4.1: Intervals

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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 \(a<b\),

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

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

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.