10.21 Supremum and the Completeness Axiom

Supremum and the Completeness Axiom

Definition

Facts of a set and an opposite set

Facts: If \(A\neq \emptyset\). Let \(B=\{-a:a\in A\}\)

The upper bounds of \(A\) are the opposites of lower bounds of \(B\).

The lower bounds of \(A\) are the opposites of upper bounds of \(B\).

The maximum of \(A\) is the opposite of minimum of \(B\).

The minimum of \(A\) is the opposite of maximum of \(B\).

The infimum of \(A\) is the opposite of supremum of \(B\).

The supremum of \(A\) is the opposite of infimum of \(B\).

Example

\(A=\{1,1.5,7\},B=\{-1,-1.5,-7\}\)

\(A=(-3,7.2],B=[-7.2,3)\)

Corollary of Completeness Axiom

Theorem: If \(A\subseteq \R\), \(A\neq \emptyset\) and \(A\) has lower bounds, then \(A\) has \(inf(A)\).

We need to use If \(A\subseteq \R\), \(A\neq\emptyset\) and \(A\) has upper bounds, then \(A\) has \(sup(A)\).

Then we can take the opposite if \(A\)

Proof

Since \(A\) has lower bounds, then \(-A\) has upper bounds by above facts

Since \(-A\subseteq\mathbb{R},-A\neq\emptyset\), then \(-A\) has a supremum which is \(sup(-A)\)

Again by facts, since \(A\) is the opposite of \(-A\), then \(sup(-A)=inf(A)\)

Thus \(A\) has \(inf(A)\)

Consequences

Archimedes property

1. \(\forall x\in\mathbb{R},\exists n\in\mathbb{N}\) such that \(n\geq x\) ( \(\N\) is not bounded above)

   By contradiction: Suppose \(\exists x\in \R\) such that \(x>n,\forall n\in\N\Rightarrow x\) is UB of \(\N\neq \emptyset\)

   By the axiom of completeness, \(\N\) has supremum. Let us call it \(s=sup(\N)\)

   By definition we have, \(\begin{cases}s\geq n,\forall n\in\mathbb{N}\\ s\leq s^{\prime},\forall s^{\prime}\in UBs~of~\N\end{cases}\)

   Since \(n\in\mathbb{N}\Rightarrow n+1\in\mathbb{N}\Rightarrow s\geq n+1\Rightarrow s-1\geq n\), then \(s-1\) is a upper bound of \(\N\)

   Then \(s-1\geq s\) which is a contradiction!

   Also can be proved by well ordering principle (WOP)

2. \(\forall x>0,\exists n\in \N\) such that \(\frac1n<x\)​

   Let \(x>0\), by 1 \(\exists n\in \N\) such that \(n\geq\frac{1}{x}+1\Rightarrow n>\frac{1}{x}\Rightarrow\frac{1}{n}<x\)

Density in \(\mathbb{R}\) of \(\mathbb{Q}\) (\(\mathbb{Q}\) is dense in \(\R\)​)

Recall: Density in Q

Density of Q in R

\(\forall x,y\in \R,x<y~~~~\exists q\in \mathbb{Q},x<q<y\)

3 cases

1. \(x<y<0\)​

2. \(x<0<y\), take \(q=0\)​

3. \(0<x<y\)

   A rational number is a quotient of integers, so we must produce \(m ∈ Z\) and \(n ∈ N\) so that \(x<\frac{m}{n}<y\)

   The first step is to choose the denominator n large enough so that consecutive increments of size \(1/n\) are too close together to “step over” the interval \((x, y)\).

