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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 possess 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 cant exactly picture the successive bisection method on a line, so if anyone could please somehow explain it in more clear manner; since i cant 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?

Elliti123
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2 Answers2

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Let's say you have a bounded sequence $a_n$ that lies between the bounds $[-M, M]$, then you bisect this to get two intervals $[-M, 0]$ and $[0,M]$. One of those intervals contain a finite number of terms and the other, infinite, as picture below:

enter image description here

We labelled $I_1$ as the interval with the infinite number of terms and let $a_{n_1} \in I_1$. Now we bisect $I_1$ again, as can be seen above. $I_2$ is the interval with the infinite number of terms. Select a $a_{n_2}$ where $n_2 > n_1$ and $a_{n_2} \in I_2$.

In general, you can continue this construction of the closed interval $I_k$ by bisecting the interval $I_{k-1}$ and labelling the half that contains an infinite number of terms $I_k$. Then select $n_k > n_{k-1} > \cdots > n_2 > n_1$, such that $a_{n_k} \in I_k$.

Zain Patel
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  • When you say that one of those intervals contains a finite number of terms and the other infinite i get confused. Does this happen is bounded sequences ? Or we just generally choose the half that has infinite terms in it? And what does it exactly mean for an interval to contain infinite terms? – Elliti123 Jul 21 '15 at 17:01
  • It happens in bounded sequences, yes. For example, the sequence $a_n = 1/2^n$ has a finite number of terms in the interval $[1/4, 1]$ and an infinite number of terms in the interval $[0, 1/4]$. It quite literally means the interval that has an infinite number of terms as $n \to \infty$. – Zain Patel Jul 21 '15 at 17:04
  • Is it only for bounded sequences or does it also work for other sequences as well? – Elliti123 Jul 21 '15 at 17:10
  • @Elliti123 Only bounded sequences. – Zain Patel Jul 21 '15 at 17:10
  • Could you provide a proof? – Elliti123 Jul 21 '15 at 17:11
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    Thanks it seems more clear now. – Elliti123 Jul 21 '15 at 17:12
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Suppose $[0,1]$ contains infinitely many $x_n$. Then either $[0,1/2]$ contains infinitely many $x_n$ or $[1/2,1]$ contains infinitely many $x_n$ (or both).

since by this method each interval should have a clear beginning and end which we do know, right? Not sure exactly what you mean. The proof is not about what we know, it's about what's true.

Say $I_1=[0,1]$ contains infinitely many $x_n$. If $[0,1/2]$ contains infinitely many $x_n$ let $I_2=[0,1/2]$; otherwise let $I_2=[1/2,1]$. At this point we don't "know" what $I_2$ is. But we do know that $I_2$ contains infinitely many $x_n$.

David C. Ullrich
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