Consider a non-empty set $X$ and its complete lattice of topologies (see also this thread).

The discrete topology is Hausdorff. Every topology that is finer than a Hausdorff topology is also Hausdorff. A minimal Hausdorff topology is such that no strictly coarser topology is Hausdorff.

The trivial topology is compact. Every topology that is coarser than a compact topology is also compact. A maximal compact topology is such that no strictly finer topology is compact.

A compact Hausdorff topology is both minimal Hausdorff and maximal compact. A minimal Hausdorff topology need not be compact and a maximal compact topology need not be Hausdorff (Steen & Seebach, Examples 99 and 100).

Question: It seems that there is some duality between Hausdorffness and compactness? Can this kind of duality be stated more explicitly (e.g. in some category theoretic formulation)? Is a compact topology on a fixed set $X$ some "dual" of a Hausdorff topology on $X$?

Here it is stated that a Hausdorff topology need not contain a minimal Hausdorff topology. If there is some duality these notions then we could also automatically deduce that a compact topology must not be contained in a maximal compact topology.

  • 1
    $\begingroup$ Considering the last paragraph: A compact topology is always contained in a maximal compact topology (Kovár, in "On Maximality of Compact Topologies" (2005), Theorem 2, drops.dagstuhl.de/opus/volltexte/2005/118). Similarly, a sequentially compact topology is always contained in a maximal sequentially compact topology (Künzi, van der Zypen, "Maximal (sequentially) compact topologies" (2003), Theorem 1, arxiv.org/abs/math/0306082v1) $\endgroup$ – yadaddy Jan 14 '16 at 12:58

There are several ways I think of expressing this 'duality'. But before describing this, maybe it would help to explain a sense in which 'existence' (at least one element, or totality of a relation) is dual to 'uniqueness' (at most one element, or well-definedness of a relation).

So let's consider the category whose objects are sets and whose morphisms are relations $R: A \to B$: let $R(a, b)$ denote the truth value of $(a, b) \in R$, and for relations $R: A \to B$ and $S: B \to C$, define the composite $SR: A \to B$ by the rule $SR(a, c) = \exists_{b: B} R(a, b) \wedge S(b, c)$. The identity relation $1_A: A \to A$ is the diagonal subset of $A \times A$, defined by $1_A(a, b) \Leftrightarrow a = b$.

Technically this "category" is a 2-category, where 2-cells are given by inclusions $R \subseteq S$ between relations of the same type $A \to B$; we will write 2-cells as $R \leq S$. Even more, we have a $\dagger$-operation which takes a relation $R: A \to B$ to its opposite $R^{op}: B \to A$, where $R(a, b) \Leftrightarrow R^{op}(b, a)$. Altogether the structure one obtains is what is known as in categorical literature as an allegory, or as a bicategory of relations: there are various axiomatic frameworks for describing categories of relations.

There are also various notions of 'duality' in such a situation. One is by reversing the direction of 1-cells (called 'op'), another is by reversing the direction of 2-cells (called 'co'), and a third is by reversing directions of both (called 'co-op').

In this 2-categorical context, we may categorically express the condition that a relation $R: A \to B$ is well-defined or functional (for all $a \in A$ there exists at most one $b \in B$ such that $R(a, b)$ is true) by the condition $R \circ R^{op} \leq 1_B$.

The co-op dual of this condition is $1_A \leq R^{op} \circ R$, which translates to saying that to each $a \in A$ there exists at least one $b \in B$ such that $R(a, b)$, or that $R$ is a total relation.

If we have both conditions $R \circ R^{op} \leq 1_B$ and $1_A \leq R^{op} \circ R$, then the relation is a function; in other words, a relation $R$ is a function iff considered as a 1-cell in the 2-category $\mathbf{Rel}$, it has a right adjoint (which is necessarily $R^{op}$; this is a good exercise).

Onto the topology: if $X$ is a topological space and $\beta X$ is the set of ultrafilters on the underlying set of $X$, then there is a convergence relation $\gamma: \beta X \to X$ where $\gamma(U, x)$ (an ultrafilter $U$ converges to $x$) means that the filter of neighborhoods of $x$ is contained in $U$. In fact the very notion of topological space can be expressed in terms of ultrafilter convergence, as explored in the notion of 'relational $\beta$-module' for which you can find an account at the nLab.

A space $X$ is compact iff every ultrafilter on $X$ converges to at least one point. This is the same as saying that the convergence relation $\gamma: \beta X \to X$ is total. A space $X$ is Hausdorff iff every ultrafilter converges to at most one point. This is the same as saying that $\gamma: \beta X \to X$ is well-defined. A space is compact Hausdorff iff its convergence relation $\gamma$ is a function: this is an important ingredient in the theorem that compact Hausdorff spaces are exactly algebras of the ultrafilter monad.


