# Proof of Nested Interval Theorem: insufficient

I have to prove that if $\mathrm{length}(I_{n}) \to 0$, then $\bigcap_{n=1}^{\infty} I_{n}$ contains precisely one point.

The proof given in the text says that if $b_n>a_n$, $b_{n}$ converges to $b$ and $a_{n}$ converges to $a$ as $n\to\infty$, and $b_{n}-a_{n}\to 0$, then $a=b$.

Convergence says that for any $\epsilon$, we can find $a_{n}$ and $b_{n}$ such that $|b_{n}-a_{n}|<\epsilon$. This is fine. It is however, also true that an infinite sequence can be made on every $\epsilon$- one of the many such sequences is $\epsilon, \frac {\epsilon}{2}, \frac {\epsilon} {2^2}\dots$

However small the value of $\epsilon$ may be, I don't see it ever containing 1 point. Reiterating, although we know $\epsilon\to0$, a length of $\epsilon$ will still contain an infinite number of points.

Where am I going wrong?

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I have edited your second paragraph since it made no sense (e.g. one says "$x_n$ converges to $x$" not "$x$ converges to $x_n$") –  Zev Chonoles Apr 8 '13 at 7:54
The intervals $(0,\frac{1}{n})$ are nested, and satisfy the length condition, but the intersection is empty. You need more conditions. Since the length goes to zero, the intersection can have at most one point. This is easy to prove by contradiction. If the intersection has two points, then the length condition must be violated. –  copper.hat Apr 8 '13 at 8:08
The set of naturals also is the union of {n}'s -- and at no point you have an infinite set..But when you take infinite union , you have an infinite set as you will agree.This is similar.. –  Halil Duru Apr 8 '13 at 15:31

Perhaps it would be useful to test your intuition on an example.

Let the interval $I_n$ be $[0,\frac{1}{n}]$, so that $a_n=0$ for all $n$, and $b_n=\frac{1}{n}$ for all $n$. Then $a_n\to 0$ and $b_n\to 0$ as $n\to\infty$, and $|b_n-a_n|\to 0$, so the theorem applies. And we see that $\bigcap_{n=1}^\infty I_n=\{0\}$.

Do you agree that $\bigcap_{n=1}^\infty I_n=\{0\}$? Remember, given subsets $A_1,A_2,\ldots$ of $\mathbb{R}$, by definition we have $$\bigcap_{n=1}^\infty A_n=\{x\in\mathbb{R}\mid x\in A_n\text{ for all }n\}.$$ Are there any numbers $x$, other than $0$, that are in the interval $[0,\frac{1}{n}]$ for every $n$?

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Perhaps the question deserves a little more attention. I"m not saying I can give you a number or a point that's there in the intersection of the infinite number of nested sets. I can't. But proving the existence of such a point is an entirely different thing. Let's take the interval $[0,\frac{1}{n}]$ as $n\to\infty$. I say the point $\frac{1}{2n}$ is also there apart from $0$. Now you take $[0,\frac{1}{2n}]$. I say $\frac{1}{4n}$ is there in this interval. Do you not think this process can move in a never-ending cycle? I say never-ending instead of 'infinitely'. –  Ayush Khaitan Apr 8 '13 at 8:17
Sure, for any fixed choice of endpoint $\frac{1}{n}$, e.g. $\frac{1}{17}$, you can find infinitely many numbers in $[0,\frac{1}{17}]$. And if I choose a smaller endpoint, you can still find infinitely many numbers. But that is simply irrelevant to the question of what elements are in the intersection, and the intersection is what the theorem is talking about. –  Zev Chonoles Apr 8 '13 at 8:22
There is no "process", even though that may (sometimes) be a useful intuition; it is a static definition. In the same way, $1$ is simply equal to $0.999\ldots$; it is not a "process" or a "converging to", it is literally $$1=\sum_{n=1}^\infty (9\cdot 10^{-n})$$ –  Zev Chonoles Apr 8 '13 at 8:25
You talked about $\bigcap^{\infty}[0,\frac{1}{n}]$. The intersection of any number of these will be the smallest set $[0,\frac{1}{n}]$. Hence the need to talk about the number of points in this interval- this interval is the intersection itself. Your contention is that the infinite intersection of such sets is $0$. Mine is that although this may be true, I'm not convinced with the argument given –  Ayush Khaitan Apr 8 '13 at 8:39
You are also right in emphasizing 'fixed choice of endpoint'. But any converging sequence is made up only of such 'fixed points', and if I can prove a property for all such fixed points, then the whole of the sequence too has to have this property. Basically, I expect to be proven wrong by the end of this discussion. What you're saying has been stated in every analysis textbook. I hope you can help that to happen. –  Ayush Khaitan Apr 8 '13 at 8:40
Let (In)  0 and let A be the set of all points in ⋂∞ In.


