The Closure of a Set and the Distance from Points to a Set

# The Closure of a Set and the Distance from Points to a Set

Recall from the The Distance Between Points and Subsets in a Metric Space page that if $(S, d)$ is a metric space, $A \subseteq S$, and $x \in S$ then we can define a function $f_A : S \to \mathbb{R}$ for all $x \in S$ by:

(1)
\begin{align} \quad f_A(x) = \inf \{ d(x, y) : y \in A \} \end{align}

We said that the distance from $x$ to $A$ is defined to be the number $f_A(x)$. We will now look at a nice theorem which gives us an alternative definition for the closure of a set in terms of collection of all points in $S$ that are of a distance of $0$ from $A$.

 Theorem 1: Let $(S, d)$ be a metric space and let $A \subseteq S$. Then the closure of $A$ is the set of all points $x \in S$ whose distance to $A$ is equal to $0$, that is, $\bar{A} = \{ x \in S : f_A(x) = 0 \}$.
• Proof: Let $x \in \bar{A}$. Then $x$ is either an isolated point of $A$ or an accumulation point of $A$. If $x$ is an isolated point of $A$ then $x \in A$, and so:
(2)
\begin{align} \quad f_A(x) = \{ \inf d(x, y) : y \in A \} = d(x, x) = 0 \end{align}
• Therefore $x \in \{ x \in S : f_A(x) = 0 \}$. If $x$ is not an isolated point of $A$ then $x$ is an accumulation point of $A$ and so for all $r > 0$ the open ball centered at $x$ with radius $r$ contains some points of $A$ different from $x$, that is, for all $r > 0$:
(3)
\begin{align} \quad B(x, r) \cap A \setminus \{ x \} \neq \emptyset \end{align}
• So for every $r > 0$ there exists a $y \in A$ such that $d(x, y) = r$. So $f_A(x) = 0$ and therefore $x \in \{ x \in S : f_A(x) = 0 \}$. So:
(4)
\begin{align} \quad \bar{A} \subseteq \{ x \in S : f_A(x) = 0 \} \end{align}
• Now suppose that $x \in \{ x \in S : f_A(x) = 0 \}$. Then $f_A(x) = 0$. So either $x \in A$ or $x \not \in A$. If $x \in A$ then since $A \subseteq bar{A}$ we see that $x \in \bar{A}$. If $x \not \in A$ then since $x \in S$ and $f_A(x) = 0$ we see that for every open ball centered at $x$ with radius $r > 0$ there exists a $y \in A$ such that $d(x, y) = r$. So for all $r > 0$ we have that $B(x, r) \cap A \setminus \{ x \} \neq \emptyset$ which implies that $x$ is an accumulation point of $A$, so $x \in A’ \subseteq A \cup A’ = \bar{A}$. Therefore:
(5)
\begin{align} \quad \bar{A} \supseteq \{ x \in S : f_A(x) = 0 \} \end{align}
• So we conclude that $\bar{A} = \{ x \in S : f_A(x) = 0 \}$. $\blacksquare$