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This is related to Problem $4.6$ in ``The Probabilistic Method'' by Alon and Spencer, where one essentially has to prove the following:

Let $p$ be a prime, and $A$ be any subset of $\mathbb{F}_p$. Then for a uniformly chosen $z \in \mathbb{F}_p$, we have that the dilate $z \cdot A$ intersects every interval of width $\frac{2p}{\sqrt{|A|}}$ with probability at least $0.1$. Here, `intervals' are with respect to the (cyclic) additive group.

My question: What is the correct order of $t$ such that $z\cdot A$ almost surely intersects every interval of width $t$?

For concreteness, let $A$ be any subset of $\mathbb{F}_p$ of size $p^{0.5}$. Then, for a uniformly chosen $z \in \mathbb{F}_p$, does $z \cdot A$ intersect every interval of width $p^{0.5 + \epsilon}$ with high probability?

Here, high probability means probability approaching $1$ as $p$ goes to infinity.

Of course, $p^{0.5}$ is arbitrary. A general statement would say something for intervals of width $\frac{p}{|A|} \cdot \omega(1)$, for some slow growing $\omega(1)$.

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  • $\begingroup$ It seems that you may not quite grasp the notion of a random dilate. It is not some particular dilate which can intersect or not intersect some interval. One can only speak about the probability that a random dilate intersects every interval of given width etc. $\endgroup$ – Seva Jan 9 at 15:45
  • $\begingroup$ @Seva I am sorry, I meant that the probability approaches $1$ as $p$ approaches infinity, but forgot to type. I have edited accordingly. Thanks for pointing this out. $\endgroup$ – Aditya Jan 9 at 15:55

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