9.3

1. Consider \(S=\{1-\frac{1}{n^2}:n\in N\}\) and determine \(sup(S)\) if it exists

   Solution

   Note that \(M=1\) is an upper bound of S since \(1-\frac{1}{n^2}<1\) for all \(n\in N\)​

   Now if \(M'<M\), then take \(n_0\in N\) such that \(\sqrt{\frac{1}{1-M'}}<n_0\) and thus \(1-\frac{1}{n^2}\in S\) such that \(M'<1-\frac{1}{n_0^2}\) since \(\frac{1}{1-M'}<n_0^2\)​

   \(\frac{1}{n_0^2}<1-M'\), since \(M'<M=1\), then \(1-M'>0\)​

   \(M'<1-\frac{1}{n_0^2}\) therefore by the previous theorem \(1=M=sup(S)\)​

   ​

2. Consider \(E_1=\{x\in Q:0\leq x<1\}~and~E_2=\{(1+\frac{1}{n})^2:n\in N\}\)​

   • Prove that each of the sets is bounded below. Which of them has a max and min element

   • E1

     • Max

     Note that E is bounded below by 1 because \(x<1\) for evey \(x\in E_1\)

     Suppose that \(E_1\) has a max \(M\), then \(M\in E_1\) and thus \(M\in Q\) with \(0\leq M<1\)

     Note \(M<\frac{M+1}{2}<1\)

     \(\frac{M+1}{2}\in E\) with \(M<\frac{M+1}{2}\) which is a contradiction because M is an upper bound of E.

     Therefore \(E_1\) does not have a max. * Min

     Note that \(E_1\) is bounded below by 0 since \(0\leq x\) for all \(x\in E\) and as \(0\in E_1\), we have that \(min(E_1)=0\)

   • E2

     • Max

     Note that \(n\geq m\) then \(\frac{1}{n}\leq \frac{1}{m}\) and \(\frac{1}{n^2}\leq \frac{1}{m^2}\)

     Hence \((1+\frac{1}{n})^2=1+\frac{2}{n}+\frac{1}{n^2}\leq 1+\frac{2}{m}+\frac{1}{m^2}=(1+\frac{1}{m})^2\)

     In particular, if \(n\geq 1\), we have \((1+\frac{1}{n})^2\leq (1+\frac{1}{1})^2=4\) * Min

     Note that \((1+\frac{1}{n})^2>1\) for all \(n\in N\) and thus 1 is a lower bound of \(E_2\).

     Suppose that \(E_2\) has a min m, that is \(m=(1+\frac{1}{k})^2\) for some \(k\in N\)​

     But since \(k+1>k\) we have that \((1+\frac{1}{k+1})^2<(1+\frac{1}{k})^2=m\)​

     That is \((1+\frac{1}{k+1})^2\in E_2\) such that \((1+\frac{1}{k+1})^2<m\) which is a contradiction because m is a lower bound of \(E_2\).

     Therefore \(E_2\) does not have a min element. * Determine the sup of each sets exist

   • \(E_1\)

     We have that 1 is an upper of \(E_1\). Now if \(M<1\) we have \(x\in Q\) such that \(M<x<1\) (Density)

     Then by the previous theorem: \(1=sup(E_1)\)​ * \(E_2\)

     Since 4 is an upper bound of \(E_2\) and \(max(E_2)=4\) then \(4=sup(E_2)\)​

3. For all \(z,w\in C\) and \(x\in R\), prove that

   • \(|z|=0\) if and only if \(z=0\)​

   Let \(z=a+bi\)​

   \(|z|=0\Leftrightarrow \sqrt{a^2+b^2}=0\Leftrightarrow a^2+b^2=0 \Leftrightarrow a=b=0\Leftrightarrow z=0\)​ * \(|xz|=|x||z|\)​

   \(|xz|=|x(a+bi)|=|xa+xbi|=\sqrt{(xa)^2+(xb)^2}=\sqrt{x^2(a^2+b^2)}=\sqrt{x^2}\sqrt{a^2+b^2}=|x||z|\)​ * \(|z+w|\leq |z|+|w|\)​

   \(|z+w|^2=(z+w)\overline{(z+w)}=(z+w)(\overline{z}+\overline{w}) =z\bar z+z\bar w+\bar z w+w\bar w=|z|^2+z\bar w+ \overline{z \bar w}+|w|^2=|z|^2+2Re(z \bar w)+|w|^2\leq|z|^2+2|z\bar w|+|w|^2=|z|^2+2||z||w|+|w|^2=(|z|+|w|)^2\)

   (\(Re(z)\leq|z|\), \(|zw|=|z||w|\), \(|\bar z|=|z|\))

   Therefore \(|z+w|\leq |z|+|w|\)

4. Find \(z\in C\) such that \(Re(\frac{1+i}{z})=0\)​

   Easy: \(z=a-ai,a\neq 0\)

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