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

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 $g<b-a$ 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.

HallsofIvy

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: a_n= b, then one those intervals will contain b and so contain an "infinite number" (in fact, all) of the terms of the sequence.

led5v

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.

Vargo

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/(b-a).

led5v

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 sub-interval 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. "

Vargo

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/(b-a).