How to proof that nested intervals are an equivalent relation?

Question

I want to show that a relation on the space of all sequences of nested intervals is an equivalence relation.

Definition: Let $[a_n,b_n]_{n\in\mathbb{N}}$ and $[c_n,d_n]_{n\in\mathbb{N}}$ be two "sequences of nested intervals".

Define: $[a_n,b_n]_{n\in\mathbb{N}}\simeq [c_n,d_n]_{n\in\mathbb{N}} :\Leftrightarrow \forall n\in\mathbb{N}: a_n\leq d_n \wedge c_n\leq b_n $

with the Definiton for nested Intervalls:

Definition: Let $[a_n,b_n]$ be sequence of Intervalls. We call it nested Intervall $:\Leftrightarrow$a) and b)

a) $\forall n\in\mathbb{N}: a_n\leq a_{n+1}$ and $b_{n+1}\leq b_n $

b) $\{b_n-a_n\}_{n\in\mathbb{N}}$ is a nullsequence

Proof:

Reflexive relation:

Let $[a_n,b_n]_{n\in\mathbb{N}}$ be a sequence of nested intervals. Because $\forall n\in\mathbb{N}: a_n\leq b_n$ follows $[a_n,b_n]_{n\in\mathbb{N}}\simeq [a_n,b_n]_{n\in\mathbb{N}}$

Symmetric Relation:

Let $[a_n,b_n]_{n\in\mathbb{N}}\simeq [c_n,d_n]_{n\in\mathbb{N}}$. $\Leftrightarrow \forall n\in\mathbb{N}a_n\leq d_n \wedge c_n\leq b_n$

We have to show, that $[c_n,d_n]_{n\in\mathbb{N}}\simeq [a_n,b_n]_{n\in\mathbb{N}}$, wich meens $c_n\leq b_n \wedge a_n\leq d_n$. This follows immediately.

Transitive relation:

Let $[a_n,b_n]_{n\in\mathbb{N}}\simeq [c_n,d_n]_{n\in\mathbb{N}}$ and $[c_n,d_n]_{n\in\mathbb{N}}\simeq [e_n,f_n]_{n\in\mathbb{N}}$

$\Leftrightarrow \forall n\in\mathbb{N}: a_n\leq d_n \wedge c_n\leq b_n \wedge c_n\leq f_n \wedge e_n\leq d_n $

...?

//edit// I can show that there is at least one $a_n \leq s \leq b_n$ for every 'nested intervall' quite simple... maybe i should use that fact?

...

I just dont see how to mange the transitive part. I think it cant be too hard. Thanks for your help!

Answer 1

It's true. I'll show transitivity. Define the relation between sequences of closed intervals as: $$\begin{align} ([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N} \iff & \forall n\in\Bbb N: \\ &\quad a_n \le d_n, \text{ and }\tag{i} \\ &\quad c_n\le b_n.\tag{ii} \end{align}$$ A sequence of nested intervals is a sequence with the properties (a) and (b). These properties imply that there is a unique point $L$ in the intersection of the intervals, which moreover equals the sup of the left endpoints, the inf of the right endpoints, and the limit of both endpoint sequences:

Any sequence of nested intervals $([a_n,b_n])_{n\in\Bbb N}$ converges to a limit $L$, in several ways: $$\begin{align}\tag{$\dagger$}\\ L &= \lim_n a_n = \sup_n a_n\\ &= \lim_n b_n = \inf_n b_n \\ &= \text{the unique element of }\bigcap_n [a_n,b_n]. \end{align}$$

This is fairly well known; proof or reference on request.

Claim: $\simeq$-related sequences of nested intervals converge to the same limit:

For sequences of nested intervals $([a_n,b_n])_{n\in\Bbb N}$, $([c_n,d_n])_{n\in\Bbb N}$, $$if ([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N} ,\, then \, \lim_n a_n = \lim_n c_n = \lim_n b_n = \lim_n d_n.\tag{*}$$

Indeed, given such sequences, let $L=\lim_n a_n$; then we have: $$\begin{align} L = \lim_n a_n &\le \lim_n d_n\tag{by (i)} \\ &= \lim_n c_n\tag{by ($\dagger$)} \\ &\le \lim_n b_n\tag{by (ii)} \\ &= L.\tag{by ($\dagger$)} \\ \end{align}$$

Finally,

Given sequences of nested intervals $$([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N} \simeq ([e_n,f_n])_{n\in\Bbb N},\, $$ we have, for every $m\in\Bbb N$, $$ a_m \le f_m \text{ and } e_m\le b_m\tag{$result$}; $$ therefore, $$ ([a_n,b_n])_{n\in\Bbb N} \simeq ([e_n,f_n])_{n\in\Bbb N}. $$

By hypothesis, all six endpoint sequences converge to the same limit $L$, which also equals the $\sup$s and $\inf$s of left-and right-hand endpoint sequences respectively. Given $m$, we have, by (*), $$a_m \le \sup_n a_n = L = \inf_n f_n \le f_m,$$and similarly, $$e_m \le \sup_n e_n = L = \inf_n b_n \le b_m;$$hence the result. But that result, true for all $m$, says exactly that $\simeq$ holds between the sequences $([a_n, b_n])$ and $([e_n, f_n])$, so the conclusion follows.