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10.17 Absolute value and Supremum

Absolute value

Definition

For \(x\in R\) define \(|x|=\begin{cases}x~~~~~~~~~\text{if }x\geq 0\\-x~~~~~~\text{if }x\leq 0\end{cases}\) absolute value of \(x\)

Example

  1. If \(|x|=0\Rightarrow x=0\)

  2. \(\forall x,|-x|=|x|\)​

Proposition

  1. \(|x\cdot y|=|x||y|\)​

  2. \(|x|+|y|\geq|x+y|\) (Triangular inequality)

    Analysis: It's seems complex to prove it case-by-case because we need to consider positivity and negativity of \(x,y,x+y\), but it works and is the easiest way!

    Proof

    If \(x+y\geq 0\Rightarrow |x+y|=x+y\), and since \(|x|\geq x\) and \(|y|\geq y\)

    Then \(|x+y|=x+y\leq |x|+|y|\)

    If \(x+y<0\Rightarrow\left|x+y\right|=-\left(x+y\right)=-x+\left(-y\right)\leq\left|x\right|+\left|y\right|\)​

Bounds, Max, Min, Sup, Inf

Definition

Let \(A\neq \emptyset\) be a set of real numbers (that is \(A\subset \R\))

We say:

  1. If \(\forall a\in A,\alpha\geq a\), \(\alpha\) is an upper bound of \(A\)

    If \(\forall a\in A,\alpha\leq a\), \(\alpha\) is an lower bound of \(A\)

  2. If \(\begin{cases}\alpha \text{ is upper bound of }A\\\alpha \in A\end{cases}\)​\, \(\alpha\) is a maximum of \(A\)

    If \(\begin{cases}\alpha \text{ is lower bound of }A\\\alpha \in A\end{cases}\)\, \(\alpha\) is a minimum of \(A\)

  3. Intuition definition of sup and inf

    If \(s\) is the minimum of the upper bounds\, \(s\) is the supremum of \(A\)

    If \(i\) is the maximum of the lower bounds\, \(i\) is the infimum of \(A\)

  4. Formal definition of sup and inf

    Supremum:

    \(1)a\leq s~~~\forall a\in A\left(s\text{ is an upper bound for A}\right)\\2)If~s^{\prime}\geq a~~~\forall a\in A\Rightarrow\) \(s^{\prime}\geq s,\forall s^{\prime}\in UBs~of~A\)

    This means if we find any other upper bounds, they are all greater than sup

    Infimum

    \(1)i\leq a~~~\forall a\in A\left(i\text{ is an lower bound for A}\right)\\2)If~i^{\prime}\leq a~~~~\forall a\in A\Rightarrow i^{\prime}\leq i,\) \(\forall i^{\prime}\in LBs~of~A\)

  5. Another definition of sup and inf

    \(s=sup(A)\Leftrightarrow\)​\(\begin{cases}a\leq s,\forall a\in A\\if~s'<s\Rightarrow \exists a\in A, s'<a\leq s\end{cases}\)

    This means if we find any element less than sup, there always exists a element in the set between them

    \(i=inf(A)\Leftrightarrow\)​\(\begin{cases}i\leq a,\forall a\in A\\if~i'>i\Rightarrow \exists a\in A, i\leq a<i'\end{cases}\)

  6. The third definition of sup and inf​

Remark

  1. If \(\max(A)\) exists, then the \(\sup(A)=\max(A)\)

    If \(\min(A)\) exists, then the \(\inf(A)=\min(A)\)

  2. Sometimes there is no maximum.

    It is the same with the supremum such as \(\N\)

  3. If \(A\neq 0\) and has upper bounds, it definitely has a \(sup(A)\)

    Because Axiom of completeness: If \(A\subseteq \R\), \(A\neq\emptyset\) and \(A\) has upper bounds, then \(A\) has \(sup(A)\).

    Remark

    Axiom of completeness is about bounded set. Thus, if you want to use it, you should consider construct it.

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