106 lines
3.4 KiB
Markdown
106 lines
3.4 KiB
Markdown
# Lecture 2
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Ordered sets, least upper bounds and fields.
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## Warm up
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(a) The statements says: $\forall a\in A, \exists s\in a$ such that $s\geq 7$.
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The negation is $\exist a\in A,\forall s\in a$, such that $s<7$.
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## Ordered sets
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Let $S$ be a set. An order on $S$ is a relation satisfying:
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1. "trichotomy". If $x,y\in S$, then exactly on eof the these statements are hold: $x<y,x=y,x>y$.
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2. "transitivity". If $x,y,z\in S$, then $x<y,y\implies x<z$.
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An ordered set is a set with order.
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### Definition 1.7
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Let $S$ be an ordered set and let $E\subset S$ ($E$ is the "universe", all the element you can ever think of...)
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1. $\beta\in S$ is an upper bound of $E$ if $\forall x\in E,x\leq \beta$
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2. $E$ is bounded above if $\exist \beta \in S$ such that $\beta$ is the upper bound of $E$.
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#### Example
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1. $S=\mathbb{Q}, E=\{1,2,3\}$ ($E$ is bounded above)
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* 3,4,3.5 are all upper bounds of $E$.
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* 2,2.5 is not upper bounds of $E$.
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* $\pi$ is not upper bound of $S$ since $\pi\notin S$
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2. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$ ($E$ is bounded above)
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* The upper bound is $1$.
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3. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x\}$ ($E$ is not bounded above)
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* Sad
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4. $S=\mathbb{Q}, E=\phi$.
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* $\beta\in S$ is an upper bound of $E$ if $\forall x\in E,x\leq \beta\equiv\beta\in S$ is not an upper bound of $E$ if $\exists x\in E,x> \beta$
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* So this statement is true for any rational numbers since $\cancel{\exist} a\in E$ such that $x>\beta$.
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### Definition 1.8
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Least upper bound, LUB, supremum, SUP
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Let $S$ be an ordered set and $E\subset S$. We say $\alpha\in S$ is the LUB of $E$ if
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1. $\alpha$ is the UB of $E$. ($\forall x\in E,x\leq \alpha$)
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2. if $\gamma<\alpha$, then $\gamma$ is not UB of $E$. ($\forall \gamma <\alpha, \exist x\in E$ such that $x>\gamma$ )
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#### Lemma
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Uniqueness of upper bounds.
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If $\alpha$ and $\beta$ are LUBs of $E$, then $\alpha=\beta$.
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Proof:
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Suppose for contradiction $\alpha$ and $\beta$ are both LUB of $E$, then $\alpha\neq\beta$
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WLOG $\alpha>\beta$ and $\beta>\alpha$.
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EOP
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We write $SupE$ to denote the LUB of $E$.
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This also applies to $GLB$ (greatest lower bound) and infinum of $E$
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#### Example
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1. $S=\mathbb{Q}, E=\{1,2,3\}$ ($E$ is bounded above)
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* $SupE=3$, $Inf E=1$
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2. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$ ($E$ is bounded above)
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* $SupE=3$, $Inf E=1$
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$SupE$ and $Inf E=1$ don't have to $\in E$
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3. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x\}$ ($E$ is not bounded above)
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* $SupE=\infty$ or not defined, $Inf E=0$
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4. $S=\mathbb{Q}, E=\phi$.
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* $SupE=-\infty$ or not defined, $Inf E=\infty$ or not defined, we don't put $\infty$ in $\mathbb{Q}$
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Important example
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5. $S=\mathbb{Q}, A=\{p\leq \mathbb{Q}:p>0, p^2<2\}$.
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* $A$ is not empty and bounded above. However, $Sup A$ des not exists.
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If $S=\mathbb{R}, A=\{p\leq \mathbb{Q}:p>0, p^2<2\}$.
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* $A$ is not empty and bounded above. However, $Sup A=\sqrt{2}$.
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#### Least upper bound property (LUBP)
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if $\forall E\subset S$ that tis non-empty and bounded above, $\exist Sup E\in S$.
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#### Greatest upper bound property (GLBP)
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S has greatest lower bound property (GLBP) if $\exist E\subset S$ that is non-empty and bounded below, $\exists Inf E\in S$
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$\mathbb{Q}$ does not have LUBP and GLBP.
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#### Theorem 1.11
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Let $S$ be an ordered set. Then $S$ has the LUBP $\iff$ $S$ has the GLBP
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Proof:
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Let $S$ be a set with LUBP. (we want to show $S$ has GLBP) |