The Interior Points of Sets in a Topological Space Examples 2

The Interior Points of Sets in a Topological Space Examples 2

Recall from The Interior Points of Sets in a Topological Space page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then a point $a \in A$ is called an interior point of $A$ if there exists an open set $U \in \tau$ such that:

(1)
\begin{align} \quad a \in U \subseteq A \end{align}

We also proved some important results for a topological space $(X, \tau)$ with $A \subseteq X$:

• $A$ is open if and only if every $a \in A$ is an interior point of $A$, i.e., $A = \mathrm{int} (A)$.
• If $U \in \tau$ is such that $U \subseteq A$ then $U \subseteq \mathrm{int} (A)$.
• $\mathrm{int} (A)$ is the largest open subset of $A$.

We will now look at some examples regarding interior points of subsets of a topological space.

Example 1

Consider the topological space $(\mathbb{R}, \tau)$ where $\tau = \{ (-n, n) : n \in \mathbb{Z}, n \geq 1 \}$ and let $A = (-\pi, e)$. What is $\mathrm{int} (A)$?

It's not hard to see that $(-1, 1) \subseteq A$, $(-2, 2) \subseteq A$, and $(-3, 3) \not \subseteq A$ since $2.9 \in (-3, 3)$ and $2.9 \not \in (-\pi, e)$. Therefore the largest open subset of $A$ is $(-2, 2)$ and hence:

(2)
\begin{align} \quad \mathrm{int} (A) = (-2, 2) \end{align}

Example 2

Let $(X, \tau)$ be a topological space. Prove that if for every $A \subseteq X$ we have that $A = \mathrm{int} (A)$ then $\tau$ is the discrete topology on $X$.

Suppose that for every $A \subseteq X$ we have that $A = \mathrm{int} (A)$. Then every $A \subseteq X$ is open, i.e., every subset of $X$ is open, so $\tau = \mathcal P(X)$ and hence $\tau$ is the discrete topology on $X$.

Example 3

Let $(\mathbb{R}, \tau)$ be a topological space where $\tau$ is the topology of open intervals. What is $\mathrm{int} (\mathbb{N})$?

Consider the point $1 \in \mathbb{N}$. Notice that for any open interval $(a, b)$ such that $1 \in (a, b)$ we have that $(a, b) \not \subseteq \mathbb{N}$. For example, $1 \in (0, 2)$ but $(0, 2) \not \subseteq \mathbb{N}$. In fact, for all $a, b \in \mathbb{R}$ such that $a < 1 < b$ we have that there exists a number $x \in \mathbb{R}$ such that $x$ is not a natural number and where $a < 1 < x < b$. So $x \in (a, b)$ and $x \not \in \mathbb{N}$.

The same applies for all $n \in \mathbb{N}$, and hence $\mathrm{int} (\mathbb{N}) = \emptyset$.