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I am asked to prove the nested interval principle by using the axiom of completeness. That is, for a decreasing sequence of nested closed intervals $I_1, I_2, I_3,...$, there exists exactly one $x ∈ R$ such that $∀n ∈ N : x ∈ I_n $.

The way I understand the question, I need to prove that the intersection of all these nested intervals is a singleton set. Is my intuition correct? If so, I think I know how to prove this. Please do not give any hints for the proof if my guess is correct. Thanks in advance.

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    The intersection of a decreasing sequence of nested closed intervals is non-empty, but it is not a singleton in general . Consider $I_n= [0,1+\frac{1}{n}]$ for $n\geq 1$. See https://en.wikipedia.org/wiki/Nested_intervals – Robert Z Oct 06 '19 at 13:53
  • If we write $I_n=[a_n,b_n]$ then you also need the condition $\lim_{n\to\infty}(b_n-a_n)=0$, otherwise the intersection might be more than $1$ point. – Mark Oct 06 '19 at 13:54
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    I probably should have not omitted the following clause in the definition for nested closed intervals that was given to us: $∀ε > 0 ∃n ∈ N : |I_n| < ε$, where $|I_n| = b_n−a_n$ is the length of the interval. –  Oct 06 '19 at 13:58

2 Answers2

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It is better to understand exactly what is at play at each stage.

Let $I_n=[a_n,b_n]$. Then $I_1 \supseteq I_2 \supseteq \cdots$ implies $$ a_1 \le a_2 \le \cdots a_n \le \cdots \le b_n \le \cdots \le b_2 \le b_1 $$ Thus, $A=\{a_1,a_2,\dots\}$ is a non empty set that is bounded above and $B=\{b_1,b_2,\dots\}$ is a non empty set that is bounded below. Let $a=\sup A$ and $b=\inf B$, using completeness. Let $I=[a,b]$. Then $I$ is non empty and $I \subseteq \bigcap I_n$, which proves the nested interval principle. (Actually, $I=\bigcap I_n$, but this not needed at this point.)

Now, if $b_n-a_n \to 0$, then $I=\bigcap I_n$ is a singleton.

lhf
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Your intuition is correct - the intersection of decreasing (to length $0$) nested intervals is a singleton.

Also, you might consider this additional intuitive explanation of what is happening:

A 'small' closed interval $[a_i,b_i]$ can be viewed as an approximation of some real number. Indeed a scientist may even announce that the 'actual' number $x$ satisfies

$\quad a_i \le x \le b_i$

As these nested approximations 'telescope down', and you 'keep going',
you 'pick out' or 'point out' the real number number $x$ of interest (it can't be 'missing in action').

A fun exercise is to see if you can come up with a definition of nested decreasing open intervals so that the intersection is also guaranteed to be a singleton.

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