Why is there a bijection between the ultrafilters that converge and a topology If we call $\mathcal{UF}(X)$ the set of ultrafilters on a set $X$. I read here that there is a bijection between topologies on a set $X$ and $\{0,1\}^{\mathcal{UF}(X)}$. As I am unfamiliar with category theory, I do not have enough background to understand the proof. Can one give me a simple proof ?
 A: It's not quite the set of subsets of $UF(X)$. That is, it's not enough to know which ultrafilters converge: in any compact space every ultrafilter converges, so all compact topologies would be indistinguishable in this way. Instead what one gets from a topology is a map $UF(X)\to P(X)$ the powerset of $X$ that sends an ultrafilter to the set (possibly empty) of its limit points. Such a map is called a convergence structure on $X$.
To get a topology from a map $c:UF(X)\to P(X)$, the basic idea is to construct the neighborhood filter $N_{x,c}$ of $x$ as the intersection of all ultrafilters converging to $x$. This will construct an inverse to the operation of the preceding paragraph: the neighborhood filter is by definition the intersection of all filters convering to $x$ and any proper filter is given by the intersection of all ultrafilters refining it (assuming the choice principle that every proper filter is contained in an ultrafilter.) 
But if our map $c$ didn't come from a topology, the $N_{x,c}$ might not fit together as a topology. We have to know that for every $A\in N_{x,c}$ that $x\in A$ and that there's $B\subset A$ such that $A\in N_{y,c}$ for every $y\in B$. (This is one of the standard equivalent definitions of a topology: see here.) The simplest thing to do is to just add these as requirements on $c$. That is, $c$ is a topological convergence structure on $X$ if whenever $A\in \cap U: x\in c(U), x\in A$ and there's $B\subset A$ with $A\in \cap_{y\in B}(\cap U:y\in c(U))$. Now the operation assigning $(N_{x,c})_{x\in X}$ to a topological convergence structure $c$ is the two-sided inverse of the map sending a topology to its associated convergence structure $c$. All that we still have to check is that $c$ is the convergence structure for $(N_{x,c})$, which is immediate since ultrafilters converging to $x$ are exactly those containing $N_{x,c}$. 
So this works, but the extra requirements on $c$ were cooked in just to get the correspondence with topologies. If you want an explanation for why topological spaces are natural objects, this is cheating, which is the motivation for Barr's categorical work you saw referenced at your link. The upshot is that we can recover topological spaces as objects associated to the mapping $X\mapsto UF(X)$, so that if we think ultrafilters are more natural than topologies we can arguably base the latter on the former.
