Interior point in topology

• Definition
• Related concepts

Given a subset AA of a topological space XX, a point x∈Ax \in A is an interior point of AA it it is contained in the interior of AA, hence if AA is a neighborhood of xx in XX, i.e., if there is some open subset containing xx that is included in AA.

The Interior Points of Sets in a Topological Space Examples 1

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:

\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 set $X = \{ a, b, c \}$ and the nested topology $\tau = \{ \emptyset, \{ a \}, \{a, b \}, X \}$. Let $A = \{ a, c \} \subset X$. What are the interior points of $A$?

We note that all interior points of $A$ must be contained in $A$ by the definition of an interior point, so we need to only check whether $a \in A$ is an interior point and whether $c \in A$ is an interior point.

For $a \in A$, does there exists an open set $U \in \tau$ such that $a \in U \subseteq A$? Yes! The set $U = \{ a \} \in \tau$ and:

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

Therefore $a \in A$ is an interior point of $A$.

For $c \in A$, does there exist an open set $U \in \tau$ such that $a \in U \subseteq A$? No! The only set in $\tau$ containing $c$ is the wholeset $X = \{ a, b, c \}$ and $X \not \subseteq A$ since $b \in X$ and $b \not \in A$. Therefore $c$ is not an interior point of $A$.

Example 2

Consider an arbitrary set $X$ with the discrete topology $\tau = \mathcal P [X]$. Let $S \subseteq X$. What are the interior points of $S$?

Let $x \in S$. Since $S \subseteq X$, we have that $S \in \tau = \mathcal P[X]$. Let $U = S$. Then for each $x \in S$ we have that:

\begin{align} \quad x \in U = S \subseteq S \end{align}

Therefore every point $x \in S$ is an interior point of $S$.

Example 3

Consider an arbitrary set $X$ with the indiscrete topology $\tau = \{ \emptyset, X \}$. Let $S$ be a nontrivial subset of $X$. What are the interior points of $S$?

Let $S$ be a nontrivial subset of $X$. Then:

\begin{align} \quad \emptyset \subset S \subset X \end{align}

For all $x \in S$, we see from the nesting above that there exists no open set $U \in \tau$ such that $x \in U \subseteq S$. Therefore, every point $x \in S$ is not an interior point of $S$.

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 $A$.

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

• 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, $\mathrm{int} [\emptyset] = \emptyset$. ||

• 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:

\begin{align} \quad U \subseteq \mathrm{int} [A] \quad \blacksquare \end{align}

• 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:

\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 $A$. $\blacksquare$

• 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 $\mathrm{int}[\mathrm{int}[A]] \subseteq A$. $\blacksquare$

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:

\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:

\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:

\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 $\mathrm{int} [A] = [a, b] \times [c, d]$: