8.28 Completeness and density

1.Equivalent definitions of supremum and infimum=8=

Proposition: \(A\subset R,~A\neq\emptyset\)

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

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   Proof

   \(\Rightarrow\)) Clearly \(s\geq a\) \(\forall a\in A\). So the first condition in "Equiv" is satisfied

   Now, take \(s'<s\). We need to prove that \(\exists a\in A|s'<a\leq s\)​

   Suppose it is not true, then \(\forall a\in A|a\leq s'\). Then, \(s'\) is an upper bound for A

   But now, Since \(s=sup(A)\Rightarrow s\leq s'\). Contradiction!

   Then \(\exists a\in A|s'<a\leq s\)​

   \(\Leftarrow\)) We need to prove that \(s=sup(A)\).

   Clearly, \(a\leq s,\forall a\in A\), so s is an upper bound

   Now, take \(s'\) another upper bound. We need to prove that \(s'\geq s\)​

   Suppose \(s'<s\), then by condition in "Equiv" \(\exists a\in A, s'<a\leq s\) means \(s'\) cannot be an upper bound

   Contradiction!

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   Corollary

   \(A\subset R,A\neq\emptyset,s=sup(A)\Rightarrow\forall\varepsilon>0,\exists a_{\varepsilon}\in A|s-\varepsilon<a_{\varepsilon}\leq s\)

   Strategy in analyse

   move a tiny bit and let element in it

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

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2.Axiom of completeness (Equivalent to the axiom of continuity)

If \(A\subseteq R,A\neq\emptyset,A\) is bounded above, then \(\exists s\in R: s=sup(A)\)

Or equivalently, if \(A\subset R,A\neq\emptyset,\)A bounded below, then \(\exists i\in R,i=inf(A)\)

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Proposition: \(N\) is not bounded above in \(R\)

Suppose not: \(N\) is bounded above, \(N\neq\emptyset\)

Let \(M\in R\) be the sup of N, then \(m\leq M~~~\forall m\in N\)

Now, \(m+1\in\N~~\forall m\in N\Rightarrow m+1\leq M~~~\forall m\in \N\Rightarrow m\leq M-1~~~\forall m\in N\Rightarrow M-1\) is upper bound for N

Contradiction

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3.Archimedean property of N in R

Let \(x,y\in R,~x,y>0\) Then, \(\exists m\in N\) such that \(m\cdot x>y\)

Proof: \(mx>y\Rightarrow m>\frac{y}{x}\) True!! since the previous proposition

Corollary: Given any positive \(\varepsilon>0\) there exist \(m\in N|0<\frac{1}{m}<\varepsilon\Rightarrow m>\frac{1}{\varepsilon}\)

In fact if \(m_0\)​ is such \(\frac{1}{m_0}<\varepsilon\Rightarrow \frac{1}{m}<\varepsilon~~\forall m\geq m_0\)

We can use this when we see \(\frac{1}{n}\) and n is a natural number

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4.Density of Q in R

Given \(x,y\in R|0<x<y\) there exist \(q\in Q|x<q<y\)

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Proof we cannot use average due to cannot ensure it in Q

Note that \(0<y-x\). Then choose \(m\in N|\frac{1}{m}<y-x\) 3.Archimedean property of N in R

\(A=\{j\in N|\frac{j}{m}>x\}\neq\emptyset~~A\subseteq N\) due to the property

WOP:\(\exists k\in N|k=min(A),k\in N\)

If \(k=1\) then \(x<\frac{1}{m}<y-x<y\Rightarrow x<\frac{1}{m}<y~~q=\frac{1}{m}\)

If \(k\neq 1\Rightarrow k>1\Rightarrow k-1\in N\Rightarrow k-1\notin A\Rightarrow \frac{k-1}{m}\leq x\Leftrightarrow \frac{k}{m}-\frac{1}{m}\leq x \Leftrightarrow \frac{k}{m}\leq x+\frac{1}{m}<x+y-x=y\)

Then \(x<\frac{k}{m}<y\) \(q=\frac{k}{m}\)