Assume A contains multiple points, a0,a1,a2… Then choose a0 such that |a0| < |ak| for all k ≠ 0, choose a1 such that |a1| < |ak| for all k ≠ 0, 1, etc. so that |a0| < |a1| <|a2| < |a3| < … But (In)  0 implies that for any ε > 0, ∃ n s.t. ∀x with |x|> ε, x does not lie in In. Choosing ε = |a0|/2 implies that for some n, ε = |a0|/2 < |a0| < |a1| <…<|ak| and therefore ak does not lie in In for all k natural. This contradicts, so there cannot be multiple points in A. Further, it is clear that no matter how ε > 0 is chosen, -ε < 0 < ε. This means if (In)  0, then 0 lies in In for all n. Therefore 0 ∊ A, and combined with the previous result we conclude A = {0}.

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Think you would be a big fan of non-standard analysis. In this form of calculus, the nested intervals contain points besides 0, and these points are defined in terms of a decreasing sequence. So the answer depends on the kinds of math you use.

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So you are given a sequence of intervals $I_n:=[a_n,b_n]$ with $I_{n+1}\subset I_n$ for all $n\geq0$.

In order to proceed we have to agree on the fact that the system ${\mathbb R}$ is complete. This implies, e.g., that any nonempty and bounded set $S\subset{\mathbb R}$ has a least upper bound $\sup(S)\in{\mathbb R}$, called supremum of $S$, and a largest lower bound $\inf(S)\in{\mathbb R}$, called infimum of $S$. (This is a statement about the fine structure of ${\mathbb R}$; it rules out holes in places where we hoped to find a number.)

The nesting of the intervals $I_n$ implies $$a_0\leq a_m\leq a_{\max\{m,n\}}\leq b_{\max\{m,n\}}\leq b_n\leq b_0\qquad \forall\ m,\ \forall n\tag{1}\ .$$ Therefore the set $A$ of left endpoints $a_n$ is bounded and therefore has a supremum $\alpha\in{\mathbb R}$, and similarly the set $B$ of right endpoints $b_n$ has an infimum $\beta\in{\mathbb R}$.

Consider a fixed $b_n$. By $(1)$ this $b_n$ is $\geq a_m$ for all $m$; therefore $b_n$ is an upper bound for $A$, and this implies $b_n\geq\alpha$, since $\alpha$ is the smallest upper bound of $A$. Since this is true for each individual $b_n$ the number $\alpha$ is a lower bound of $B$, and $\beta$ being the largest such bound it follows that $\alpha\leq\beta$.

We now have $a_n\leq\alpha\leq\beta\leq b_n$ for all $n\geq0$, and this implies that any number $x$ with $\alpha\leq x\leq\beta$ is contained in all $I_n$. Assume that there are two such numbers $x<x'$. Then $[x,x']\subset I_n$ for all $n$, and this would imply that all $I_n$ have a length $\geq\delta:=x'-x>0$, contrary to assumption.

