1
$\begingroup$

In a lecture we covered the nested intervals theorem, which was formulated that way (I try to translate from french, so please forgive me if I make mistakes):

Let $\{I_n\}_{n\in \mathbb{N}}$ be nested, closed (and bounded) sets. Then there exists $x \in \mathbb{R}$ such that $x \in I_n$ for all $n \in \mathbb{N}$.

My first question is the following: why does this theorem require the sets to be bounded? (I understand why they need to be closed, because there are counterexamples if they are open sets, but I don't really understand why they need to be bounded, and what would exactly happen if they were not bounded.)

While looking for informations on Wikipedia about this, another (probably dumb) question arised: the article says:

The rational number line does not satisfy the nested interval theorem. For example, the sequence (whose terms are derived from the digits of pi in the suggested way) $$[3,4] \supset [3.1,3.2] \supset [3.14,3.15] \supset [3.141,3.142] \supset \cdots$$ is a nested sequence of closed intervals in the rational numbers whose intersection is empty. (In the real numbers, the intersection of these intervals contains the number pi.)

But why is the intersection of these intervals empty? I know that a "basic" answer could be "because $\pi$ belongs to the real numbers but not to the rationals", but I don't really get why we couldn't have, for example, a non-empty interval as an intersection of all the sets. For instance, considering only the four sets quoted above, why couldn't we have $\bigcap_{n \in \mathbb{N}} I_n = [3.141 , 3.142]$ ? Is it because the theorem considers an infinite numbers of sets? (This being said, even if the answer to this last question should be yes, I would still wonder why we can't find a rational number in the intersection of an infinite number of sets.)

Finally, a last question arised from this other Wikipedia article, which says:

One can consider the complement of each interval, written as $(-\infty ,a_{n})\cup (b_{n},\infty )$. By De Morgan's laws, the complement of the intersection is a union of two disjoint open sets. By the connectedness of the real line there must be something between them. This shows that the intersection of (even an uncountable number of) nested, closed, and bounded intervals is nonempty.

I tried to look for what was "the connectedness" of the real line, but I didn't find anything really helpful, and as a result I didn't understand what this part of the article means.

Any help would be greatly appreciated!

$\endgroup$
  • 2
    $\begingroup$ For the unbounded case, consider $I_n=[n,\infty)$. $\endgroup$ – Andrés E. Caicedo Apr 19 '17 at 17:06
  • $\begingroup$ Note that you can fix this by extending $\mathbb{R}$ to contain $\{\pm\infty\}$ because the above interval (and all like it) are no longer closed. $\endgroup$ – Stella Biderman Apr 19 '17 at 17:44
1
$\begingroup$

Andrés E. Caicedo already gave an example of nested, closed and unbounded interval which does not have any point in the countable intersection.


About the intersection part, let $I_n=[a_n,b_n]$. Now, by the construction of $I_n$, we have $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\pi$.

Let $x\in\bigcap_{n \in \mathbb{N}}I_n$; then you have $a_n<x<b_n$, $ \forall n\in \mathbb{N}$. Since both limits converge to $\pi$, $x=\pi \notin \mathbb{Q}$. Thus, in the rationals, $\bigcap_{n \in \mathbb{N}}I_n=\emptyset$.


This definition of connectedness is probably topological. It says, in your case, that the real line cannot be represented by the union of two open and disjoint sets. For example, $(-\infty,a)\cup(b,+\infty)\neq\mathbb{R}$, even if $a=b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.