Specialization determines the topology of a sober space? Given a sober space $A$, define a preorder on it like this: $x\le y\in A$ iff $\overline{\{x\}}\subset \overline{\{y\}}$. Let $X,Y$ be sober spaces and isomorphic as sets with preorder. Are $X,Y$  homeomorphic?
 A: The answer is no. In fact, as the comments pointed out, this is very far from true. I will construct an easy counterexample.
Equip the set of real numbers $\mathbb{R}$ with the discrete topology and call the resulting topological space $D$. Every set is closed in the discrete topology, so we have $\overline{\{x\}} = \{x\}$ for each $x \in \mathbb{R}$. This forces the resulting specialization order $\leq_D$ to be equality: for all $x,y \in \mathbb{R}$, $x \leq_D y$ precisely if $x=y$. 
Now equip the set of real numbers $\mathbb{R}$ with the usual metric topology and call the resulting space $M$. The complement of $\{x\}$ is the open set $(-\infty,x) \cup (x,\infty)$, so $\{x\}$ is closed in $M$ for each $x$. Therefore, the resulting specialization order $\leq_M$ is again just equality.
Notice that every metric space is Hausdorff, so a fortiori sober. Consequently, the spaces $M$ and $D$ defined above are sober. Morever, $(\mathbb{R}, \leq_D)$ and $(\mathbb{R}, \leq_M)$ are isomorphic (as a matter of fact, equal) as preordered sets. However, the spaces $M$ and $D$ are clearly not homeomorphic.
A: While this is not true in general, there is one important special case where it is true: the topology on a Noetherian sober space is determined by its specialization preorder.  More strongly, the following is true.

Proposition: Let $X$ be a Noetherian sober space.  Then every closed subset of $X$ is the closure of a finite subset of $X$.

It follows that the topology is determined by the specialization preorder, since the specialization preorder determines the closures of singletons and thus the closures of finite sets, and thus all the closed sets by the Proposition.
Proof of Proposition: Suppose some closed subset of $X$ is not the closure of a finite set.  Since $X$ is Noetherian, there is then a minimal closed subset $A\subseteq X$ that is not the closure of a finite set.  Since $X$ is sober, $A$ cannot be irreducible, since then it would be the closure of a singleton.  So we can decompose $A=B\cup C$ where $B,C\subset A$ are strictly smaller closed sets.  By minimality of $A$, there are finite sets $S$ and $T$ such that $B=\overline{S}$ and $C=\overline{T}$.  But then $A=\overline{S\cup T}$ is the closure of a finite set, which is a contradiction.
