Help me find or prove an obvious (?) lemma

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SUMMARY

The discussion centers on a mathematical lemma concerning the relationship between a subset A and its superset B within the natural numbers, specifically when A contains "almost all" of B. The main conclusion is that if the growth of B is well-defined, the ratio of the nth elements of A and B approaches 1 as n approaches infinity. The formalization involves limits and asymptotic notation, particularly |B∩{1,2,…,n}|∼n(log n)^k, suggesting that this result may relate to concepts in Number Theory regarding the densities of integer sets.

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  • Understanding of limits and asymptotic notation in mathematics.
  • Familiarity with subsets and supersets in set theory.
  • Knowledge of Number Theory, particularly the concept of densities of sets of integers.
  • Basic proficiency in mathematical proofs and formalization techniques.
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  • Research "asymptotic analysis in Number Theory" to explore related concepts.
  • Study "density of sets of integers" for foundational knowledge on the topic.
  • Examine "Bar's Theorem" and its applications in mathematical proofs.
  • Investigate "growth rates of sequences" to understand implications on ratios of sequences.
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Mathematicians, statisticians, and students of Number Theory seeking to deepen their understanding of relationships between subsets and their supersets, particularly in the context of limits and growth rates.

CRGreathouse
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I was trying to prove something when I realized that it generalized fairly nicely. The basic idea is that if a subset contains 'almost all' of its superset (both sets inside N), and the superset has 'nice' growth, the ratio between the nth element of the two sets approaches 1 as a limit.

Here's an attempt to formalize:

Given A\subseteq B\subseteq\mathbb{N}, with
\lim_{n\to\infty}\frac{|A\cap\{1,2,\ldots,n\}|}{|B\cap\{1,2,\ldots,n\}|}=1
and
|B\cap\{1,2,\ldots,n\}|\sim n(\log n)^k for some k (perhaps this can be weakened to remaining between k1 and k2)
it follows that
\lim_{n\to\infty}a_n/b_n=1
where a_i is the ith largest member of A and likewise for b_i.

1. Is this known? Does it have a name or a standard (textbook?) reference?
2. Failing #1, is there an easy proof for this?
3. Failing #2, is the result -- or something similar -- true?
 
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I don't know a proof, or of a quick method of attack, but this does look familiar as a type of problem I saw in Number Theory many years ago. Look for a discussion of densities of sets of integers in those books - that may be a guide.
Keep checking back - I'm sure someone here, far more knowledgeable about this area than this poor statistician (:smile:) , will soon offer some advice.
 
I don't see how a_1/b_1 depends on n.
 
DeadWolfe said:
I don't see how a_1/b_1 depends on n.

:rolleyes:

Sorry, that should be a_n/b_n.

The intuition here is that if A is almost all of B, their ratio shouldn't get too large if the sequence is well-behaved and grows slowly.
 
statdad said:
I don't know a proof, or of a quick method of attack, but this does look familiar as a type of problem I saw in Number Theory many years ago. Look for a discussion of densities of sets of integers in those books - that may be a guide.

Well, I'm still looking. It does seem a rather familiar result, though, just as you mention... I can't help but think I'm missing something glaringly obvious ('a simple reduction to foo, with a judicious application of Bar's Theorem, shows that this is true for |A| > N for some N...').
 
OK, so here goes.

The parent sequence is distributed so for any \varepsilon>0, the nth member b_n is at most k_\varepsilon n^{1+\varepsilon}. The subsequence (a_i)_n is such that f(m) that maps m to the unique n such that a_m=b_n has
\lim_{m\to\infty}f(m)/m=1
or alternatively for every M there is a g(M) with
f(m)\le g(M)m for all m > M.

Now I just need to show that g(M) needn't get too big, right?
 

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