Criterion for a Subset of Euclidean Space to be Closed

Criterion for a Subset of Euclidean Space to be Closed

Recall that if $S \subseteq \mathbb{R}^n$ then:

• A point $\mathbf{x} \in \mathbb{R}^n$ is an adherent point of $S$ if every ball $B(\mathbf{x}, r)$ contains a point from $S$, and the set of all adherent points of $S$ is called the closure of $S$ denoted $\bar{S}$.
• A point $\mathbf{x} \in \mathbb{R}^n$ is an accumulation point of $S$ if every ball $B(\mathbf{x}, r)$ contains a point from $S$ different from $\mathbf{x}$, and the set of all accumulation points of $S$ is called the derived set of $S$ denoted $S'$.
• A point $\mathbf{s} \in S$ is an isolated point of $S$ if there exists a ball $B(\mathbf{x}, r)$ that contains no points from $S$ different from $\mathbf{x}$.

We will now look a theorem which gives us criterion for $S \subseteq \mathbb{R}^n$ to be closed.

 Theorem 1: Let $S \subseteq \mathbb{R}^n$. Then $S$ is closed if and only if $\bar{S} \subseteq S$.

In other words, $S$ is closed if and only if $S$ contains all of its adherent points.

• Proof: $\Rightarrow$ Suppose that $S$ is closed. Assume that that instead $\bar{S} \not \subseteq S$. Then there exists an adherent point $\mathbf{x} \in \bar{S}$ such that $\mathbf{x} \not \in S$. We will show that there exists a ball centered at $\mathbf{x}$ that contains no elements from $S$ which contradicts $\mathbf{x}$ from being an adherent point.
• Since $\mathbf{x} \not \in S$ we must have that $\mathbf{x} \in \mathbb{R}^n \setminus S$, i.e., [$\mathbf{x} \in S^c$. Since $S$ is closed, we have that $S^c$ is open, so $\mathbf{x}$ is contained in an open set. So $\mathbf{S^c} = \mathbf{\mathrm{int}(S^c)}$, i.e., $\mathbf{x} \in \mathrm{int} (S^c)$, so there exists a positive real number $r > 0$ such that:
(1)
\begin{align} \quad B(\mathbf{x}, r) \subseteq S^c = \mathbb{R}^n \setminus S \end{align}
• Therefore there exists a ball centered at $\mathbf{x}$ that contains no elements from $S$ which contradicts $\mathbf{x}$ from being an adherent point. Therefore the assumption that $\bar{S} \not \subseteq S$ is false, so $\bar{S} \subseteq S$.
• $\Leftarrow$ Now suppose that $\bar{S} \subseteq S$. We want to show that $S$ is closed. Consider the complement $S^c = \mathbb{R}^n \setminus S$. Let $\mathbf{x} \in S^c$. Then all $\mathbf{x} \in S^c$ is not an adherent point of $S$, so there exists a ball centered at $\mathbf{x}$ with radius $r_x > 0$ such that $B(\mathbf{x}, r_x)$ contains no elements from $S$, i.e., $B(\mathbf{x}, r_x) \subseteq S^c$ for all $\mathbf{x} \in S^c$. So every point $\mathbf{x} \in S^c$ is an interior point of $S^c$ so $\mathrm{int} (S^c) = S^c$, so $S^c$ is open and $S$ is therefore closed. $\blacksquare$
 Corollary 1: Let $S \subseteq \mathbb{R}^n$. Then $S$ is closed if and only if $S = \bar{S}$.
• Proof: By Theorem 1 we have that $S$ is closed if and only if $\bar{S} \subseteq S$ so we only need to show that $S \supseteq S$.
• Let $\mathbf{x} \in S$. Then every ball centered at $\mathbf{x}$ contains itself, so $\mathbf{x} \in \bar{S}$. Therefore $S \subseteq \bar{S}$ as desired. $\blacksquare$