  We show that there is some absolute constant $c>0$, such that for any
union-closed family $\mathcal{F} \subseteq 2^{[n]}$, if \mbox{$|\mathcal{F}|
\geq (\frac{1}{2}-c)2^n$}, then there is some element $i \in [n]$ that appears
in at least half of the sets of $\mathcal{F}$. We also show that for any
union-closed family $\mathcal{F} \subseteq 2^{[n]}$, the number of sets which
are not in $\mathcal{F}$ that cover a set in $\mathcal{F}$ is at most
$2^{n-1}$, and provide examples where the inequality is tight.
