## infinte number of terms from a sequence in a sub-interval

I have across the following argument, which seems wrong to me, in a larger proof (Theorem 4 on page 9 of the document available at http://www.whitman.edu/mathematics/S...athanWells.pdf). I would appreciate if someone can shed light on why this is true.

The argument is that given a sequence $a_k$ of points in [a,b], we can say that a sub-interval of [a,b] exists such that it is smaller than some value $g<b-a$ and contains an infinite number of terms from $a_k$.

I disagree with the above statement because lets say that the sequence $a_k$ always returns a constant value, say b. Then the above statement doesnt hold.
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 Quote by led5v I have across the following argument, which seems wrong to me, in a larger proof (Theorem 4 on page 9 of the document available at http://www.whitman.edu/mathematics/S...athanWells.pdf). I would appreciate if someone can shed light on why this is true. The argument is that given a sequence $a_k$ of points in [a,b], we can say that a sub-interval of [a,b] exists such that it is smaller than some value $g Yes, it does. It says "an infinite number of terms". It does NOT say "an infinite number of different terms". If your sequence were constant: an= b, then one those intervals will contain b and so contain an "infinite number" (in fact, all) of the terms of the sequence.  Thanks for the clarification and I understand the point. But still I am not very convinced with the proof of Theorem 4 in the cited document above. The proof utilizes the above statement to conclude that two distinct values of the sequence ($a_k$), say$a_K$and$a_K′$would exist in a sub-interval of [a,b] such that$a_K−a_K′$is greater than some value. Now, I think the above may not be possible and there is nothing in the original statement that guarantees that two distinct values of the sequence$(a_k)\$ would fall in a subinterval of [a,b]. This may happen but we cannot say that it would happen for sure? Thanks again.
 The statement is this. We are given an interval I which is partitioned into a finite number of subsets I_1, ...., I_M (which I refer to as bins). Let a_k be any sequence of numbers in I. Then there exists at least one bin, call it I_1, for which there exists an infinite set of indices $K\subset \mathbb{N}$ such that for all k in K, a_k belongs to I_1. That statement is true on the level of sets, meaning it has nothing to do with the fact that I is an interval and that I_1, ..., I_M are subintervals of a partition. And it has nothing to do with the nature of the sequence a_k. In other words, we can construct any sequence a_k and this fact will be true. In this case the author constructs the sequence a_k by using the hypothesis that f is unbounded on the interval and requiring that $|f(a_k)|>k$. Given that construction, let I_1 refer to a subinterval for which there an infinite number of indices k such that a_k belongs to I_1. Let K denote the set of indices for which a_k belongs to I_1. Choose k_1 in K. Now let B >0 be any positive number. K is an infinite subset of the positive integers, so it contains an element k_2 for which $k_2 > B+ |f(a_{k_1})|$ Then we conclude that a_{k_2} belongs to the same bin I_1 and $|f(a_{k_2})| > k_2 > B+ |f(a_{k_1})$ Now let B = n/(b-a).