cup (A^c \setminus \partial A) \quad (*) \end{align}

- We therefore want to show that $A \setminus \partial A$ and $A^c \setminus \partial A$ are both open sets.

- Let $x \in A \setminus \partial A$. Then $x \in A$ and $x$ is not on the boundary of $A$. Since $x$ is not on the boundary of $A$ we have that there exists an open neighbourhood $U$ ($U \in \tau$) that intersects $A^c$ trivially, i.e., $U \cap A^c = \emptyset$. Therefore $U \subseteq A$ and $x \in U \subseteq A$, so $x \in \mathrm{int}(A \setminus \partial A)$ and $\mathrm{int} (A \setminus \partial A) = A \setminus \partial A$ so $A \setminus \partial A$ is open.

- Let $x \in A^c \setminus \partial A$. Then $x \in A^c$ and $x$ is not on the boundary of $A$. Since $x$ is not on the boundary of $A$ we have that there exists an open neighbourhood $U$ of $x$ ($U \in \tau$) that intersects $A$ trivially, i.e., $U \cap A = \emptyset$. Therefore $U \subseteq A^c$ and $x \in U \subseteq A^c$, so $x \in \mathrm{int}(A^c \setminus \partial A)$, so $\mathrm{int} (A^c \setminus \partial A) = A^c \setminus \partial A$ so $A^c \setminus \partial A$ is open.

- From $(*)$ we see that $(\partial A)^c = X \setminus \partial A$ is the union of two open sets and so $(\partial A)^c$ is open. Therefore $\partial A$ is closed. $\blacksquare$