# On an exercise in The Probabilistic Method : random dilate of a set in a finite field

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 $$\omega(1)$$.

• 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. – Seva Jan 9 at 15:45
• @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. – Aditya Jan 9 at 15:55