The Closedness of Finite Sets in a Metric Space

Recall from the Open and Closed Sets in Metric Spaces page that a set $S \subseteq M$ is said to be open in $M$ if $S = \mathrm{int} (S)$ and $S$ is said to be closed if $S^c$ is open.

We will now see that every finite set in a metric space is closed. We will first prove a useful lemma which shows that every singleton set in a metric space is closed.

Lemma 1: Let $(M, d)$ be a metric space. Then for each $x \in M$ the singleton set $\{ x \}$ is closed in $M$.

• Proof: Let $x \in M$ and consider the singleton set $\{ x \}$. The complement of $\{ x \}$ is $S \setminus \{ x \}$ which we want to prove is an open set.

• Let $y \in M \setminus \{ x \}$. Then there exists a distance $d(x, y) > 0$ between the points $y$ and $x$ with respect to the metric $d$. Let $r = \frac{d(x, y)}{2}$. Then the open ball centered at $y$ with radius $r$ is:

(1)

\begin{align} \quad B(y, r) = \left \{ z \in M : d(y, z) < r = \frac{d(x, y)}{2} \right \} \end{align}

• Then we have that the open ball $B(y, r)$ is fully contained in $M \setminus \{ x \}$. Hence $y \in \mathrm{int}(M \setminus \{ x \}$ and so $\mathrm{int} (M \setminus \{ x \}) = M \setminus \{ x \}$ so $M \setminus \{ x \}$ is an open set and so the singleton set $\{ x \}$ is closed. $\blacksquare$

Theorem 1: Let $(M, d)$ be a metric space. If $S \subseteq M$ is a finite set then $S$ is closed in $M$.

• Proof: Let $S \subseteq M$ be a finite set. Then $S$ has $n$ elements for some $n \in \mathbb{N}$ and can be written as:

(2)

\begin{align} \quad S = \{ x_1, x_2, ..., x_n \} \end{align}

• From Lemma 1, each singleton set $\{ x_j \}$ for $j \in \{1, 2, ..., n \}$ is closed. Note that:

(3)

\begin{align} \quad S = \bigcup_{j=1}^{n} \{ x_j \} \end{align}

• Therefore $S$ is a finite union of closed sets and is hence closed. $\blacksquare$