The Real Line and the Epsilon Neighbourhood of a Real Number

The Real Number Line

One way to represent the real numbers $\mathbb{R}$ is on the real number line as depicted below.

We will now state the important geometric representation of the absolute value with respect to the real number line.

Definition: If $a$ and $b$ are real numbers, then we say that the distance from $a$ to the origin $0$ is the absolute value of $a$, $\mid a \mid$. We say that the distance between $a$ and $b$ is the absolute value of their difference, namely $\mid a - b \mid$.

For example consider the numbers $-2$ and $2$. There is a distance of $4$ in between these numbers because $\mid -2 - 2 \mid = \mid -4 \mid = 4$.

Epsilon Neighbourhood of a Real Number

Definition: Let $a$ be a real number and let $\epsilon > 0$. The $\epsilon$-neighbourhood of the number $a$ is the set denoted $V_{\epsilon} (a) := \{ x \in \mathbb{R} : \: \mid x - a \mid < \epsilon \}$. Alternatively we can define $V_{\epsilon}(a) := \{x \in \mathbb{R} : a - \epsilon < x < a + \epsilon \}$.

For example, consider the point $1$, and let $\epsilon_0 = 2$. Then $V_{\epsilon_0} (1) = \{ x \in \mathbb{R} : \mid x - 1 \mid < 2 \} = (-1, 3)$.

We will now look at a simple theorem regarding the epsilon-neighbourhood of a real number.

Theorem 1: Let $a$ be a real number. If $\forall \epsilon > 0$, $x \in V_{\epsilon} (a)$ then $x = a$.

• Proof of Theorem 1: Suppose that for some $x$, $\forall \epsilon > 0$, $\mid x - a \mid < \epsilon$. We know that then $\mid x - a \mid = 0$ if and only if $x - a = 0$ and therefore $x = a$. $\blacksquare$