Let $X$ be an ordered set. For all $A\subseteq X$, define $A^u$ to be set of all upper bounds of $A$ and similarly denote by $A^l$ the set of all lower bounds. Let $\mathcal{D}$ be the set off all downsets of $X$. Denote by $c\colon \mathcal{D}\to \mathcal{D}$ the closure operator given by $c(A)=A^{ul}$.
Let $D$ be a downset. I was trying to prove that the supremum of the set $$\{\downarrow x \mid x\in D\}$$ in $\mathcal{D}$ is precisely $c(D)$. It seems to me that it has to be true, but I need a hint to prove this...
this is false. consider the poset given by $a \leq x$ and $b \leq x$. The set $\{a,b\}$ is a downset. The supremum of $\{\{a\},\{b\}\}$ in $\mathcal{D}$ is $\{a,b\}$. However, $\{a,b\}^{ul}=\{a,b,x\}$.