The Interior Points of Sets in a Topological Space

Recall from the The Open Neighbourhoods of Points in a Topological Space page that if $(X, \tau)$ is a topological space and $x \in X$ then an open neighbourhood of $x$ is any open set $U$ ($U \in \tau$) such that $x \in U$.

Given a subset $A \subseteq X$, we will give a special name to the points $a \in A$ that contain an open neighbourhood $U$ fully contained in $A$.

Definition: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. A point $a \in A$ is called an Interior Point of $A$ if there exists an open neighbourhood $U$ ($U \in \tau$) of $a$ such that $a \in U \subseteq A$. The set of all interior points of $A$ is called the Interior of $A$ and is denoted $\mathrm{int} (A)$.

Let's now look at some simple results regarding interior points of a subset of $X$.

Proposition 1: Let $[[$ (X ,\tau)$ be a topological space. a) The interior of the whole set $X$ is $X$, that is, $\mathrm{int} (X) = X$. b) The interior of the empty set is the empty set, that is $\mathrm{int} (\emptyset) = \emptyset$.

• Proof of a) $X$ is an open set. For each $x \in X$, $X$ is an open neighbourhood of $x$, and so every $x \in X$ is an interior point of $X$. Thus $\mathrm{int} (X) = X$.

• Proof of b) $\emptyset$ is an open set. Since $\emptyset$ has no points, it vacuously satisfies the definition above, and thus, $\mathrm{int} (\emptyset) = \emptyset$. ||

Proposition 2: Let $(X, \tau)$ be a topological space. If $U \subseteq A \subseteq X$ and $U$ is open, then $U \subseteq \mathrm{int} (A)$.

• Proof: Let $U \subseteq A \subseteq X$ and let $U$ be an open set. Then for all elements in $a \in U$, we have that $a \in A$. So each $a \in U$ is such that $a \in U \subseteq A$, so $a \in \mathrm{int} (A)$. Therefore:

(1)

\begin{align} \quad U \subseteq \mathrm{int} (A) \quad \blacksquare \end{align}

Proposition 3: Let $(X, \tau)$ be a topological space. If $A \subseteq X$ then $\mathrm{int} (A)$ is the largest open subset of $A$.

• Proof: Suppose not. Then for some $p \not \in \mathrm{int} (A)$ we have that $\mathrm{int} (A) \cup \{ p \}$ is a larger open subset of $A$. But then if $U = \mathrm{int} (A) \cup \{ p \} \in \tau$ then:

(2)

\begin{align} \quad p \in U = \mathrm{int} (A) \cup \{ p \} \subseteq A \end{align}

• Therefore $p \in \mathrm{int} (A)$, a contradiction. Therefore our assumption that a larger open subset of $A$ exists was false. Hence $\mathrm{int}(A)$ is the largest open subset of $A$. $\blacksquare$

Proposition 4 (Idempotency of the Interior of a Set): Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then the interior of the interior of $A$ is equal to the interior of $A$, that is, $\mathrm{int}(\mathrm{int}(A)) = \mathrm{int}(A)$.

• Proof: By definition, $\mathrm{int} (\mathrm{int}(A))$ is the set of all interior points of $\mathrm{int}(A)$. By proposition 2, $\mathrm{int}(A)$ is open, and so every point of $\mathrm{int}(A)$ is an interior point of $\mathrm{int}(A)$. Therefore $\mathrm{int}(A) \subseteq \mathrm{int}(\mathrm{int}(A))$. But also by proposition 2 we have that $\mathrm{int}(\mathrm{int}(A)) \subseteq A$. $\blacksquare$

Example 1

Consider the set $X = \{ a, b, c \}$ with the nested topology $\tau = \{ \emptyset, \{ a \}, \{ a, b \}, \{a, b, c \} \}$. If we choose the set $A = \{ a, c \} \subset X$, we note that $a \in A$ is an interior point of $A$ if we let $U = \{ a \} \in \tau$ since:

(3)

\begin{align} \quad a \in U = \{ a \} \subset \{ a, c \} = A \end{align}

However, the point $c \in A$ is not an interior point with respect to the topology $\tau$. The only subset $U \in \tau$ that contains $c$ is $U = \{ a, b, c \}$ and:

(4)

\begin{align} \quad c \in U = \{a, b, c \} \not \subseteq \{a, c \} = A \end{align}

Therefore $\mathrm{int} (A) = \{ a \}$.

Example 2

For another example, consider the set $\mathbb{R}^2$ with the topology induced by the standard metric $d(x, y) = \| \mathbf{x} - \mathbf{y} \| = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$ for all $\mathbf{x} = (x_1, x_2), \mathbf{y} = (y_1, y_2) \in \mathbb{R}^2$. A set $S \subseteq \mathbb{R}^2$ if for every $x \in S$ there exists a positive real number $r > 0$ such that the open disk centered at $x$ with radius $r$ denoted $B(\mathbf{x}, r) = \{ \mathbf{y} \in S : d(\mathbf{x}, \mathbf{y}) < r \}$ is contained in $S$, that is:

(5)

\begin{align} \quad S \: \mathrm{open} \Leftrightarrow \exists r > 0 \: : x \in B(\mathbf{x}, r) \subseteq S \end{align}

If $a < b$ and $c < d$ then graphically we can represent the subset $A = [a, b) \times [c, d) \subseteq \mathbb{R}^2$ as:

The interior points of $A = [a, b) \times [c, d)$ are the points $\mathbf{x} = (x_1, x_2)$ such that $a < x_1 < b$ and $c < x_2 < d$. Any points with $x_1 = a$ and/or $x_2 = c$ cannot be interior points since there would then exist no positive real number $r > 0$ such that the disk centered at $\mathbf{x}$ with radius $r$ would be a subset of $A$ as illustrated in the following image:

Therefore the set of interior points is $\mathrm{int} (A) = (a, b) \times (c, d)$: