# Prove that $\xi\leq\eta$ and $\displaystyle\cap_{n=1}^{\infty}I_n=[\xi,\eta]$.

Let $I_1\supset I_2\supset\cdots \text{be a sequence of nested closed finite intervals,where}$ $I_n=[a_n,b_n]$.Let $\xi=\sup\{a_n:n\in\mathbb N\}$, $\eta=\inf\{b_n:n\in\mathbb N\}$ then how can we prove that $\xi\leq\eta$ and $\displaystyle\cap_{n=1}^{\infty}I_n=[\xi,\eta]$.

My try:I tried but did not solve it correctly.Thank you.

• It's quite similar to another question Limit of a monotonically increasing sequence and decreasing sequence which already has an accepted answer, but without the assumption about $\eta-\xi=0$. From that answer, we see that $\forall i,j \in \Bbb N, a_i \le b_j$, so we take supremum and infimum over $\Bbb N$ on LHS and RHS respectively to conclude that $\xi \le \eta$. Then it would be easy to see that $[\xi,\eta] \subseteq \bigcap\limits_{n=1}^{+\infty} I_n$. Use the definition of supremum and infimum to conclude the other inclusion Feb 9 '17 at 19:45

Obviously $a_i\le b_j\quad\forall i,j.$
Thus $\{a_1,a_2,\dots\}$ is bounded above so it has a $\sup$. Similarly $\eta$ exists. Furthermore more we see that $a_i\to\xi$ as $i\to\infty$ and similarly for $b_i.$ consider the limit of $b_i-a_i\ge0$ to get $\xi\le\eta$
Now write down your definitions for $\sup$ and $\inf$ with quantifiers ($\forall,\exists$) and write down a statement with quantifiers which is equivalent to $x\in\bigcap I_i$ and stare at it a bit and then you should be able to write down the rest of the answer