Topology on partially ordered set Let $(X, \leq )$ be a partially ordered set. How would you define a topology on $X$ such that the closed sets are precisely the order-closed sets? 
Where $B \subset X$ is order-closed if $\mathcal{I}B = B = \mathcal{D}B$. Where
$$\mathcal{I}B := \{ a: \exists C \subset B,~~C \neq \emptyset,~~a = \sup C \},$$
$$\mathcal{D} B := \{a: \exists C \subset B, ~~C \neq \emptyset, ~~~ a = \inf C \}. $$
I know that for a totally ordered set $X$, we can consider the order topology defined by subbasis sets of the form $(a, \infty)$ and $(-\infty,b)$ for $a,b \in X$. Could I just get a hint, not a full answer. 
Thanks for any assistance.
 A: Let $\tau$ be the collection of order-closed subsets of $X$, we show that this defines a topology on $X$. 


*

*We have that $\emptyset \in \tau$ vaculously and $X \in \tau$ trivially.

*$\tau$ is closed under arbitrary intersection, this can easily be shown.

*We show that $\tau$ is closed under finite union. Let $B_{1}, B_{2} \in \tau$. We want to show that $B_{1} \cup B_{2} \in \tau$. Thus we first want to show that  $\mathcal{I}(B_{1} \cup B_{2}) = B_{1} \cup B_{2} = \mathcal{D}(B_{1} \cup B_{2})$. 


We show the first equation. One inclusion is easy. We show $\mathcal{I}(B_{1} \cup B_{2}) \subset B_{1} \cup B_{2}$. Take $x \in \mathcal{I} (\mathcal{B_{1} \cup B_{2}})$, there exists a $C \subset B_{1} \cup B_{2}$ such that $\sup C = x$. We want to show that $$x = \sup (C \cap B_{1})~~~\text{or}~~~x = \sup(C \cap B_{2}).$$
Assume neither is true. We know that $c \leq x$ for all $c \in C \cap B_{1}$ or $c \in C \cap B_{2}$. Assume that there is some $y_{1}$ and $y_{2}$ such that $$y_{1} = \sup (C \cap B_{1})~~\text{ and }~~y_{2} = \sup (C \cap B_{2}).$$ Since $B_{1}$ and $B_{2}$ are order-closed, it follows that $y_{1} \in B_{1}$ and $y_{2} \in B_{2}$, therefore $y_{1} \leq x$ and $y_{2} \leq x$. Hence $x = \sup((C \cap B_{1}) \cup \{y_{1}\})$ and $x = \sup((C \cap B_{2})\cup \{y_{2}\})$. If there are other upper bounds then we get that $x = \sup((C \cap B_{1}) \cup (\bigcup_{i}  y_{i}))$ and $x = \sup((C \cap B_{2})\cup (\bigcup_{j} y_{j}))$. But then $x = \sup D_{1}$ for some $D_{1} \subset B_{1}$ and $x = \sup D_{2}$ for some $D_{2} \subset B_{2}$. Hence $x \in B_{1} \cup B_{2}$. 
Hence $\mathcal{I}(B_{1} \cup B_{2}) \subset B_{1} \cup B_{2}$.
It can similarly be shown that $\mathcal{D}(B_{1} \cup B_{2}) = B_{1} \cup B_{2}$.
$\square$
