
# 4.1: Intervals


## Definition: intervals

Given any two extended real numbers $$a<b,$$ we call the set

$(a, b)=\{x: x \in \mathbb{R}, a<x<b\}$

an open interval.

Given any two finite real numbers $$a \leq b,$$ we call the sets

$[a, b]=\{x: x \in \mathbb{R}, a \leq x \leq b\},$

$(-\infty, b]=\{x: x \in \mathbb{R}, x \leq b\},$

and

$[a,+\infty)=\{x: x \in \mathbb{R}, x \geq a\}$

closed intervals.

We call any set which is an open interval, a closed interval, or is given by, for some finite real numbers $$a<b$$,

$(a, b]=\{x: x \in \mathbb{R}, a<x \leq b\}$

or

$[a, b)=\{x: x \in \mathbb{R}, a \leq x<b\},$

an interval.

## Proposition $$\PageIndex{1}$$

If $$a, b \in \mathbb{R}$$ with $$a<b,$$ then

$(a, b)=\{x: x=\lambda a+(1-\lambda) b, 0<\lambda<1\}.$

Proof

Suppose $$x=\lambda a+(1-\lambda) b$$ for some $$0<\lambda<1 .$$ Then

$b-x=\lambda b-\lambda a=\lambda(b-a)>0,$

so $$x<b .$$ Similarly,

$x-a=(\lambda-1) a+(1-\lambda) b=(1-\lambda)(b-a)>0,$

so $$a<x .$$ Hence $$x \in(a, b)$$.

$$\quad$$ Now suppose $$x \in(a, b) .$$ Then

$x=\left(\frac{b-x}{b-a}\right) a+\left(\frac{x-a}{b-a}\right) b=\left(\frac{b-x}{b-a}\right) a+\left(1-\frac{b-x}{b-a}\right) b$

and

$0<\frac{b-x}{b-a}<1.$

Q.E.D.