Open and Closed Sets in the Discrete Metric Space

Open and Closed Sets in the Discrete Metric Space

Recall from the Open and Closed Sets in Metric Spaces page that if $(M, d)$ is a metric space then a subset $S \subseteq M$ is said to be either open if $S = \mathrm{int} (S)$. Furthermore, $S$ is said to be closed if $S^c$ is open, and $S$ is said to be clopen if $S$ is both open and closed.

We also noted that a set $S$ may be neither open, closed, or clopen.

We will now look at the open and closed sets of a particular interesting example of a metric space - the discrete metric space - which can be obtained by taking a nonempty set $M$ with The Discrete Metric defined for all $x, y \in M$ by:

\begin{align} \quad d(x, y) = \left\{\begin{matrix} 0 & \mathrm{if} \: x = y\\ 1 & \mathrm{if} \: x \neq y \end{matrix}\right. \end{align}

We will now show that for every subset $S$ of a discrete metric space is both closed and open, i.e., clopen.

Theorem 1: Let $(M, d)$ be the discrete metric space where for all $x, y \in M$, $d(x, y) = \left\{\begin{matrix} 0 & \mathrm{if} \: x = y\\ 1 & \mathrm{if} \: x \neq y \end{matrix}\right.$. Then every subset $S \subseteq M$ is clopen.
  • Proof: We first show that every singleton set is open. Consider the set $\{ x \}$ and consider the ball $B \left ( x, \frac{1}{2} \right )$. Since $d(x, y) = 1$ if and only if $x \neq y$, we see that the ball centered at $x$ with radius $r = \frac{1}{2} > 0$ contains only $x$ and that:
\begin{align} \quad B \left ( x, \frac{1}{2} \right ) \subseteq \{ x \} = \{ x \} \end{align}
  • Therefore $\{ x \}$ is open.
  • Now more generally, for any subset $S \subseteq M$ where $x \in S$ we see that $B \left ( x, \frac{1}{2} \right ) = \{ x \} \subseteq S$, so $S$ is open.
  • Now consider the complement $S^c = M \setminus S$. If $y \in S^c$ then once again, $B \left (y, \frac{1}{2} \right ) = \{ y \} \subseteq S^c$ so $S^c$ is open. But then by definition, $S$ is also closed, so $S$ is clopen. $\blacksquare$
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