In the 2-category of sets and relations, the condition of compactness on the convergence relation of a topological space is co-op dual to the condition of Hausdorffness.

There are various other ways in which the duality between compactness and Hausdorffness manifests itself. One is that $X$ is compact if every projection map $X \times Y \to Y$ is closed, whereas $X$ is Hausdorff if the diagonal $X \to X \times X$ is closed (projections and diagonals being the two ingredients of product structures) -- although it would take some time to elaborate a sense in which these properties should be seen as "dual".

  • $\begingroup$ Thank you for such a nice insight. I am not deeply involved into category theory in order to fully grasp the $\beta$-relational module concept. I take with me that a topology on $X$ can be equivalently defined in terms of ultrafilter convergence, i.e. a topology on $X$ is a particular relation $\gamma \subseteq \beta X \times X$ which specifies convergence in $X$ by $F \to x$ iff $(F, x) \in \gamma$. (As an example, on $X = \{ 0, 1 \}$ we have 16 relations $R \subseteq \beta X \times X$ but only 4 of them are convergence relations: trivial, discrete and the two Sierpinski relations.) ... $\endgroup$ – yadaddy Nov 5 '15 at 10:50
  • $\begingroup$ ... The co-op duality is not between the topologies on $X$ as sets, resp. not between the convergence relations: we clearly cannot have a bijection between the set of compact topologies and the set of Hausdorff topologies (and with possibly fixed points being compact Hausdorff topologies). It is rather a duality between the properties of being a compact topology vs. being a Hausdorff topology (i.e. the conditions on the corresponding convergence relations are co-op dual). $\endgroup$ – yadaddy Nov 5 '15 at 10:50
  • $\begingroup$ @yadaddy It occurred to me that that's what you might have wanted: a duality either between the category of compact spaces and the category of Hausdorff spaces, or between the poset of Hausdorff topologies on a set and the poset of compact topologies (maybe on a different set). The first seems unlikely to me, and I haven't found a good opportunity yet to pursue the second possibility (or even just a Galois connection). $\endgroup$ – Todd Trimble Nov 5 '15 at 11:49
  • $\begingroup$ In addition to the co-op duality of the properties "compact" and "Hausdorff", it would be also of interest to know, whether on the lattice of topologies on a fixed set $X$ there is a "duality" between compact and Hausdorff topologies. As you mentioned, I think, we should not fix $X$ but consider two sets $X$ and $Y$ and give a duality between these. It would be good if such a duality could dualize the statement "There is a Hausdorff topology on $X$ that does not contain a minimal Hausdorff topology" to "There is a compact topology on $Y$ that is not contained in a maximal compact topology". $\endgroup$ – yadaddy Nov 5 '15 at 12:52
  • 1
    $\begingroup$ @yadaddy In the case where $X$ is finite, there is just one Hausdorff topology, but all topologies on $X$ are compact. So you can't expect such a duality (a Galois correspondence) without changing the underlying set. In fact, I think this observation rules out even such a duality where we have two sets $X, Y$ to play with (e.g. the set of compact topologies on a finite set can never be in bijection with the Hausdorff topologies on another set). $\endgroup$ – Todd Trimble Nov 5 '15 at 13:09

Hausdorff is dual to discrete. Compact is dual to overt.

A space $X$ is Hausdorff if and only if the diagonal $\Delta_X = \{(x,x) \mid x \in X\}$ is closed in $X \times X$. A space $X$ is discrete if and only if $\Delta_X$ is open in $X \times X$.

Given a space $X$ let $\mathcal{O}(X)$ be its topology, seen as a topological space equipped with the Scott topology. Let $\Sigma = \{\bot, \top\}$ be the Sierpinski space.

Now, a space $X$ is compact if and only if the map $\forall_X: \mathcal{O}(X) \to \mathcal{O}(1)$, defined by $$\forall_X (U) = \begin{cases} \top & \text{if $U = X$} \\ \bot & \text{else} \end{cases}$$ is continuous. A space is overt if and only if the map $\exists_X : \mathcal{O}(X) \to \mathcal{O}(1)$, defined by $$\exists_X (U) = \begin{cases} \top & \text{if $U \neq \emptyset$} \\ \bot & \text{else} \end{cases}$$ is continuous. See Part II of Martin Escardo's Synthetic topology of data types and classical spaces for a good tutorial on these ideas, or for a deep dive into the subject Paul Taylor's work on Abstract Stone Duality, and in particular his Foundations for Computable Topology.

