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I am doing Exercise I.10.16 from textbook Analysis I by Amann/Escher.

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Could you please verify if my attempt contains logical gaps/errors?


My attempt:

Let $I_n := [a_n, b_n]$ for all $n \in \mathbb N$.

(a)

It follows from $a_n \le b_n$ for all $n \in \mathbb N$ that $a_n$ is a lower bound of $\{n \in \mathbb N \mid b_n\}$ and that $b_n$ is an upper bound of $\{n \in \mathbb N \mid a_n\}$. So $\sup a_n, \inf b_n$ exist and $a_n \le \inf b_n$, $\sup a_n \le b_n$ for all $n \in \mathbb N$. On the other hand, $\inf b_n \le b_n$, $a_n \le \sup a_n$ for all $n \in \mathbb N$. As a result, $\inf b_n, \sup a_n \in [a_n, b_n]$ and thus $|\inf b_n - \sup a_n| \le b_n - a_n$ for all $n \in \mathbb N$.

If $\inf b_n - \sup a_n \neq 0$ then $|I_n| =b_n -a_n \ge |\inf b_n - \sup a_n| > 0$ for all $n \in \mathbb N$. This contradicts $(ii)$. Thus $\sup a_n = \inf b_n \in [a_n,b_n]$.

If $x \in \cap_n I_n$ then $a_n \le x \le b_n$ for all $n \in \mathbb N$. So $\sup a_n \le x \le \inf b_n$ and thus $x = \sup a_n = \inf b_n$.

(b)

For $x \in \mathbb R$, there exist $m,M \in \mathbb Q$ such that $m < x < M$. Let $A:= \{p \in \mathbb Q \mid m < p < x\}$ and $B:= \{p \in \mathbb Q \mid x < p < M\}$. Then $|A| = |B| = \aleph_0$. Let $(c_n)_{n \in \mathbb N}$ and $(d_n)_{n \in \mathbb N}$ be enumerations of $A$ and $B$ respectively.

We define $(a_n)_{n \in \mathbb N}$ recursively by $a_0 = m$ and $a_{n+1} = c_{i_0}$ where $i_0 := \min \{i \in \mathbb N \mid c_i > a_n\}$. Similarly, We define $(b_n)_{n \in \mathbb N}$ recursively by $b_0 = M$ and $b_{n+1} = d_{i_0}$ where $i_0 := \min \{i \in \mathbb N \mid d_i < b_n\}$.

It is easy to verify that $(I_n)_{n \in \mathbb N}$, where $I_n := [a_n, b_n]$, is a nest of intervals that satisfies the conditions.


Update: From Asaf Karagila's elegant suggestion here, I present a shorter approach for (b).

For $x \in \mathbb R$, there exist $m,M \in \mathbb Q$ such that $m < x < M$. Let $(c_n)_{n \in \mathbb N}$ be an enumerations of $\mathbb Q$.

We define $(a_n)_{n \in \mathbb N}$ recursively by $a_0 = m$ and $a_{n+1} = c_{i_0}$ where $i_0 := \min \{i \in \mathbb N \mid a_n < c_i < x\}$, and $(b_n)_{n \in \mathbb N}$ by $b_0 = M$ and $b_{n+1} = c_{i_0}$ where $i_0 := \min \{i \in \mathbb N \mid x < c_i < b_n\}$.

It is easy to verify that $(I_n)_{n \in \mathbb N}$, where $I_n := [a_n, b_n]$, is a nest of intervals that satisfies the conditions.

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