Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.9

Solution:

### Part a

We argue it by contradiction. Suppose $\lim s_n=s < a$. Then $a-s >0$. Since $s_n\ge a$ for all but finitely many $n$, there exist $N_1>0$ such that $s_n\ge a$ for all $n > N_1$.

Let $\epsilon =\dfrac{a-s}{2}$. Then there exists $N_2>0$ such that if $n > N_2$ then

$$

s_n-s<\epsilon.

$$ Take $N=\max\{N_1,N_2\}$. If $n>N$, then we have

$$

s_n-s<\epsilon,\quad s_n\ge a.

$$ Therefore, we have

$$

a \le s_n < s+\epsilon=s+\frac{a-s}{2}=\frac{a+s}{2}.

$$ Thus, we obtain

$$

a < \frac{a+s}{2}

$$ which implies that $a < s$. This contradicts with the assumption $ s < a$ at the beginning. Hence such an assumption is impossible. We argued that $s\ge a$.

### Part b

We argue it by contradiction. Suppose $\lim s_n=s > a$. Then $s -a >0$. Since $s_n\le a$ for all but finitely many $n$, there exist $N_1>0$ such that $s_n\le a$ for all $n>N_1$.

Let $\epsilon =\dfrac{s-a}{2}$. Then there exists $N_2>0$ such that if $n> N_2$ then

$$

|s_n-s|<\epsilon\Longrightarrow s_n - s>-\epsilon.

$$ Take $N=\max\{N_1,N_2\}$. If $n>N$, then we have

$$

s_n-s>-\epsilon,\quad s_n\le a.

$$ Therefore, we have

$$

a \ge s_n > s-\epsilon=s-\frac{s-a}{2}=\frac{a+s}{2}.

$$ Thus, we obtain

$$

a > \frac{a+s}{2}

$$ which implies that $a > s$. This contradicts with the assumption $ s > a$ at the beginning. Hence such an assumption is impossible. We argued that $s\le a$.

### Part c

By assumption, we have $s_n\le b$ and $s_n\ge a$ for all but finitely many $n$. Using Part (a) and Part (b), we have $a\le \lim s_n\le b$. Therefore $\lim s_n$ belongs to $[a,b]$ as well.