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Let have at least elements, for some (we say that has density at least ), and very large .

We know, by Meshulam’s theorem (Roth’s theorem in vector spaces), and even more strongly by Ellenberg-Gijswijt, who used the Croot-Lev-Pach method, that must contain a 3-term arithmetic progression, in other words an affine line.

But does it have to contain planes or even affine subspaces of larger dimensions?

First we examine the case of a random set , thus each is taken in with probability independently of each other.

This implies that any given subspace has probability

to be included in .

Now how many affine subspaces of dimension are there in a space of dimension ? This is where is the number of linear subspaces of dimension , and is easily seen to equal

One gets the obvious bound

Thus the expected number of subspaces of dimension in is at most .

This is of the order of magnitude of a constant

when is of the order of .

So when , there may be no single subspace of dimension .

However, using

the lower bound

may be used to show similarly that when

the set is likely (in fact, almost sure) to contain subspaces of dimension .

Now we prove that if a set has positive density and is large enough, then must contain a subspace of dimension .

The key is basically to use Varnavides averaging argument along with the Ellenberg-Gijswijt bound.

Let be the minimum (if it exists) such that any set of density must contain a subspace of dimension . The Ellenberg-Gijswijt bound amounts to ; indeed, just has to be at least to ensure a line. Then Varnavides argument is an averaging trick that says that a dense set must contain many lines. So many that it must contain many parallel lines. So many that the set of starting points of these lines must in turn contain a line. In which case we obtain a plane. And so on. This was used by Brown and Buhler to derive the recursive relationship

where and is the number of linear spaces of dimension in a space of dimension . We can bound by , which yields .

And so

Inducting, one obtains

Thus . In other words , which is exactly the claimed result.

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