topology of kolmogroff space Let $X$ be a Kolmogoroff space. Show that $x\in \overline{\{x'\}}$ is an order relation between $x$ and $x'$ in $X$. and that if this relation is written as $x\leq x'$ then given topology on $X$ is identical to right topology ($[x,\rightarrow[$) determined by this ordering. I have proved the first part($x\in \overline{\{x'\}}$ is ordered), but how to show the identical relation with right topology. 
 A: That is impossible.  Consider the set S with two different Hausdorff topologies.  They are both Kolmogroff spaces.
They have the same order, namely x <= y iff x = y.
As they are different topologies, one of them cannot be
homeomorphic to the right topology of that order.    
A: Your question is incompletely formulated, I think. If $X$ is a Kolmogorov (a.k.a $T_0$) space then indeed $x \le y$ iff $x \in \overline{\{y\}}$ is a partial order (the so-called specialisation order). In fact, it's a pre-order (reflexive and transitive) in any space but the $T_0$ property ensures the antisymmetry: $x \le y$ and $\le x$ implies $x=y$.
But the Bourbaki problem quoted in the comments (General Topology, Chapter 1, p. 117 problem 2c) is for an Alexandrov space, a space where the intersection of any family of open sets is open! (it's somewhat hidden in the beginning and Bourbaki does not call it this). In that case the stated result does hold:
Define for each $x \in X$, the set $M_x = \bigcap \{O\mid x \in O \text{ open } \}$ the minimal open set containing $x$ (it's open because we have an Alexandrov space). $x \le y$ iff $y \in M_x$ (iff every open set containing $x$ contains $y$) and the $T_0$ property ensures that $x \neq y$ implies that $M_x \neq M_y$. In that case it is indeed not too hard to show that $X$ has the upper topology wrt this partial order $\le$. E.g. $[x,\rightarrow)=M_x$ is open and every open set $O$ of $X$ is the union of the minimal neighbourhoods of its elements, so upper-open too.
The arbitrary intersection property makes a big difference...
