4.1: Intervals
- Page ID
- 22657
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.
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.