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Sep 30, 2012 at 10:09 comment added Jakob Douglas: You are right that Ramsey's Theorem gives ridiculously bad bounds. The trivial proof that Bill Johnson pointed out for the Lemma in the original question seems to give something like $M=2\ell/\delta$ and $\varepsilon = \delta/(2 M)^\ell$. So we get a "good" lower bound for $M$, but if we choose a very large $M$ then $\varepsilon$ will be "bad". The Ramsey proof for the Lemma in my "answer" below gives a $\varepsilon$ independent of $M$, e.g. $\varepsilon(\delta,k)=\delta/(10)^k$, and something ridiculously large for $M$ (which we get out of Ramsey's Theorem).
Sep 29, 2012 at 21:23 answer added Jakob timeline score: 3
Sep 29, 2012 at 21:13 comment added Douglas Zare What bounds are you supposed to get using Ramsey theory? The ones I get from Ramsey theory seem really bad compared to what I expect is true.
Sep 29, 2012 at 8:20 comment added Ed Dean Jakob, you might find the closely related Theorem 2.2 of the following paper to be of interest: dx.doi.org/10.4115/jla.2012.4.3
Sep 29, 2012 at 7:12 answer added Robin Tucker-Drob timeline score: 1
Sep 29, 2012 at 0:13 comment added Bill Johnson Sure, but then some collection of $\ell$ sets contains many points (not that this gives a very good estimate).
Sep 28, 2012 at 23:28 comment added Goldstern @bill johnson: I think this only gives you a large set of points that are in at least $\ell$ sets, but Jakob wants that they are all in the same $\ell$ sets.
Sep 28, 2012 at 23:22 comment added Bill Johnson Analyse the sum of the characteristic functions of the sets, which has integral at least $|M| \delta$ and sup norm at most |M|$.
Sep 28, 2012 at 22:28 comment added Andrés E. Caicedo Welcome, Jakob.
Sep 28, 2012 at 22:16 history edited Goldstern
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Sep 28, 2012 at 22:13 comment added Goldstern Welcome to mathoverflow! What a coincidence, I was just thinking about this very question :-)
Sep 28, 2012 at 21:53 history asked Jakob CC BY-SA 3.0