  • $\begingroup$ What's an example of something that's not overt? (I'm aware that in the category of topological spaces, everything is overt, which is why I say "something") $\endgroup$ – André Henriques Nov 5 '15 at 23:26
  • 2
    $\begingroup$ Let $f : X \to S$ be a continuous map of Hausdorff spaces (or, more generally, locales). Then, following Grothendieck's relative point of view, we have a space "relative to $S$", and it is overt if and only if $f : X \to S$ is an open map. $\endgroup$ – Zhen Lin Nov 6 '15 at 0:03
  • $\begingroup$ Non-trivial overt spaces appear in computable topology, for example, as there the map $\exists_X$ needs to be computable. For instance, take $X = (0,a)$, an open interval with Euclidean topology where $a < 1$ is a real number such that for rational $q$ it is not semi-decidable whether $q < a$. If this space were overt then we could semi-decide whether $q < a$ by calculating $\exists_X((q, 1) \cap X)$. $\endgroup$ – Andrej Bauer Nov 6 '15 at 7:04
  • $\begingroup$ Excuse my ignorance (I've looked at a couple of references but could not figure out an answer): What are the objects of the category of "computable topological spaces"? Is it the case that a computable topological space is a topological space equipped with some extra structure? Is there a functor from computable topological spaces to topological spaces? Is a computable topological space some kind of algorithm, written in some language (in which case there would be presumably only countably many computable topological spaces)? $\endgroup$ – André Henriques Nov 6 '15 at 9:53
  • $\begingroup$ @AndréHenriques: there are several approaches to computable topology, but yes, in most of them a computable space is a space with an extra computable function. For instance, you could pick an enumeration of basic opens (assuming we have countably-based spaces only) and that will induce computability on points and continuous maps. Another possibiity is that you represent the computable structure of a space $X$ by a continuous (partial) surjection $\mathbb{N}^\mathbb{N} \to X$, and this will induce computability on $X$. I can give references if desired. $\endgroup$ – Andrej Bauer Nov 6 '15 at 14:44

I used to think of "compact" as dual to "discrete".

Here are two instances of compact/discrete dualities:

Pontryagin duality:
Given an abelian group $A$, its Pontryagin dual $A^*:=\mathrm{Hom}(A,S^1)$ is compact iff $A$ is discrete (and vice-versa).

Six functor formalism:
The analogue of "compact" in algebraic geometry is the concept of a proper morphism. Similarly, the analogue of "discrete" is the notion of an étale morphism. For proper maps $f:X\to Y$ we have $f_*=f_!$, whereas for étale maps we have $f^*=f^!$.

  • 3
    $\begingroup$ A space $X$ is Hausdorff if the diagonal $X \to X \times X$ is a closed embedding. It is discrete if the diagonal is an open embedding. :-) $\endgroup$ – Todd Trimble Nov 5 '15 at 18:03
  • $\begingroup$ In particular, being discrete is a kind of separation condition – a very strong one! $\endgroup$ – Zhen Lin Nov 5 '15 at 18:19
  • 6
    $\begingroup$ Actually, come to think of it, discreteness is one of the duals of compactness (as in my answer). It's the co-dual (reversing direction of 2-cells), i.e., the condition $\gamma^{op} \circ \gamma \leq 1_{\beta X}$, which says that whenever two ultrafilters $U, V$ converge to the same point, they must be the same (in other words, the only ultrafilter that can converge to a point is the principal ultrafilter at that point; this condition is equivalent to discreteness). $\endgroup$ – Todd Trimble Nov 5 '15 at 19:57
  • 3
    $\begingroup$ IMHO the six functor formalism example is more from topology than algebraic geometry. Indeed, you are not using (quasi-)coherent sheaves ($f_!$ is not defined for them). But what I really wanted to write was: sick analogies bro. $\endgroup$ – pro Nov 6 '15 at 1:00
  • $\begingroup$ @ Todd Trimble. Btw, I do know the projective Limit Theorem (due to Hochschild & Mostow) for compact T_1 spaces with closed maps> It has the desired 2 conclusions exactly as the well-known theorem for inverse limits of compact Hausdorff spaces. Is this known as it ought to be ? In fact, the main application was to linear algebraic groups with the Coset topology (or f.d vector spaces with the coset topology whose inverse limits are the linearly compact vector spaces). Thank you. $\endgroup$ – Nazih Nahlus Nov 22 '16 at 20:00

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.