A basis for a finite topology. I am preparing for my first course in Topology.  I read a lot of confusing articles about the correspondence between the number of topologies on a finite set X and the number of preorders on X.  A preorder on a set X is a binary relation on X that is reflexive and transitive.  I would like to know if the following statement is true.:
A basis for a topology on a finite set X is the collection of sets: 
{ {y|y is in every open set containing x}|x in X }.  In other words, given a finite topology we can form a basis by finding the intersection of all open sets containing x for each x in X.  We may need to delete some duplicate sets.  Then the statement would imply that an upperbound for the minimum number of basis elements is the cardinality of X. 
 A: Yes, you are correct. The weight of a topology is the smallest cardinality for a basis.
Your argument proves that the weight of a topology on a finite set $X$ is always less than or equal to $|X|$.
I don't see the relevance of your first paragraph though.
A: Yes, define $O_x = \cap \{O \in \mathcal{T}: x \in O \}$ for every $x \in X$ which is well-defined for any finite topology (in particular if $X$ is finite). This is the minimal base for the topology $\mathcal{T}$, for obvious reasons (any base must contain all $O_x$, and these sets certainly form a base almost by definition. This indeed shows that $w(X) \le \left|X\right|$. 
As to the relation with the pre-order statement: if we have a topology then $x \le y$ iff $x \in \overline{\{y\}}$ is a pre-order, as clearly $x \in \{x\} \subseteq \overline{\{x\}}$ so that $x \le x$ and if $x \le y$ and $y \le z$ we have $\{y\} \subseteq \overline{\{z\}} (\text{from } y \le z)$, so $\overline{\{y\}} \subseteq \overline{\overline{\{z\}}}= \overline{\{z\}}$, and as $x \in \overline{\{y\}}$ we have $x \in \overline{\{z\}}$ and so $x \le z$, giving transitivity. Now by definition $x \in \overline{\{y\}}$ iff $x$ is in all open sets that contain $y$, so $x \le y$ iff $x \in O_y$, by definition of $O_y$, and then we also have $O_x \subseteq O_y$, as $O_y$ is then one of the open sets that contains $x$ and $O_x$ is the minimal such set. So this pre-order is also reflected in this minimal base, by inclusion. 
