# Disjoint union of affine subspaces contains a larger affine subspace

I'd like to say that a large structured subset of the $n$-dimensional Boolean cube $\{0,1\}^n$ contains a non-trivial affine subspace. To be more specific, I want to prove/disprove that for some function $d=d(n)\gg\log(n)$ the following holds. If one partitions the $n$-dimensional Boolean cube into $2^{n-d}$ affine subspaces of dimension $d$, and then takes a half of these subspaces, then their union must contain an affine subspace of dimension $d+1$.

For $d < \log(n)-O(1)$ the statement holds just because each large subset of the Boolean cube always contains an affine subspace of dimension $\log(n)-O(1)$. For large subsets without any structure this logarithmic bound is essentially tight, since a counting argument shows that a random subset doesn't contain larger affine subspaces.

I'm wondering if the aforementioned structure of the set lets us find larger subspaces. I would appreciate any references to similar results (e.g., a disjoint union of arithmetic progressions of length d contains a longer arithmetic progression etc).

• Let $e=n-d.$ Then the property you want to prove or disprove always fails for $e=1$ and sometimes (in $4/5$ of the cases?) for $e=2.$ – Aaron Meyerowitz Jan 20 '18 at 19:19
• Right, this property trivially holds for $d<\log{n}$ and doesn't necessarily hold when $d$ is close to $n$. I'm wondering about the intermediate regime. For example, does it always hold for $d=\mathrm{poly}(\log{n})$? – Alex Golovnev Jan 23 '18 at 23:11

• Hi Shachar, thanks! Is there a simple way to analyze random parity decision trees? I currently don't see even how to handle the case $d=(1-\varepsilon)n$. – Alex Golovnev Jan 23 '18 at 23:09