A note on transitive union-closed families

We show that the Union-Closed Conjecture holds for the union-closed family generated by the cyclic translates of any fixed set.


Introduction
If X is a set, a family F of subsets of X is said to be union-closed if the union of any two sets in F is also in F. The celebrated Union-Closed Conjecture (which has been a well-known 'folklore' conjecture since the mid-1970's, before appearing as a conjecture of Frankl in [2]) states that if X is a finite set and F is a union-closed family of subsets of X (with F = {∅}), then there exists an element x ∈ X such that x is contained in at least half of the sets in F. Despite the efforts of many researchers over the last forty-five years, and a recent Polymath project [5] aimed at resolving it, this conjecture remains wide open. It has only been proved under very strong constraints on the ground-set X or the family F; for example, Balla, Bollobás and Eccles [1] proved it in the case where |F| ≥ 2 3 2 |X| ; more recently, Karpas [4] proved it in the case where |F| ≥ ( 1 2 − c)2 |X| for a small absolute constant c > 0; and it is also known to hold whenever |X| ≤ 12 or |F| ≤ 50, from work of Vučković andŽivković [8] and of Roberts and Simpson [7]. We note that Reimer [6] proved that the average size of a set in an arbitrary finite union-closed family F is at least 1 2 log 2 (|F|); this yields (by averaging) a good approximation to the Union-Closed Conjecture in the case where F is large, e.g. it implies that there is an element contained in at least an Ω(1)-fraction of the sets in F, in the case where |F| = 2 Ω(n) .
In this note, we prove the conjecture in the special case where X is Z n , the cyclic group of order n, and F consists of all unions of cyclic translates of some fixed set. This is a question asked in the Polymath project [5].  We remark that it is possible to deduce a slightly weaker form of Theorem 2 from a theorem of Johnson and Vaughan (Theorem 2.10 in [3]). In fact, the result of Johnson and Vaughan, after applying a quotienting argument, yields that there is an element of G contained in at least (|F| − 1)/2 of the sets in F. (Since F may have odd size, for example when G is Z 3 and R = {0, 1}, this is not quite enough to deduce Theorem 2.) We are indebted to Zachary Chase for bringing this paper of Johnson and Vaughan to our attention.
A short explanation of our notation and terminology is in order. As usual, if G is an Abelian group, and A, B ⊂ G, we write A + B = {a + b : a ∈ A, b ∈ B} for the sumset of A and B. Similarly, if a ∈ G and B ⊂ G, we define a + B = {a + b : b ∈ B}. For any x ∈ G, we let −x denote the inverse of x in G, and for any set A ⊂ G, we let −A = {−a : a ∈ A}. We say a subset A ⊂ G is symmetric if A = −A. If X is a finite set, we write P(X) for the power-set of X.

Proof of Theorem 2.
Before proving Theorem 2, we introduce some useful concepts and notation. Let G be a fixed, finite Abelian group, and let R ⊂ G be fixed. For any set A ⊂ G, we define its R-neighbourhood to be We note that, if R is symmetric and contains the identity element 0 of G, then the R-neighbourhood of any set A is precisely the graph-neighbourhood of A in the Cayley graph of G with generating-set R \ {0}, and similarly, the R-interior of A is precisely the graph-interior of A with respect to this Cayley graph.
Proof of Theorem 2. Let G be a fixed, finite Abelian group and let R ⊂ G be a fixed, nonempty subset of G. Let F = {A + R : A ⊂ G} be the union-closed family consisting of all unions of translates of R.
We define a function f : P(G) → P(G) by It is clear that for any set S ⊂ G, |Int R (S)| ≤ |S|, since for any element r ∈ R, the function x → x + r is an injection from Int R (S) into S. Hence, Next, we observe that f (S) = (−(G \ S)) + R for all S ⊂ G.
Indeed, for any x ∈ G, it holds that It follows that f (P(G)) ⊂ F.
Finally, we observe that the restriction f | F is an injection. This might seem surprising at first glance, but it follows immediately from the fact that To see (3), let S = A + R and observe that N R (Int R (S)) ⊂ S holds by definition (in fact for any set S). On the other hand, if S = A + R, then we have A ⊂ Int R (S) and therefore S = A + R ⊂ N R (Int R (S)). Hence, S = N R (Int R (S)), as required.
Putting everything together, we see that f | F is a bijection from F to itself and satisfies |S| + |f (S)| ≥ |G| for all S ∈ F.