## How should one think about non-Hausdorff topologies?

In most basic courses on general topology, one studies mainly Hausdorff spaces and finds that they fit quite well with our geometric intuition and generally, things work "as they should" (sequences/nets have unique limits, compact sets are closed, etc.). Most topological spaces encountered in undergraduate studies are indeed Hausdorff, often even normed or metrizable. However, at some point one finds that non-Hausdorff spaces do come up in practice, e.g. the Zariski topology in algebraic geometry, the Fell topology in representation theory, the hull-kernel topology in the theory of C*-algebras, etc.

My question is: how should one think about (and work with) these topologies? I find it very difficult to think of such topological spaces as geometric objects, due to the lack of the intuitive Hausdorff axiom (and its natural consequences). With Hausdorff spaces, I often have some clear, geometric picture in my head of what I'm trying to prove and this picture gives good intuition to the problem at hand. With non-Hausdorff spaces, this geometric picture is not always helpful and in fact relying on it may lead to false results. This makes it difficult (for me, at least) to work with such topologies.

As this question is somewhat ambiguous, I guess I should make it a community wiki.

EDIT: Thanks for the replies! I got many good answers. It is unfortunate that I can accept just one.

-
Regarding the Zariski topology, I think about it as what's left if you forget "most" of the open sets that really "should" be there. This point of view makes it clear why it is not Hausdorff and it fits in both with the classical topology on complex varieties and the étale topology and its variations. – Dan Petersen Oct 29 2010 at 12:09
awesome question! – B. Bischof Oct 29 2010 at 20:57
Atleast on $\mathbb{A}^n$, I like to think of the open sets as the sets on which regular functions do not vanish. The fact that there are not enough regular functions to make the Zariski topology hausdorff just says to me that algebraic varieties are more like holomorphic manifolds than differential manifolds – Daniel Barter Oct 29 2010 at 23:39
Thank you! I've been thinking of asking essentially the same question for a while. – Darsh Ranjan Oct 30 2010 at 0:08
Cool question! How, for example, am I supposed to think about topologies (like the unitary dual of a discrete group $G$) where elements are in the closures of singletons comprised of other elements...I work with these, but am certainly not "used to" it! Is my "net" thinking off on this? Thanks for asking this! – Jon Bannon Oct 30 2010 at 14:40

For a variety of reasons, it's often useful to develop an intuition for finite topological spaces. Since the only Hausdorff finite spaces are discrete, one will have to deal with the non-Hausdorff case almost all the time.

The fact of the matter is that the category of finite spaces is equivalent to the category of finite preorders, i.e., finite sets equipped with a reflexive transitive relation. In terms of a picture, draw an arrow $x \to y$ between points $x$ and $y$ whenever $x$ belongs to the closure of $y$ (or the closure of $x$ is contained in the closure of $y$). This defines a reflexive transitive relation.

Two points $x$, $y$ have the same open neighborhoods if and only if $x \to y$ and $y \to x$. It follows that the topology is $T_0$ (the topology can distinguish points) if and only if the preorder is a poset, where antisymmetry of $\to$ is satisfied.

The closure of a point $y$ is the down-set {$x: x \to y$}, and a set is closed iff it is downward closed in the preorder. In the finite case, I believe it is true that every closed irreducible set (one that isn't the union of two proper closed subsets) is the closure of a point = principal ideal; if the point is unique, the space is called sober. Sober spaces are the kinds of spaces that arise as underlying topological spaces of schemes, and it seems to be true that a finite space is sober iff it is $T_0$.

