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?