The Interior Points of Sets in a Topological Space

# 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 \tauthen: (2) \begin{align} \quad p \in U = \mathrm{int} (A) \cup \{ p \} \subseteq A \end{align} • Thereforep \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$U = \{ a \} \in \tausince: (3) \begin{align} \quad a \in U = \{ a \} \subset \{ a, c \} = A \end{align} However, the pointc \in A$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$S, that is: (5) \begin{align} \quad S \: \mathrm{open} \Leftrightarrow \exists r > 0 \: : x \in B(\mathbf{x}, r) \subseteq S \end{align} Ifa < b$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)\$: 