Basic Thms. Regarding the Boundary of a Set in a Topological Space

Basic Theorems Regarding the Boundary of a Set 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 called a boundary point of $A$ if $x$ is contained in the closure of $A$ but not in the interior of $A$, that is, $x \in \bar{A} \setminus \mathrm{int}(A)$.

Equivalently, $x$ is a boundary point of $A$ if every $U \in \tau$ with $x \in U$ intersects $A$ and $A^c$ nontrivially. Furthermore, we said that the set of all boundary points of $A$ is called the boundary of $A$ and is denoted $\partial A$.

We also proved a very important theorem in that the boundary of any set $A$ is a closed set, i.e., $\partial A$ is closed.

Important Note!

Before we look at a theorem regarding the boundary of sets in a topological space, we will first make a very important point. Let $(X, \tau)$ be a topological space and suppose that $A, B \in X$ where $A \subseteq B$. Does $\partial A \subseteq \partial B$? The answer is NO as illustrated with the following diagram:

For example, consider the topological space $(\mathbb{R}^2, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}^2$ of open disks. Consider the following subsets:

(1)

\begin{align} \quad A = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 < 1 \} \quad B = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 < 4 \} \end{align}

Then $A$ is the open disk centered at the origin with radius $1$ and $B$ is the open disk centered at the origin with radius $2$. Clearly $A \subseteq B$. The sets $A$ and $B$ are illustrated below:

However, notice that the boundaries of $A$ and $B$ are:

(2)

\begin{align} \quad \partial A = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \} \quad \partial B = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 = 4 \} \end{align}

In other words, the boundary of $A$ is the circle centered at the origin with radius $1$ and the boundary of $B$ is the circle centered at the origin with radius $2$.

These circles are concentric and do not intersect at all. Hence, $\partial A \not \subseteq \partial B$ and $\partial B \not \subseteq \partial A$.

We will now look at an important theorem regarding the boundary of sets in a topological space.

Theorem 1: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $\partial A \cup \partial B \supseteq \partial (A \cup B)$.

• Proof: Let $x \in \partial (A \cup B)$. Then $x \in \overline{A \cup B} \setminus \mathrm{int} (A \cup B)$.

• Since $x \in \overline{A \cup B}$ we have that $x \in \bar{A} \cup \bar{B}$ by one of the theorem on the Basic Theorems Regarding the Closure of Sets in a Topological Space page. Therefore $x \in \bar{A}$ or $x \in \bar{B}$ $(*)$

• Similarly, since $x \not \in \mathrm{int} (A \cup B)$ we have that $x \not \in \mathrm{int} (A) \cup \mathrm{int} (B)$ (since if so then $x \in \mathrm{int} (A \cup B)$ since $\mathrm{int} (A) \cup \mathrm{int} (B) \subseteq \mathrm{int} (A \cup B)$ by one of the theorems on the Basic Theorems Regarding the Interior Points of Sets in a Topological Space page). Hence $x \not \in \mathrm{int} (A)$ and $x \not \in \mathrm{int} (B)$ $(**)$.

• Combining $(*)$ and $(**)$ we see that:

(3)

\begin{align} \quad x \in (\bar{A} \setminus \mathrm{int} (A) ) \cup (\bar{B} \setminus \mathrm{int} (B)) = \partial A \cup \partial B \end{align}

• Therefore $\partial A \cup \partial B \supseteq \partial (A \cup B)$. $\blacksquare$

Important Note!

Note that in general we have that $\partial A \cap \partial B \not \subseteq \partial (A \cap B)$. For example, consider the sets $A = [0, 1)$ and $B = (1, 2]$. The boundaries of these sets are:

(4)

\begin{align} \quad \partial A = \{ 0, 1 \} \quad \partial B = \{1, 2 \} \end{align}

Therefore $\partial A \cap \partial B = \{ 1 \}$.

Now we have that $A \cap B = \emptyset$. So:

(5)

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

Clearly $\partial A \cap \partial B = \{ 1 \} \not \subseteq \emptyset = \partial (A \cap B)$.

Furthermore, we also have that in general $\partial A \cap \partial B \not \supseteq \partial (A \cap B)$. Once again consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}$. Let $A = [0, 2]$ and $B = [1, 3]$. Then $\partial A = \{ 0, 2 \}$ and $\partial B = \{1, 3 \}$ so:

(6)

\begin{align} \quad \partial A \cap \partial B = \{0, 2 \} \cap \{1, 3 \} = \emptyset \end{align}

Furthermore $A \cap B = [1, 2]$ and so the boundary of $A \cap B$ is given as:

(7)

\begin{align} \quad \partial (A \cap B) = \{ 1, 2 \} \end{align}

Clearly $\emptyset \not \supseteq \{ 1, 2 \}$ and so in this case, $\partial A \cap \partial B \not \supseteq \partial (A \cap B)$.