Understand Successive Bisection for Theorem Proof

[elliti123]

I have come across the proof of a theorem and i am unsure of some specific points in the proof so i hope someone could enlighten me. Here is the theorem and the proof straight from the book :

Theorem. Every bounded sequence possesses at least one limiting point.

Proof : We again determine the number in question by a suitable nest of intervals. By hypothesis there exists an interval $J_0$ which contains all the terms of the given sequence $(x_n)$. To this interval we apply the method of successive bisection and designate as $J_1$ its left or right half according as the left half contains an infinite number of the terms of the sequence or not. By the same rule we designate half of $J_1$ as $J_2$, and so on. Then the intervals of the nest $(J_n)$ so formed all have the property that an infinite number of the terms is contained in each, whilst to the left of their left endpoints there is always at most a finite number of points of the sequence. The point L thus defined is obviously a limiting point; for if $\epsilon$>0 is given arbitrarily, choose from the succession of intervals $J_n$ one, say $J_q$, whose length is < $\epsilon$. The terms of $(x_n)$, in number infinite, which belong to the interval $J_q$ then lie ipso facto in the $\epsilon$-neighbourhood of L, - which proves all that we require.

Now i can't exactly picture the successive bisection method on a line, so if anyone could please somehow explain it in more clear manner; since i can't exactly seem to get it from the text because it seems a bit vague. Secondly how exactly do the intervals $J_n$ contain infinite number of terms from the sequence $(x_n)$; is it a property of the method or do i not know something since by this method each interval should have a clear beginning and end which we do know, right?

 

[RUber]

First, remember that you have an infinite sequence contained on a finite interval, so if you cut off a section of the interval that has a finite number of points, the remaining interval will necessarily have an infinite number of points.
You don't need to cut the interval in half each time, but it is convenient to say that the length of your n'th cut of the interval is $\frac{1}{2^n}L$ where L is the length of the original interval.

For example, you have the sequence x_n = 1/n defined on the interval [0,1].
Your first cut would be at .5, and you would see that there are a finite number of points in [.5,1] and infinite points in [0,.5].
Second cut would be applied to the interval [0,.5] and would be at .25. Similarly, the lower half would contain infinite points, so you would keep the interval [0,.25].
In this case it is clear that since $x_n \to 0$, you will always keep the interval from [0, 2^n], and by successive bisection of the interval, you could get to a small enough neighborhood that there are infinite points in the sequence contained in an epsilon neighborhood.

Is there a specific example you are working on?

 

[RUber]

Also, what do you mean by:

Now i can't exactly picture the successive bisection method on a line

A line is not a bounded sequence. It is a collection of points which are all infinitely close to one another, so every point on a line is a limit point, since there are infinite points located within an epsilon neighborhood of any point on the line.

 

[Mark44]

I have come across the proof of a theorem and i am unsure of some specific points in the proof so i hope someone could enlighten me. Here is the theorem and the proof straight from the book :

Theorem. Every bounded sequence possesses at least one limiting point.

Proof : We again determine the number in question by a suitable nest of intervals. By hypothesis there exists an interval $J_0$ which contains all the terms of the given sequence $(x_n)$. To this interval we apply the method of successive bisection and designate as $J_1$ its left or right half according as the left half contains an infinite number of the terms of the sequence or not. By the same rule we designate half of $J_1$ as $J_2$, and so on. Then the intervals of the nest $(J_n)$ so formed all have the property that an infinite number of the terms is contained in each, whilst to the left of their left endpoints there is always at most a finite number of points of the sequence. The point L thus defined is obviously a limiting point; for if $\epsilon$>0 is given arbitrarily, choose from the succession of intervals $J_n$ one, say $J_q$, whose length is < $\epsilon$. The terms of $(x_n)$, in number infinite, which belong to the interval $J_q$ then lie ipso facto in the $\epsilon$-neighbourhood of L, - which proves all that we require.

Now i can't exactly picture the successive bisection method on a line, so if anyone could please somehow explain it in more clear manner; since i can't exactly seem to get it from the text because it seems a bit vague. Secondly how exactly do the intervals $J_n$ contain infinite number of terms from the sequence $(x_n)$; is it a property of the method or do i not know something since by this method each interval should have a clear beginning and end which we do know, right?

First off, your single $ characters aren't doing anything but clutter up what you've written. For inline LaTeX, use two # characters at each end of what you want to be rendered. E.g, ##J_0##.

Here's an example that might help you visualize what's going on, with $a_n = \frac 1 n, n \ge 1$. This sequence is contained in the interval [0, 1], so is bounded. The subinterval J₁ = [0, 1/2] contains an infinite number of terms of this sequence. The other subinterval contains only two of the sequence values: 1/2 and 1. Continue dividing the subintervals into smaller and smaller half intervals.

 

[elliti123]

Is there a specific example you are working on?

No i am not it was just a theorem, but it seemed odd until now thanks it seems more clear. And about i mentioning the line; i just wanted to know how does the bisection looks like which is clear already by your explanation.

 

[elliti123]

And if i have correctly understood the proof assumes that the bounded sequences actually have some part of their terms, let's say that an interval of the sequence could be finite and there is another which is infinite. Why is that?

 

[RUber]

There are an infinite number of terms in the sequence. So if any interval contains a finite number, the remaining points are still infinite and must be contained in the other interval.

 

[elliti123]

This would only be true for a bounded sequence right?

 

[Mark44]

And if i have correctly understood the proof assumes that the bounded sequences actually have some part of their terms, let's say that an interval of the sequence could be finite and there is another which is infinite. Why is that?

No. If a sequence is bounded, then all of its terms are contained in a closed interval. Think about the example I gave - the sequence {1, 1/2, 1/3, ..., 1/n, ...}. All of the terms in this sequence are in the interval [0, 1].

 

[RUber]

Right. Applying the same logic, an unbounded sequence would be one that for any finite interval, there exist a infinite number of points outside the interval.
An example is $x_n = n$. For any finite interval [0, K] where $K<\infty$, there are a finite number of terms inside the interval and infinite points contained in the infinite interval $[K, \infty)$.

There is not a lot you can do with unbounded sequences.

 

[disregardthat]

Interesting side note: Some sequences bounded to an interval contain all of its points. Simply order the rationals in $[0,1] \cap \mathbb{Q}$. Every real number in [0,1] will be a limit point.