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Jun 20, 2015 at 18:33 comment added YCor Ah ok I was misleaded by the fact the first $I$ could have been written $I=I_k$...
Jun 20, 2015 at 18:31 comment added Will Sawin @YCor Exactly, in limit $I$ is fixed. The statement is for each $I$, if $D(I) \geq 2$, then $\lim_{ n \to \infty} D(I^n)/n>1$. This is not contradicted by the first statemnt of my post, in which $I$ varies.
Jun 20, 2015 at 18:24 comment added YCor If you vary $I$ the notation $\lim_{n\to\infty}$ is quite misleading, and anyway I can't guess the statement you have in mind if you don't want to write it down.
Jun 20, 2015 at 18:15 comment added Will Sawin @YCor The limit for any fixed $I$. In that I am varying $I$.
Jun 20, 2015 at 17:28 comment added YCor I'm confused by the last section where you say $D(I)\ge 2$ implies $\lim D(I^n)/n>1$, since it contradicts the very first statement of your post.
Jun 20, 2015 at 16:56 comment added Will Sawin @BorisBukh Did everything you two wanted and more.
Jun 20, 2015 at 16:56 history edited Will Sawin CC BY-SA 3.0
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Jun 20, 2015 at 16:24 vote accept Boris Bukh
Jun 20, 2015 at 16:20 comment added Boris Bukh Nice! I believe that your deleted argument shows that if $D(I)\geq 2$, then $\lim D(I^n)/n>1$. So, as @YCor says, it might be worth undeleting it.
Jun 20, 2015 at 16:10 comment added YCor I think that the other part (namely that if $V(I)$ is not contained in a line then there is a good inequality, which you just erased) is worth keeping in the post.
Jun 20, 2015 at 15:58 history edited Will Sawin CC BY-SA 3.0
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Jun 20, 2015 at 15:27 history edited Will Sawin CC BY-SA 3.0
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Jun 20, 2015 at 13:43 history answered Will Sawin CC BY-SA 3.0