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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.) 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. 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}}.

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The maximal antichain question has a negative answer for infinite X. Split X into two infinite halves put the discrete topology on one half and the indiscrete topology on the other. –  François G. Dorais Feb 19 '10 at 21:52
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Is it obvious that there exists a compact Hausdorff topology on every set? –  Qiaochu Yuan Feb 20 '10 at 0:11
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Is it obvious that there exists a compact Hausdorff topology on every set? Yes (using the well-ordering theorem) ... the order topology on the set of ordinals up to and including a given ordinal. –  Gerald Edgar Feb 20 '10 at 0:17
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Steen & Seebach 99. –  Gerald Edgar Feb 20 '10 at 0:21
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About Qiaochu's question: is it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology? –  Joel David Hamkins Feb 20 '10 at 4:19

4 Answers 4

up vote 24 down vote accepted

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)

  • There is a maximal compact topology on a countable space which is not Hausdorff. See Steen & Seebach 99. (Comment by Gerald Edgar)

  • There is a minimal Hausdorff topology on a countable space which is not compact. See Steen & Seebach 100. (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)

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Great! This is exactly the sort of answer for which I was hoping (although I don't like the restriction to countable spaces---I wonder if it can be generalized?). Thanks François, Gerald and Qiaochu for the great information (and especially Gerald for the Steen and Seebach reference). But now I observe that since you all answered in comments, we are unable to bestow the deserved reputation on any of you.... –  Joel David Hamkins Feb 20 '10 at 4:18
    
Thanks for the edit, Todd! –  Joel David Hamkins Oct 12 at 17:18
    
For purposes of bestowing reputation, ask only one question per post. You are more likely to get answers rather than comments. –  Gerald Edgar Oct 12 at 18:03

I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:

"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"

In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number, not Dedekind-finite. Obviously, finite sets can be given the discrete topology. Suppose $X$ is an infinite set, fix $x\in X$, and let $Y:=X\setminus\{x\}.$ Now let $\mathcal T$ be the set of all subsets $U$ of $X$ such that either (1) $U\subseteq Y$ or (2) $x\in U$ and $X\setminus U$ is finite.

Then $\mathcal T$ is a compact Hausdorff topology on $X$, and in particular, $\langle X,\mathcal T\rangle$ is homeomorphic to the Alexandrov one-point compactification of $Y$ in the discrete topology.

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+1. Thanks very much for this observation. –  Joel David Hamkins Jun 27 '13 at 11:45
    
No problem at all. I had this realization while I was learning about Alexandrov compactifications, and I'm pleased that my musings came to some use. –  Cameron Buie Jul 4 '13 at 17:35

Most of this is classical, starting with the memoir by Alexandrov and Urysohn, in which they introduced their notion of the compact Hausdorff space (as bicompact), and also of the absolutely closed space (closed in any Hausdorff superspace) including an extensive discussion of them. This and the minimal Hausdorff spaces, and the related stuff, is very nicely presented as exercises in the Bourbaki General Topology; also Engelking covers these topics in his classical monography (which had several editions). Needless to say, a number of research papers was devoted to minimal Hausdorff spaces and similar.

It's easy to see why the standard Euclidean topology in the space of rational numbers cannot be weakened to a compact topology. The key is: Baire property.

A pretty general result of this type appeared in my paper, Minimal Hausdorff Spaces and $T_1$-Bicompacta, Soviet DAN 1968, v.178, pp 24-26. Let's talk about $T_1$-spaces only, so that complete regularity implies Hausdorff. Theorem 1' states:

Let a completely regular space be a countable union of its nowhere dense (i.e. having empty interior) compact subsets. Then its topology does not dominate any minimal Hausdorff topology.

Still more general (but easier to prove) is Theorem 1 there:

Let a Hausdorff space   $X$ be a countable union of its nowhere dense (i.e. closed and having empty interior) compact subsets. Assume also that it is a dense subspace of a Hausdorff space which has the Baire property. Then topology of   $X$   does not dominate any minimal Hausdorff topology.

The formulation of Theorem 1 suggests how to prove Theorem 1'.

NOTE:   in the published paper the formulation of Theorem 1' missed word countable (it appears in Theorem 1).

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I didn't state any results about the relation between $T_1$- and $T_2$-compacta to keep my answer reasonably short. One can find some results and references in the paper mentioned above. –  Włodzimierz Holsztyński Jun 29 '13 at 20:15

In fact there are spaces which are "minimal Hausdorff" -- they have no coarser Hausdorff topology -- but are not compact. It turns out that these spaces are "H-closed" (every open cover has a finite subfamily whose closures cover) and semi-regular (the collection of regular open sets form a base). A minimal Hausdorff space is compact exactly when it is Urysohn. Spaces which have coarser minimal Hausdorff topologies are called Katĕtov. A "nice" example of a space which is not Katĕtov is the space of rational numbers $\mathbb{Q}$.

I'm not sure about compact spaces, but I suspect that a Hausdorff space has a unique coarser minimal Hausdorff topology exactly when it is H-closed. One direction I'm sure of -- the semi-regularization of an H-closed space is minimal Hausdorff.

By the way, (one of) THE BOOK(s) on this topic is Extensions and absolutes of Hausdorff spaces by Porter and Woods, however it discusses Hausdorff spaces almost exclusively.

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