Discrete Set/Examples/Rational Numbers

Example of Discrete Set

Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.

The rational numbers $\Q$ do not form a discrete set within $T$.

Proof

Let $x \in \Q$ be an arbitrary rational number.

Let $\epsilon \in \R_{>0}$ be an arbitrary (strictly) positive rational number.

Let $\map {B_\epsilon} x \subseteq \R$ be the open ball of radius $\epsilon$ on $\R$ whose center is $x$.

That is:

$\map {B_\epsilon} x := \openint {x - \epsilon} {x + \epsilon}$

is the open interval as described.

Consider the rational number $y = x + \dfrac \epsilon 2$.

It is seen that $y \in \map {B_\epsilon} x$.

As $\epsilon$ is arbitrary, it follows that no such $\map {B_\epsilon} x$ contains only $x$ from $\Q$.

Hence by definition $x$ is not isolated in $\Q$.

As $x$ is arbitrary, the result follows.

$\blacksquare$

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): discrete set
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): discrete set