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Infinte number of terms from a sequence in a subinterval 
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#1
Nov1212, 04:18 AM

P: 3

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 subinterval of [a,b] exists such that it is smaller than some value $g<ba$ 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. 


#2
Nov1212, 05:13 AM

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#3
Nov1212, 05:37 AM

P: 3

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 subinterval 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. 


#4
Nov1212, 08:39 PM

P: 350

Infinte number of terms from a sequence in a subinterval
Nah, the proof is fine except for one or two typos. It is not a_k  a_k' that is supposed to be large, but f(a_k)f(a_k'). The point is this. The interval is partitioned into subintervals. Given the sequence a_k, an infinite number of these must fall into one of the bins (subintervals). That is true for any sequence in the original interval. Then he uses the fact that f(a_k) > k for each k to show that it is possible to select indices k and k' such that f(a_k)f(a_k') is larger than n/(ba).



#5
Nov1312, 02:46 AM

P: 3

My understanding is that in the original theorem, we want to prove
∃y.∀x.a≤x∧x≤b⇒fx≤y If we contradict this statement, it becomes: ∀y.∃x.(a≤x∧x≤b)∧y<fx which can be further simplified to the following form: ∀k.(a≤(g m)∧(g m)≤b)∧m<(f(g m)) where m is a positive integer and (g_m) represents the sequence (a_k) in my original question. Now, the main concern that I have is that how can I mathematically prove that two distinct values of (g_m) would occur in a subinterval of [a,b]. So basically, I dont understand the mathematical reasoning that allows us to reach the following statement that Vargo made from the last equation above. "Given the sequence a_k, an infinite number of these must fall into one of the bins (subintervals). That is true for any sequence in the original interval. " 


#6
Nov1312, 12:45 PM

P: 350

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 [itex] K\subset \mathbb{N}[/itex] 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 [itex] f(a_k)>k [/itex]. 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 [itex] k_2 > B+ f(a_{k_1}) [/itex] Then we conclude that a_{k_2} belongs to the same bin I_1 and [itex] f(a_{k_2}) > k_2 > B+ f(a_{k_1})[/itex] Now let B = n/(ba). 


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