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Mathematics LibreTexts

4.1: Intervals

  • Page ID
    22657
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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 \(a<b\),

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

    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.

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