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Well , first of all $I$ = $\bigcap_{n=1}^\infty I_n$ can't contain two or more points : say I includes $x_1$ and $x_2$ then

$||I||\leq||I_n||$ < $|x_2-x_1|$ for sufficiently large n -- so we can't have both.

It contains precisely the unique point $x$ which is both the limit of $a_n$ and $b_n$.

Since $\forall$n $a_n < x <b_n$ then $x \in I_n$ hence is an element of the infinite intersection by definition.

[Note that any finite intersection will have infinitely many points--but this is irrelevant ..]

We have our definition : $\bigcap_{n=1}^\infty A_n =\{x\in\mathbb{R}\mid x\in A_n\text{ for all }n\}$

As an example --to test your understanding-- what would the intersection $\bigcap_{n=1}^\infty A_n$ be

if $A_n$=$[n,\infty)$ ??

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The point I've tried to make in this thread is that there can be no intervals on $R$ containing only a finite number of points. However small the interval is, it still contains an infinite number of points.

Someone answered that let's assume an interval containing two points $x_{1} and x_{2}$. This assumption itself is faulty, as there can be no such interval. If we take an interval $[x_{1},x_{2}]$, one of the many infinite sequences in it is $x_{1}+\frac{x_{2}-x_{1}}{2}, x_{1}+\frac{x_{2}-x_{1}}{2^2}, \dots$.

If we have an infinite decreasing sequence of intervals (in length), and we start out with an interval containing infinite points, we don't necessarily reach intervals with a finite number of points at some point in the sequence.

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The statement "assume an interval contains two points $x_1$ and $x_2$" is not faulty, because it did not say exactly two points. The interval $[x_1,x_2]$ does contain the two points $x_1$ and $x_2$; it just contains other points as well (in the same way, mathematicians say "$P$ or $Q$", and mean "$P$, $Q$, or both"). Additionally, a closed interval can contain finitely many points; specifically, the closed interval $[a,a]$ is precisely the set $$\{x\in\mathbb{R}\mid a\leq x\leq a\}=\{a\},$$ which has exactly one element. –  Zev Chonoles Apr 9 '13 at 8:03
Also, it is simply not an issue of "reaching intervals with a finite number of points at some point in the sequence" - there is no "point" where we transition from infinite to finite. For any natural number $N$, the set $$\bigcap_{n=1}^N[0,\tfrac{1}{n}]=[0,\tfrac{1}{N}]$$ is infinite, and this is not contradictory in any way with the (also true) statement that $$\bigcap_{n=1}^\infty[0,\tfrac{1}{n}]=[0,0]=\{0\}$$ has one element. –  Zev Chonoles Apr 9 '13 at 8:07
Note that I am not arguing via some sort of handwaving "$\frac{1}{\infty}=0$" nonsense, just the 100% precise definition $$\bigcap_{n=1}^\infty [0,\tfrac{1}{n}]=\{x\in\mathbb{R}\mid x\in[0,\tfrac{1}{n}]\text{ for all }n\in\mathbb{N}\}=\{x\in\mathbb{R}\mid 0\leq x, \text{ and }x\leq\tfrac{1}{n}\text{ for all }n\in\mathbb{N}\}$$ –  Zev Chonoles Apr 9 '13 at 8:09
On a non-math related note: Go to this page to merge your current account with the one you used to ask this question. –  Zev Chonoles Apr 9 '13 at 8:12
@ZevChonoles- Thank you for your answer. This might seem like another stupid question, but shouldn't there be a transition point from infinite intervals to finite intervals? When we talk of a monotonically decreasing sequence converging to $0$, we say that for any $\epsilon>0$, we have $x_{n}$ such that $|x_{n}-0|<\epsilon$. An interval containing 3 points is longer than an interval containing 1 point. If the infinite intersection of the intervals is indeed $0$, shouldn't we be able to prove the existence of an interval containing two points, by the very definition of a converging sequence. –  Ayush Khaitan Apr 9 '13 at 8:58