TITLE: Union closed conjecture induction
QUESTION [10 upvotes]: I would like an example showing that one of the most basic induction approaches to the union-closed conjecture fails. If, for any union-closed family $\mathcal{A}$ of subsets of a finite set $X$, there is some $x \in X$ such that each $y \in X$ has $|\{A \in \mathcal{A} : A \ni y \text{ and } A \ni x\}| \ge \frac{1}{2}|\{A \in \mathcal{A} : A \ni x\}|$, then we can merely use induction applied to the union-closed family $\{A \in \mathcal{A} : A \not \ni x\}$ to get some $y \in X$ in at least half of the sets of $\{A \in \mathcal{A} : A \not \ni x\}$, and by our choice of $x$, we then see that $y$ is in at least half the sets of $\mathcal{A}$.
I have to think that there is a known example showing this approach doesn't work, i.e., there is an $\mathcal{A}$ with no such $x$. But I couldn't think of an example. So,:

Give an example of a finite set $X$ and a union-closed family $\mathcal{A} \subseteq \mathcal{P}(X)$ such that, for each $x \in X$, there is some $y \in X$ with $$|\{A \in \mathcal{A} : A \ni y \text{ and } A \ni x\}| < \frac{1}{2}|\{A \in \mathcal{A} : A \ni x\}|.$$ (Or prove the union-closed conjecture!)

I avoid degenerate cases, like $X = \emptyset$, $\mathcal{A} = \emptyset$, or $\mathcal{A} = \{\emptyset\}$.

REPLY [10 votes]: Let $X = 123456$, and $\mathcal{A} = \{134, 1345, 1346, 13456, 256, 2356, 2456, 23456, 123456\}$. Let $f_x$ and $f_{x, y}$ be the number of sets in $\mathcal{A}$ containing $x$, or both $x$ and $y$ respectively. Say that $y$ is rare for $x$ if $f_{x, y} < f_x / 2$. Then:

$f_1 = f_2 = 5$, but $f_{1, 2} = 1$, thus $1$ and $2$ are rare for each other;

$f_3 = f_4 = 7$, but $f_{3, 2} = f_{4, 2} = 3$, thus $2$ is rare for $3$ and $4$;

similarly, $1$ is rare for $5$ and $6$.