The Boundary of a Set in a Topological Space Examples 1

The Boundary of a Set in a Topological Space Examples 1

Recall from The Boundary of a Set in a Topological Space that if $(X, \tau)$ and $A \subseteq X$ then a point $x \in X$ is said to be a boundary point of $A$ if for every $U \in \tau$ with $x \in U$ we have that $U$ intersects $A$ and $A^c$ nontrivially, i.e., $A \cap U \neq \emptyset$ and $A^c \cap U \neq \emptyset$. We noted that the set of all boundary points of $A$ is denoted $\partial A$.

We also proved an important fact in that $\partial A$ is a closed set.

We will now look at some examples regarding the boundary of a set.

Example 1

Let $X = \{ a, b, c \}$ and let $\tau = \{ \emptyset, \{ a \}, \{ b, c \}, X \}$. Consider the set $A = \{ a \}$. Determine $\partial A$.

Since $A = \{ a \}$ we see that $A^c = \{b, c \}$.

First consider the point $a \in X$. The open neighbourhoods of this point are $\{ a \}$ and $X$. We have that $A^c \cap \{ a \} = \emptyset$. Therefore $a \not \in \partial A$.

Now consider the point $b \in X$. The open neighbourhoods of this point are $\{ b, c \}$ and $X$. We have that $A \cap \{ b, c \} = \emptyset$, so $b \not \in \partial A$.

Lastly, consider the point $c \in X$. The open neighbourhoods of this point are $\{b, c \}$ and $X$. We have that $A \cap \{ b, c \} = \emptyset$, so $c \not \in \partial A$.

Therefore $\partial A = \emptyset$.

Example 2

Let $X = \{ a, b, c, d \}$ and let $\tau = \{ \emptyset, \{ a \}, \{ b \}, \{ a, b \}, \{ b, d \}, \{a, b, d \}, X \}$. Consider the set $A = \{ a, b, c \}$. Determine $\partial A$.

Since $A = \{ a, b, c \}$ we have that $A^c = \{ d \}$.

First look at the point $a \in X$. The open neighbourhoods of $a$ are $\{ a \}$, $\{ a, b \}$, $\{a, b, d \}$, and $X$. Notice that $A^c \cap \{ a \} = \emptyset$. Therefore $a \not \in \partial A$.

Now look at the point $b \in X$. The open neighbourhoods of $b$ are $\{ b \}$, $\{a, b \}$, $\{ b, d \}$, $\{a, b, d \}$, and $X$. Notice that $A^c \cap \{ b \} = \emptyset$. Therefore $b \not \in \partial A$.

Now look at the point $c \in X$. The only open neighbourhood of $c$ is $X$. We have that $A \cap X = \{a, b , c \}$ and $A^c \cap X = \{ d \}$ - both of which sets are nonempty. Therefore $c \in \partial A$.

Lastly look at the point $d \in X$. The open neighbourhoods of $d$ are $\{ b, d \}$, $\{a, b, d \}$, and $X$. We have that $A \cap \{ b, d \} = \{ b \}$ and $A^c \cap \{ b, d \} = \{ d \}$; $A \cap \{a, b, d \} = \{a, b \}$ and $A^c \cap \{a, b, d \} = \{ d \}$; and $A \cap X = \{a, b, c \}$ and $A^c \cap X = \{ d \}$. All of these sets are nonempty. Therefore $d \in \partial A$.

Hence $\partial A = \{ c, d \}$.