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:
- 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:
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:
- By lemma 1 we have that:
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:
- And since $X \setminus (X \setminus A) = A$, we also have by proposition that:
- 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)Furthermore, the interior of $A$ is:
(8)Hence we see that the boundary of $A$ is as expected:
(9)Example 2
For another example, consider the set $B = [0, 1) \cup (2, 3) \subset \mathbb{R}$. The closure of $A$ is:
(10)The interior of $B$ is:
(11)Hence we see that the boundary of $B$ is:
(12)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: