This question is about the space of all topologies on a
fixed set X. We may order the topologies by refinement, so
that τ ≤ σ just in case every τ open set is open in σ.
Equivalently, we say in this case that τ is *coarser*
than σ, that σ is *finer* than τ or that
σ *refines* τ. (See [wikipedia on comparison of
topologies](http://en.wikipedia.org/wiki/Comparison_of_topologies).)
The least element in this order is the indiscrete topology and the largest topology is the discrete topology. 


One can show that the collection of all topologies on a fixed set is a complete lattice. In the downward direction, for example, the intersection of any collection of
topologies on X remains a topology on X, and this intersection
is the largest topology contained in them all. Similarly,
the union of any number of topologies generates a smallest
topology containing all of them (by closing under finite
intersections and arbitrary unions). Thus, the collection of all topologies on X is a complete
lattice. 


Note that the compact topologies are closed downward in
this lattice, since if a topology τ has fewer open sets than
σ and σ is compact, then τ is compact.
Similarly, the Hausdorff topologies are closed upward,
since if τ is Hausdorff and contained in σ, then
σ is Hausdorff. Thus, the compact topologies inhabit
the bottom of the lattice and the Hausdorff topologies the
top.


These two collections kiss each other in the compact
Hausdorff topologies. Furthermore, these kissing points,
the compact Hausdorff topologies, form an antichain in the
lattice: no two of them are comparable. To see this,
suppose that τ subset σ are both compact
Hausdorff. If U is open with respect to σ, then the
complement C = X - U is closed with respect to σ and
hence compact with respect to σ in the subspace
topology. Thus C is also compact with respect to τ in
the subspace topology. Since τ is Hausdorff, this
implies (an elementary exercise) that C is closed with respect to τ, and so U is
in τ. So τ = σ. Thus, no two distinct compact Hausdorff topologies are comparable, and so these topologies are spread out sideways, forming an antichain of the lattice.


My first question is, do the compact Hausdorff topologies
form a maximal antichain? Equivalently, is every topology
comparable with a compact Hausdorff topology? [Edit: François points out an easy counterexample in the comments below.] 


A weaker version of the question asks merely whether every
compact topology is refined by a compact Hausdorff
topology, and similarly, whether every Hausdorff topology
refines a compact Hausdorff topology. Under what
circumstances is a compact topology refined by a unique
compact Hausdorff topology? Under what circumstances does a
Hausdorff topology refine a unique compact Hausdorff
topology?


What other topological features besides compactness and
Hausdorffness have illuminating interaction with this
lattice?


Finally, what kind of lattice properties does the lattice
of topologies exhibit? For example, the lattice has atoms,
since we can form the almost-indiscrete topology having
just one nontrivial open set (and any nontrivial subset
will do). It follows that every topology is the least upper
bound of the atoms below it. The [lattice of topologies is
complemented](http://www.smc.math.ca/cjm/v20/cjm1968v20.0805-0807.pdf).
But the lattice is not distributive (when X has at least
two points), since it embeds N5 by the
topologies involving {x}, {y} and the topology generated by
{{x},{x,y}}.



How do the compact Hausdorff topologies sit in the lattice of all topologies on a set?
**This is a community wiki of the answers in the comments.**


* The compact Hausdorff topologies do not generally form a maximal antichain. If X is infinite, split X into two infinite halves and put the discrete topology on one half and the indiscrete topology on the other half. (Comment by François G. Dorais) **Addendum:** Without sufficient Choice, the infinite set $X$ may be [amorphous](http://en.m.wikipedia.org/wiki/Amorphous_set). Amorphous sets are *precisely* the infinite sets for which this approach doesn't work. Very little Choice is needed to ensure that no such beast exists. (Edit by Cameron Buie)
* There is a maximal compact topology on a countable space which is not Hausdorff. See [Steen & Seebach 99](http://books.google.com/books?id=DkEuGkOtSrUC&lpg=PA118&ots=3hHBTJE-k7&dq=maximal%20compact%20topology&pg=PA118#v=onepage&q=&f=false). (Comment by Gerald Edgar)
* There is a minimal Hausdorff topology on a countable space which is not compact. See [Steen & Seebach 100](http://books.google.com/books?id=DkEuGkOtSrUC&lpg=PA118&ots=3hHBTJE-k7&dq=maximal%20compact%20topology&pg=PA119#v=onepage&q=&f=false). (Comment by François G. Dorais)
* Those examples can be lifted to any cardinality space, simply by using the disjoint sum with any given compact Hausdorff space. (Comment by Gerald Edgar)
* Every set admits a compact Hausdorff topology, by topologizing it as the one-point compactification of the discrete space structure on the complement of any point. (Answer below by Cameron Buie)


(*Feel free to edit and expand*)


