Closure of a Set

For a set $A \subset \mathbb{R}$, and $x \in \mathbb{R}$, we say that $x$ is an adherent point of $A$ if for any $\epsilon > 0$, the interval $(x - \epsilon, x + \epsilon)$ contains a point of $A$. The set of all adherent points of $A$ is called the closure of $A$, denoted by $\text{Adh}(A)$.

As we know, if $i \in \mathbb{R}$ is an interior point of $A$, then there exists $\epsilon > 0$ such that $(i - \epsilon, i + \epsilon) \subset A$. Thus, $i$ is an adherent point of $A$. Also, if $b \in \mathbb{R}$ is a boundary point of $A$, then for any $\epsilon > 0$, the interval $(b - \epsilon, b + \epsilon)$ contains a point of $A$, so $b$ is an adherent point of $A$.

The set of interior points of $A$ is denoted by $\text{Int}(A)$, the set of boundary points of $A$ by $\text{Bd}(A)$, and the set of exterior points of $A$ by $\text{Ext}(A)$.

We also know that $\mathbb{R} = \text{Int}(A) \cup \text{Bd}(A) \cup \text{Ext}(A)$, so let’s check that any point $e \in \text{Ext}(A)$ is not an adherent point of $A$.

Let $e \in \text{Ext}(A)$, then there exists $\epsilon > 0$ such that $(e - \epsilon, e + \epsilon) \subset \text{Ext}(A)$. Thus, $(e - \epsilon, e + \epsilon) \cap A = \emptyset$, meaning $e$ is not an adherent point of $A$.

Finally, we can prove that $\text{Adh}(A) = \text{Int}(A) \cup \text{Bd}(A)$. Let $x \in \text{Adh}(A)$, then for any $\epsilon > 0$, the interval $(x - \epsilon, x + \epsilon)$ contains a point of $A$. If $x \in \text{Int}(A)$, then there exists $\epsilon > 0$ such that $(x - \epsilon, x + \epsilon) \subset A$, so $x \in \text{Int}(A)$. If $x \in \text{Bd}(A)$, then for any $\epsilon > 0$, the interval $(x - \epsilon, x + \epsilon)$ contains a point of $A$, so $x \in \text{Bd}(A)$.