Neighborhoods in 𝐑

The Real Number Line

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

<img src="http://mathonline.wdfiles.com/local--files/the-real-line-and-the-epsilon-neighbourhood-of-a-real-number/Screen%20Shot%202014-09-30%20at%206.46.57%20PM.png" alt="Screen%20Shot%202014-09-30%20at%206.46.57%20PM.png" class="image" />

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 $\varepsilon >0$. The $\varepsilon$-neighbourhood of the number $a$ is the set denoted Failed to parse (syntax error): {\displaystyle V_{\varepsilon} (a) := \{ x \in \mathbb{R} : \: \mid x - a \mid < \varepsilon \}} . Alternatively we can define $V_{\varepsilon }(a):=\{x\in \mathbb {R} :a-\varepsilon <x<a+\varepsilon \}$. <img src="http://mathonline.wdfiles.com/local--files/the-real-line-and-the-epsilon-neighbourhood-of-a-real-number/Screen%20Shot%202014-12-05%20at%2010.16.09%20PM.png" alt="Screen%20Shot%202014-12-05%20at%2010.16.09%20PM.png" class="image" />

For example, consider the point $1$, and let $\varepsilon _{0}=2$. Then $V_{\varepsilon _{0}}(1)=\{x\in \mathbb {R} :\mid x-1\mid <2\}=(-1,3)$.

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 $\forall \varepsilon >0$, $x\in V_{\varepsilon }(a)$ then $x=a$.

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