Let $c_n=\frac{1}{2}(a_n+b_n), r=\lim_{n\to \infty}c_n,$ and $e_n=r-c_n$. Here $[a_n,b_n]$, with $n\geq 0$, denotes the succesive intervals the arise in the bisection method when it is applied to a continuous function $f$.
(a) Show that for all $n$ and $m$, $a_m\leq b_n$
(b) Show that $r$ is the unique element in $\bigcap_{n=0}^{\infty}[a_n, b_n]$.
(c) Show that for all $n$, $[a_n, b_n]\supset [a_{n+1}, b_{n+1}].$
For (a), I'm a bit confused about how to solve this, I've thought about using induction in $n$ and $m$ but I do not know if it's the right way to solve this, could it be solved like this? How would the induction be done?
For (b), I know that because of Cantor's intersection toerema, we have $\bigcap_{n=0}^{\infty}[a_n, b_n]=\{r\}$, since $[a_{n+1}, b_{n+1}]\subset [a_n, b_n]$ and $[a_n,b_n]$ are compacts for all $n$ and $[a_n, b_n]\neq \phi$, but how do I prove that $f(r)=0$?
For (c), I know that for all $n$, we have that $a_n\leq a_{n+1}$ and $b_{n}\geq b_{n+1}$, with which $[a_{n+1}, b_{n+1}]\subset [a_n, b_n]$.
Is this argument correct? Thank you very much.
It seems easiest to me start with c). We assume that by bisection we mean that $a_0=a<b_0=b$, $f\left(a_0\right)f\left(b_0\right)\le0$ and from there we let $$c_n=\frac12\left(a_n+b_n\right)$$ And $$\left(a_{n+1},b_{n+1}\right)=\left\{\begin{array}{rl} \left(a_{n},c_{n}\right)&\text{if}\quad f\left(a_n\right)f\left(c_n\right)\le0\\ \left(c_{n},b_{n}\right)&\text{if}\quad f\left(a_n\right)f\left(c_n\right)>0\end{array}\right.$$ Then $$a_n=\frac12a_n+\frac12a_n<\frac12a_n+\frac12b_n=c_n<\frac12b_n+\frac12b_n=b_n$$ So if $f\left(a_n\right)f\left(c_n\right)\le0$ then $a_{n+1}=a_n<c_n=b_{n+1}<b_n$ and $f\left(a_{n+1}\right)f\left(b_{n+1}\right)=f\left(a_n\right)f\left(c_n\right)\le0$ and $b_{n+1}-a_{n+1}=c_n-a_n=\frac12\left(b_n-a_n\right)$.
Else if $f\left(a_n\right)f\left(c_n\right)>0$ then $a_n<a_{n+1}=c_n<b_n=b_{n+1}$ and $$f\left(a_{n+1}\right)f\left(b_{n+1}\right)=\left(\frac1{f\left(a_n\right)}\right)^2\left(f\left(a_n\right)f\left(c_n\right)\right)\left(f\left(a_n\right)f\left(b_n\right)\right)\le0$$ And $b_{n+1}-a_{n+1}=b_n-c_n=\frac12\left(b_n-a_n\right)$ So if the predicate that $a_n<b_n$ and $f\left(a_n\right)f\left(b_n\right)\le0$ is true for $n$, it is also true for $n+1$. Since it's true for $n=0$, it holds for all nonnegative integers $n$. Since $a_n\le a_{n+1}<b_{n+1}\le b_n$, $\left[a_{n+1},b_{n+1}\right]\subseteq\left[a_{n},b_{n}\right]$ and we have $\left[a_{n+1},b_{n+1}\right]\subset\left[a_{n},b_{n}\right]$ because $0<b_{n+1}-a_{n+1}<b_n-a_n$. So that establishes part c).
If $m\le n$ then $a_m\le a_n\le b_n$ else if $m>n$ then $a_m\le b_m\le b_n$, so we have part a).
We know that $\{a_n\}$ is a nondecreasing and bounded sequence so it has a limit $A$. Also $\{b_m\}$ is a nonincreasing and bounded sequence so it has a limit $B$. Since $b_{n+1}-a_{n+1}=\frac12\left(a_n-b_n\right)$ we have $$2^{n+1}\left(a_{n+1}-b_{n+1}\right)=2^{n}\left(a_n-b_n\right)=2^0\left(a_0-b_0\right)=b-a$$ So $$b_n-a_n=\frac{(b-a)}{2^n}$$ If $A<B$ then by the Archimedean property of the real numbers there is some $N$ such that $$\frac{(b-a)}{2^{n-1}}<B-A$$ So for $n\ge N$, $$b_n-a_n<\frac{B-A}2$$ Then $$\lim_{n\rightarrow\infty}\left(b_n-a_n\right)\le\frac{B-A}2$$ So $$0\le B-A\le\frac{B-A}2$$ Which contradicts the assumption that $B>A$ so it follows that $B=A=r$. Then, since also $a_n\le c_n\le b_n$ it follows by the squeeze theorem that $$\lim_{n\rightarrow\infty}c_n=r$$ If there were another element $$s\in\displaystyle{\bigcap_{n=0}^{\infty}}\left[a_n,b_n\right]$$ such that $s\ne r$, then we could find some $n$ such that $$\frac{b-a}{2^n}<|r-s|$$ So we can see that the set intersection isn't big enough to hold another element besides $r$. So that gets us to part b).
If $f\left(a_n\right)f\left(b_n\right)=0$ then either $f\left(a_n\right)=0$ or $f\left(b_n\right)=0$ so in either case there is some $x\in\left[a_n,b_n\right]$ such that $f\left(x\right)=0$. If $f\left(a_n\right)f\left(b_n\right)<0$ by the intermediate value theorem there must be some $x\in\left(a_n,b_n\right)$ such that $f\left(x\right)=0$. In any case for each interval $\left[a_n,b_n\right]$ there is some $x_n\in\left[a_n,b_n\right]$ such that $f(x)=0$. Since $r\in\left[a_n,b_n\right]$, for any $\epsilon>0$ by the continuity of $f$ there must be some $\delta(\epsilon)>0$ such that $\left|f(x)-f(r)\right|<\epsilon$ whenever $|x-r|<\delta(\epsilon)$. Choose $$N(\epsilon)=\frac{\ln\left(\frac{b-a}{\delta(\epsilon)}\right)}{\ln2}$$ Then $$|x_n-r|<\frac{b-a}{2^{N(\epsilon)}}\le\frac{b-a}{\frac{b-a}{\delta(\epsilon)}}=\delta(\epsilon)$$ So $|f(r)|=|0-f(r)|=|f(x_n)-f(r)|<\epsilon$ whenever $n>N(\epsilon)$, so $$\left|f(r)\right|=\lim_{n\rightarrow\infty}f(r)=0$$