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As noted by Seva, you're looking for Szemeredi's cube lemma, which is one of the simplest results of its kind. Let $S\subset\mathbf{F}_2^n$, and let $\|S\|_{U^k}^{2^k}$ denote the number of $e_0,e_1, …

Yes. Let $\mathbb{Z} = \bigcup_{i=1}^\infty I_i$ be a decomposition of $\mathbb{Z}$ into disjoint intervals of length $n$. We will find $A\subset\mathbb{Z}$ such that $|A\cap I_i| = 1$ for each $i$ an …

Is there any known improvement on the Kahn-Kalai-Linial inequality (on the influences of boolean functions) in the special case in which $f$ is the indicator function of an intersecting monotonic set …

The first two rows are identical with probability $2^{-n}$, so $\mathbb{P}(\det M_n = 0) \geq 2^{-n}$. Incidentally, there are $2 \times \binom{n}{2}$ events like this to consider, though not quite in …

Suppose we have some positive quantites $P$ and $Q$ which depend on some choices that we make, and we want to show that some choice makes the quotient $P/Q$ fall below some cool bound. One idea is to …

Given a group $G$ acting transitively on a set $X$ of $n$ points, consider the induced action on the set $\binom{X}{k}$ of $k$-element subsets of $X$. Obviously, if $k>n/2$, the orbit of any set is in …

This question follows up a previous question, Intersecting group orbits. Suppose a group $G$ acts transitively on a set $X$ of $n$ elements, where $n$ is even, and consider the induced action on the …

A set system $\mathcal{U}\subset P([n])$ is an upset if $B\supset A \in \mathcal{U}$ implies $B\in \mathcal{U}$, intersecting if $A,B\in\mathcal{U}$ implies $A\cap B \ne \emptyset$. Note that a no …

Suppose a group $G$ acts on a set $\Omega$. Call a subset $A \subset G$ regular (or sharply transitive or simply transitive or...) if for every two points $\omega_1, \omega_2 \in \Omega$ there is a un …

The Cheeger inequality is another example.

I have come across some references for this problem, which amount to a solution very different from the (beautiful) one given by Peter Mueller, and which also goes further. It turns out that it is pos …

Yes your claim is correct for trees. Here is a standard fact about automorphism groups of trees: Lemma: If $T$ is a finite tree then there is either a vertex or an edge fixed by every automorphism of …

This is an answer to the follow-up question about automorphisms of a subdivision. Suppose $G$ is a connected graph which is not $2$-regular. Let $G^{(k)}$ be the $k$-subdivision of $G$, i.e., the gra …

Repeating from the comments section: This (natural and beautiful) question was previously asked and answered on this site. See Collapsible group words. It also appeared recently on math.se. The questi …

Define $f_i(n) = (-1)^{b_i(n)}$ where $b_i(n)$ is the $i$th binary digit of $n$ (Speyer's example with $\lambda = e_i$). Then $(f_1, f_2, \dots)$ defines an "infinite Hadamard matrix" because the part …