# 3.10. Discrete Metric Space#

This section collects results on the discrete metric space. The discrete space is trivial and not very useful in applications. However, it helps clarify many of the theoretical underpinnings of the topology of metric spaces. Hence, its study is quite useful.

Definition 3.75

Let $$X$$ be a nonempty set:

Define:

$\begin{split} d(x,y) = \begin{cases} 0 & x = y \\ 1 & x \neq y \end{cases}\;. \end{split}$

$$(X, d)$$ is a metric space. This distance is called discrete distance and the metric space is called a discrete metric space.

In the rest of the section $$X$$ will denote the discrete metric space with the distance function defined above.

## 3.10.1. Open and Closed Sets#

Proposition 3.23

Every singleton in a discrete metric space is open.

Proof. Let $$x \in X$$. Consider the open ball $$B(x, \frac{1}{2})$$:

$B(x, \frac{1}{2}) = \left \{y \in X \ST d(x,y) < \frac{1}{2} \right \}.$

By definition of the discrete metric:

$B(x, \frac{1}{2}) = \{ x \}.$

Thus, every singleton is an open ball. Hence it is an open set.

Proposition 3.24

Every subset of a discrete space is open.

Proof. Let $$A \subseteq X$$. If $$A = \EmptySet$$ then there is nothing to prove.

For a nonempty $$A$$, write it as:

$A = \bigcup_{x \in A} \{ x \}.$

Since every singleton is open and an arbitrary union of open sets is open hence $$A$$ is open.

Proposition 3.25

Every subset of a discrete set is closed.

Proof. Let $$A \subseteq X$$. Let $$B = X \setminus A$$. By previous result, $$B$$ is open. Hence, $$A$$ must be closed.

## 3.10.2. Boundedness#

Proposition 3.26

The diameter of the open ball $$B(x, 1)$$ is 0.

Proof. Note that

$B(x, 1) = \{ x\}.$

Thus, $$B(x, 1)$$ is a singleton set.

Hence,

$\diam B(x, 1) = \sup \{ d(x,y) \ST x, y \in B(x, 1) \} = d(x,x) = 0.$

This result is a counter example to explain that while $$\diam B(x,r) \leq 2 r$$ always, it doesn’t need to be equal to $$2 r$$.

Proposition 3.27

The discrete space is bounded.

Proof. Let $$x, y \in X$$.

1. If $$x \neq y$$, then $$d(x,y) = 1$$.

2. Thus, $$\diam X = \sup d(x,y) = 1$$.

3. $$X$$ is bounded.

## 3.10.3. Rare Sets#

Proposition 3.28

The only rare (nowhere dense) subset of $$X$$ is $$\EmptySet$$.

Proof. Let $$A \subseteq X$$. $$A$$ is open as well as closed. Hence $$\closure A = A$$ and $$\interior \closure A = \interior A = A$$. Thus, $$A$$ is rare if and only if $$A = \EmptySet$$.

## 3.10.4. Cauchy Sequences#

Proposition 3.29

In a discrete metric space, a Cauchy sequence is eventually constant.

Proof. Let $$\{x_n \}$$ be a Cauchy sequence of $$X$$. Then, there exists $$k$$ such that for all $$m,n > k$$,

$d(x_m, x_n) < \frac{1}{2}.$

But then for all $$m,n > k$$, $$x_m = x_n$$. Thus, $$\{ x_n \}$$ must be eventually constant.

## 3.10.5. Completeness#

Proposition 3.30

A discrete metric space is complete.

Proof. Since every Cauchy sequence is eventually constant, hence it converges. Thus the discrete metric space is complete.

## 3.10.6. Meager Sets#

Proposition 3.31

The only meager set in $$X$$ is $$\EmptySet$$.

Proof. Recall from Theorem 3.66 that a meager set has an empty interior.

Since every nonempty subset of $$X$$ is open hence it doesn’t have an empty interior.

Thus, the only meager set is $$\EmptySet$$.

## 3.10.7. Baire Category Theorem#

Observation 3.2

Let $$X$$ be a countable set with a discrete metric.

$$X$$ is complete and it satisfies the Baire category theorem.

Let $$\{x_n\}$$ be an enumeration of $$X$$. Then, we can write $$X$$ as:

$X = \bigcup_{n=1}^{\infty} \{ x_n\}.$

Now, although it’s a countable union of singletons, the singletons themselves are not rare sets. Hence, we cannot say that $$X$$ meager.

## 3.10.8. Compactness#

Example 3.29

We can construct a closed and bounded set which is not compact.

1. Let $$X$$ be an infinite set with the discrete metric.

2. Then, $$B(x,1) = \{ x\}$$ is a singleton for every $$x \in X$$.

3. $$X$$ is closed.

4. $$X$$ is bounded since $$d(x, y) \leq 1$$ for every $$x, y \in X$$. Hence $$\diam X = 1$$.

5. Consider the open cover $$X = \bigcup_{x \in X} B(x, 1)$$.

6. This cover cannot be reduced to a finite subcover.

7. Thus, $$X$$ is not compact even though it is closed and bounded.