   ​

   First, let \(n\in \N\) such that \(\frac1n<y-x\) (Archimedes property 2)

   Let \(m\) be the first \(\N\) such that \(m>nx\Rightarrow \frac{m}{n}>x\) (Archimedes property 1)

   Let \(q=\frac{m}{n}\). We have \(x<q\)​

   Let \(m-1\leq nx\) since \(m\) is the first element (WOP) (Trick)

   Then, \(\frac{m-1}{n}\leq x\)

   Since \(\frac{1}{n}<y-x\), then \(\frac{m-1}{n}+\frac{1}{n}<y-x+x=y\) (\(q=\frac{m}{n}\)​)

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Theorem (square root)

Lemma: Let \(0<x,y\), then \(x<y\) iff \(x^2<y^2\)

Proof

\(\Rightarrow\)) \(\begin{cases}0<a<b\\0<c<d\end{cases}\Rightarrow0<ac<bd\) (Proof: \(\begin{cases}a<b\\ c>0\end{cases}\Rightarrow ac<bc\), similarly \(bc<bd\)\, by transitivity, \(ac<bd\))

Since \(0<x<y\), we multiply them twice and get \(x^2<y^2\)

\(\Leftarrow\)) Assume \(x^2<y^2\), 3 cases

\(\begin{cases}x<y~\text{must be right}\\ x=y\Rightarrow x^2=y^2\\ x>y\Rightarrow y<x\Rightarrow y^2<x^2\end{cases}\)

5.Exists of roots (Several questions arise: can it happens that some rational number works for a supremum within this set?

how can we (formally!) prove that if a supremum for that set exists, then its square is precisely 2?)

\(\forall a\in R,a\geq0,\exists t\in R\) such that \(t^2=a,t\geq 0\)

Analysis: I want to prove \(t^2=a\), it is obviously but seem to be hard to prove. Thus I want to use contradiction.

Thus I can suppose \(t^2>a\) and \(t^2<a\), but what method can I use to get a contradiction?

Always I can use WOP and Completeness axiom, but WOP is usually to deal with the problem with integer and natural number.

Completeness is usually to deal with real numbers. Thus I will use completeness axiom.

Thus I need to construct a set and use it to say that the set has a sup and get a contradiction.

But first, I need to prove the set is bounded

Proof

If \(a=0\), take \(t=0\)

If \(a>0\), \(A=\{x\geq 0:x^2<a\}\)

\(A\neq\emptyset:0\in A\) since \(0^2=0<a\)

\(A\) has UBs: Let \(n\in\N\), \(n>a\) (Archimedes property) \(\Rightarrow n^2\geq n>a\Rightarrow n^2>x^2,\forall x\in A\Rightarrow n>x,\forall x\in A\)​

Thus \(n\) is a upper bound of \(A\)

By continuity: Let \(\alpha =sup(A)\)

Claim: \(\alpha^2=a\)

3 cases: \(\begin{cases}\alpha^2>a\\ \alpha^2=a\\ \alpha^2<a\end{cases}\)

1. \(\alpha^2<a\)​

   We will find $x\in A,x>\alpha $​

   Let \(n\) such that \(\frac{1}{n}<\frac{a-\alpha^2}{2\alpha+1}\)(is positive since \(\alpha^2<a,\alpha\geq 0\)​)

   Let \(x=\alpha+\frac1n>\alpha\)​

   Let us show \(x\in A\)​

   \(x^2=(\alpha+\frac{1}{n})^2=\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2}<\alpha^2+\frac{2\alpha}{n}+\frac{1}{n}=\alpha^2+\frac{2\alpha+1}{n}<\alpha^2+a-\alpha^2=a\)​

   Contradiction!

   ​

2. Then \(x\in A\), finally \(\alpha^2>a\) is impossible: We will find \(\beta<\alpha\) such that \(\beta\) is UB of A

   Let \(n\in \N\) such that \(\frac{1}{n}<\frac{\alpha^2-a}{2\alpha}\)

   Let \(\beta=\alpha-\frac1n<\alpha\)​

   Show \(\beta\in\) UB of \(A\) \(\Leftrightarrow\beta\geq x,\forall x\in A\Leftrightarrow\beta^2\geq x^2,\forall x\in A\)

   \(\beta^2=\left(\alpha-\frac{1}{n}\right)^2=\alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2}>\alpha^2-\frac{2\alpha}{n}>\alpha^2-\left(\alpha^2-a\right)=a>x^2,\forall x\in A\)​

   Then \(\beta^2>x^2,\forall x\in A\Rightarrow\beta>x,\forall x\in A\) (take \(t=\alpha\geq 0\)​)

Definition: This \(t\) is denoted \(\sqrt a\), since \(t^2=(-t)^2\Rightarrow\left(\pm\sqrt2\right)^2=a\)​

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