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Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 1

Exercise 1.8


(a) Because $a < b$, we have $b-a\in P$. Similarly, we have $d-c\in P$. We have\begin{align*}(b+d)-(a+c)&=b+d-a-c\\ \text{ by Axiom 6 (a)} \quad&=(b-a)+(d-c)\in P.\end{align*}Hence $a+c < b+d$.

(b) We give a counterexample. Let $\mathbb F=\R$. Set $a=c=-1$ and $b=d=0$, then $a < b$ and $c < d$. But $a c=1 > bd =0$.

Exercise 1.10

Solution: We remark the following identity\[\frac{1}{k}-\frac{1}{k+1}=\frac{(k+1)-k}{k(k+1)}=\frac{1}{k(k+1)}.\]Therefore, we have\begin{align*}&\quad\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\&=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\&=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-\cdots-\left(\frac{1}{n}-\frac{1}{n}\right)-\frac{1}{n+1}\\&=1-\frac{1}{n+1}=\frac{n}{n+1}\end{align*}

Another solution: We prove it by induction on $n$.

If $n=1$, the identity is clearly true.

Suppose the identity is true for $n$, we prove it for $n+1$. By induction hypothesis, we have\begin{equation}\label{anal1.10.1}\sum_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1}.\end{equation}Then we have\begin{align*}\sum_{k=1}^{n+1}\frac{1}{k(k+1)}&=\sum_{k=1}^{n}\frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)}\\\text{use }\eqref{anal1.10.1}\quad&=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)+1}{(n+1)(n+2)}\\&=\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}\\&=\frac{n+1}{n+2}.\end{align*}Hence the identity is true for $n+1$.

By Mathematical Induction, we are done!

Exercise 1.20


(a)Let $\sup A=\alpha$ and $\sup B=\beta$. Since $\alpha < \beta$, we have $\beta -\alpha >0$. By Theorem 1.24 and setting $\e =\beta-\alpha$, there exists $b\in B$ such that \[ \beta-(\beta-\alpha) < b\Longrightarrow \alpha < b.\]Hence $b$ is an upper bound of $A$.

(b) We give an example. Let $A=\{1\}$ and $B=\{n/(n+1):n\in\mb N\}$. Then $\sup A=\sup B=1$. But no element in $B$ is an upper bound of $A$ since they are all smaller than one.

Exercise 1.21

Solution: Since $A$ is bounded below, there exists $M$ such that $M\leqslant x$ for all $x\in A$. Hence $M$ is a lower bound of $A$ and $M\leqslant \inf A$. Let $\alpha=\inf A$.

Since $A$ is non-empty, we have an element $x$ in A and hence an element $-x$ in $-A$. This implies that $-A\ne \varnothing$.

Note that $M\leqslant x$ for all $x\in A$. Multiplying both sides by $-1$, we obtain $-M\geqslant -x$ for all $x\in A$. This implies that $-M\geqslant y$ for all $y\in -A$. Therefore $-A$ is bounded above.

We then show that $\sup(-A)=-\alpha$. We use Theorem 1.24. We have

(a) $x\geqslant \alpha$ for all $x\in A$, and hence $-x\leqslant -\alpha$ for all $x\in A$. This implies $-\alpha$ is an upper bound of $-A$;

(b) for any $\e > 0$, there exists $x\in A$ such that $ x < \alpha-\e$, and thus $-x > -\alpha +\e $. This implies that $-\alpha +\e $ is not an upper bound of $-A$.

Therefore, according to Theorem 1.24, we have $\sup(-A)=-\alpha=-\inf A$.

Exercise 1.22


Exercise 1.23

Solution: Let $I_n=(1,1+1/n)$, then each $I_n$ is open and non-empty. Moreover,\[I_1\supset I_2 \supset I_3 \supset I_4 \supset \cdots.\]We show that $\cap_{n=1}^\infty I_n=\varnothing$.

We argue it by contradiction. Suppose $\cap_{n=1}^\infty I_n\ne \varnothing$, then there exist $a\in \cap_{n=1}^\infty I_n$. Hence $a\in I_n$ for all $n$, that is \begin{equation}\label{ex1-23-1}1 < a < 1+\frac{1}{n}.\end{equation} Since $a > 1$, we have $a -1 >0$. By the Archimedean Principle, we can find $m\in \mathbb N$ such that $$\dfrac{1}{m}< a-1.$$ Therefore $1+\dfrac{1}{m} < a$ which contradicts with \eqref{ex1-23-1} if we take $n=m$.

Hence $\cap_{n=1}^\infty I_n=\varnothing$.

Exercise 1.24

Solution: We use Theorem 1.24. Let $\alpha=\sup A$ and $\beta=\sup B$. Then for any $a\in A$ and $b\in B$, we have $a\leqslant \alpha$ and $b\leqslant \beta$. Adding them together, we conclude that $a+b\leqslant \alpha+\beta$ for all $a\in A$ and $b\in B$. Hence $\alpha+\beta$ is an upper bound of $A+B$. (This is part (a) of Theorem 1.24 we would like to show.)

By Theorem 1.24, for any $\e>0$, there exist $a\in A$ and $b\in B$ such that\[\alpha-\frac{\e}{2} < a \leqslant \alpha,\quad \beta-\frac{\e}{2}< b\leqslant \beta.\]Add them together, we get that\[\alpha+\beta-\e < a+b \leqslant \alpha+\beta.\]Hence we obtain part (b) of Theorem 1.24.

Using Theorem 1.24, we get that $\alpha+\beta=\sup(A+B)$, i.e. $\sup A+\sup B=\sup (A+B)$.

Exercise 1.25

Solution: We give a counter example. Let $A=B=\{-1,-2\}$, then we have $$\sup A=\sup B=-1.$$ We also have\[A\cdot B=\{1,2,4\}.\]Hence $\sup(A\cdot B)=4$, but $\sup A\cdot \sup B=1\ne \sup(A\cdot B)$.

Exercise 1.26

Solution: We shall use Definition 1.29 to show that the set is not dense in $\R$. Let \[S=\left\{\frac{m}{n}: m,n\in\mb Z, 1\leqslant n\leqslant 10\right\}.\]Then $0\in S$ and $\dfrac{1}{10}\in S$. We show that there is no element $a\in S$ satisfing $0 < a < \dfrac{1}{10}$.

We argue it by contradiction. Suppose $0 < a < \dfrac{1}{10}$ and $a=\dfrac{m}{n}$, where $1\leqslant n \leqslant 10$, then we have $m\geqslant 0$, $m\in \mathbb Z$, and\[10 m < n.\]Because $m\geqslant 1$, we have $n > 10m \geqslant 10$ which is impossible. Hence $0 < a < \dfrac{1}{10}$ cannot happen. By Definition 1.29, the set $S$ is not dense in $\R$.


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