Solutions

a) We need to prove that

$$\bigcap_{i=1}^{\infty} I_i\ne\varnothing.$$

Notice that $I_1=[a_1,b_1]\ne\varnothing$. For the induction step, suppose

$$\bigcap_{i=1}^{n} I_i\ne\varnothing$$

for $n\in\mathbb{N}$. By definition, we have

$$I_{n+1}=[a_{n+1},b_{n+1}]$$

with $I_{n+1}\subseteq I_n=[a_n,b_n]$, since

$$a_n\le a_{n+1}\le b_{n+1}\le b_n.$$

Thus,

$$\bigcap_{i=1}^{n+1} I_i\ne\varnothing.$$

b) Consider $I_n=\left(0,\frac{1}{n}\right)$. For contradiction, suppose there exists

$$x\in\bigcap_{i=1}^{\infty} I_i$$

with $x>0$. By Theorem 3, there exists a natural number $n$ such that $\frac{1}{n}<x$. This means that $x\notin I_n$, which contradicts the premise that $x\in I_i$ for all $i\ge1$.