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5.Exists of roots

Let \(x\in R,x>0\). For \(m\in N\), there exist only one \(y\in R|y^m=x\) \(y=\sqrt[m]{x}\)

Proof: \(A=\{a\in R,a>0:a^m<x\}\)

Note that: \(A\neq\emptyset,a=0,a\in A\)

A is bounded above

\(0<x<1: a^m<1\Rightarrow a<1~~~1\geq a~~\forall a\in A\)​

\(x>1:a^m<x\Rightarrow a<x\)

Then, there is a sup of A: \(y=sup(A)\)

Claim: \(y^m=x\). Suppose \(y^m\neq x\) 1. \(y^m<x\) 2.\(y^m>x\)

1. The idea is to show that we can find a tiny \(h>0|(y+h)^m<x\)

   In that case, \(y+h\in A\Rightarrow y+h\leq y\) contradiction to sup

   \((y+h)^m=\) \(\sum_{j=0}^{m}\left(\begin{matrix}m\\j\end{matrix}\right)y^{m-j}h^{j}\leq y^{m}+h\sum_{j=1}^{m}C_{m}^{j}y^{m-j}\left(0<h<1\right)\) h to make sure the item will become smaller and smaller

   \(k=\sum_{j=1}^{m}C_{m}^{j}y^{m-j}\in R\)

   Then \((y+h)^m\leq y^m+hk\) Recall that \(x-y^m>0\)​

   \(<y^m+x-y^m\) if \(hk<x-y^m\), then \(h<\frac{x-y^m}{k}\)

   \(=x\)​

   We conclude that \((y+h)^m<x\) if \(h>0\) and \(h<min\{1,\frac{x-y^m}{k}\}\)​

   \(\Rightarrow y+h\in A,y=sup(A)\).Contradiction!

2. Suppose \(y^m>x\) \((\frac{x}{y^m}<1)\)​

   Goal: find \(0<h\), small enough to make \((y-h)^m>x\)​

   If we prove this, then \(y-h\) would be upper bound. Contradiction!

   Let's find the \(h\) such that \((y-h)^m>x\)

   \((y-h)^m=y^m(1-\frac{h}{y})^m=y^m(1+(-\frac{h}{y}))^m\geq y^m(1-\frac{mh}{y})\) By Bernoulli's ineq

   Pick \(0<h<y\), \(-\frac{h}{y}>-1\).

   We want to choose \(h>0\) so small that

   \(y^m(1-\frac{mh}{y})>x\)​

   \(1-\frac{mh}{y}>\frac{x}{y^m}\) (\(\frac{x}{y^m}<1\))

   ​

   Choose \(h>0\) such that

   \(1-\frac{mh}{y}>\frac{x}{y^m}\\1-\frac{x}{y^m}>\frac{mh}{y}\\h<\frac{y}{m}\left(1-\frac{x}{y^m}\right)\)

   Conclusion with \(0<h<min\{y,\frac{y}{m}(1-\frac{x}{y^m})\}\)​

   \(\Rightarrow(y-h)^{m}\geq y^{m}\cdot(1-\frac{mh}{y})>y^{m}\cdot\frac{x}{y^{m}}=x\Rightarrow(y-h)^{m}>x.\)

   This implies that \(y-h\) is upper bound for \(A\)​

   \(y=sup(A)\) and also \(y-h\) is upper bound. Contradiction!

   Then \(y^{m}\geq x~and~y^{m}\leq x\Rightarrow y^{m}=x\)

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6.Integer part of \(x\in R\)(Also called "Floor of x")

Let \(x\in R,x>0\). Then there exists a unique \(p\in N_0|p\leq x<p+1\)

Define the set \(A_{x}=\{m\in N_0:m>x\}\neq\emptyset\) N is not bounded in R

WOP: Then there is a first element \(m\in A\)

If \(m=1\Rightarrow 0\leq x<1~~~~p=0\)

If \(m\neq 1\Rightarrow m-1\leq x<m~~p=m-1\)

1.Decimal expansions for real numbers

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