For any undirected simple graph $G=(V,E)$ we define for $v\in V$ the set $N(v) = \{w\in V: \{v,w\}\in E\}$.


Suppose $A, B$ are finite, disjoint sets, and $G = (A\cup B, E)$ is a bipartite graph with bipartition $(A,B)$ -- that is, for every $e\in E$ we have $e\cap A \neq \emptyset \neq e\cap B$. Informally speaking, I am trying to prove that if every $b\in B$ has "enough neighbors" in $A$, then $2$ members of $A$ "cover" $B$. More precisely, I would like to know whether the following statement is true:



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> > If $N(b) > |A|/2$ for all $b\in B$, then there are $a\_1, a\_2\in A$ such that $N(a\_1) \cup N(a\_2) = B$.
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On a condition concerning the number of neighbors in bipartite graphs
This is wrong. Consider, for instance, the situation where the partite sets $A$ and $B$ are two copies of $\mathbb F\_p$, with $p\equiv 1\pmod 4$ prime, and $a\in A$ is adjacent to $b\in B$ whenever $a-b$ is a square in $\mathbb F\_p$; that is, $a=b$ or $a-b$ is a quadratic residue. In this case each $b\in B$ has $(p+1)/2>|A|/2$ neighbors, but for any given $a\_1,a\_2\in A$ there are just about $\frac34p$ those $b\in B$ adjacent to either $a\_1$ or $a\_2$.