-
I think this is a good intuitive model for e.g. the Zariski topology on Spec C[x_1, ... x_n]. One has closed points, then points corresponding to one-dimensional subvarieties whose closure contains the closed points on them, then points corresponding to two-dimensional subvarieties... – Qiaochu Yuan Oct 29 2010 at 14:25
There's even finite space models for the homotopy type of spheres :) – David Carchedi Oct 29 2010 at 17:44
One computer-sciency intuition for topology is that an open set corresponds to an observable property (i.e., a property which can be semidecided). This quickly leads to topological spaces which are also partial orders and whose open sets are upward closed. Thus Todd's answer is a good one: non-Hausdorff spaces are typically partial orders in which open sets are upward closed. We are in the land of order theory here. – Andrej Bauer Oct 29 2010 at 21:07
@Todd: it is for sure true that a finite space is sober iff it is $T_0$. Here is the guarantee: the Zariski spectrum of a commutative ring is sober (hence $T_0$), and the main theorem of Mel Hochster's thesis is that a topological space is homeomorphic to some Spec R iff it is an inverse limit of finite $T_0$ spaces. (As a proof, this is of course very complicated and very likely circular, but as a sanity check it works nicely. I presume it is no problem to give a direct proof...) – Pete L. Clark Feb 3 2011 at 13:07
@Pete: yes, thanks. A direct proof is not difficult: certainly sober spaces are $T_0$ because sobriety implies distinct points have distinct closures. On the other hand, since we have an ascending chain condition on closed sets in finite spaces, we can show that every closed irreducible is the closure of some point, and that point is unique under $T_0$, and therefore we have sobriety under $T_0$. – Todd Trimble Feb 3 2011 at 14:17

One way to get non-Hausdorff spaces from Hausdorff spaces is to take quotients under mildly bad equivalence relations.

If your non-Hausdorff space comes from such a construction, then you can think of its points as being subsets of the bigger Hausdorff topological space of which it's a quotient.

-
Well, if your non-Hausdorff space comes from such a construction, not only you can think of its points in that way: they are subsets of the bigger Hausdorff topological space of which it's a quotient :) – Mariano Suárez-Alvarez Oct 29 2010 at 14:14
I like this because it makes it clear why you can't separate points- because they aren't points, they're subspaces of Hausdorff spaces and you definitely can't always separate subspaces! – Dylan Wilson Oct 29 2010 at 14:29
Let's make this very concrete. A topological group is Hausdorff iff the identity is a closed point. Therefore if N is a normal subgroup of a topological group G, G/N in the quotient topology is Hausdorff iff N is closed. Thus if you pick a non-closed normal subgroup N, G/N will not be Hausdorff. (Any open subgroup of a top. group is closed, so a non-closed subgroup of G will not be open either.) – KConrad Oct 29 2010 at 15:36
Also, every topological space $X$ has a minimal hausdorff quotient $X \to X/R$. Thus the equivalence classes may be separated in some sense. – Martin Brandenburg Oct 29 2010 at 16:01
In fact, ALL compactly generated spaces (with no separation axiom) arise as quotients of locally compact Hausdorff spaces. In fact, this is if and only if. – David Carchedi Oct 29 2010 at 17:47

Since you referred to intuition, this is possibly not too off-topic.

In a sense, the archetype of the topologic categories is, how very elementary beings perceive the world. If I was an amoeba, I'd possibly just understand space as places close, or less close to me, not otherwise structured. I'd have no particular metric idea of my own shape; I'd just feel more or less connected, &c. So, a possible answer to your question is: like a dull amoeba.

To make an example possibly closer to us, think you're in a car in the urban traffic. Due to one-way streets, metric is not the best way to organize your perception of the space: actually, the proper topology to do that is possibly not Hausdorff (usually, you can't move to A without immediately finding yourself in B, and once you are in B, you are enormously far from A, even if you changed your mind about the opportunity of the movement.)

-

I don't know if this is relevant, but here is an easy and sometimes useful remark about spaces that have only finitely many points:

The topology is determined by the relation between points "p is in the closure of q", and this may be any transitive and reflexive relation.

This applies more generally to spaces in which the union of an arbitrary set of closed sets is closed

-
Arrgh, you beat me! I'll post anyway. – Todd Trimble Oct 29 2010 at 14:09
I just said it was easy. You took the trouble to give details. – Tom Goodwillie Oct 29 2010 at 14:20
If the relation is antisymmetric (i.e. the topology is $T_0$), your space is simply a finite poset in disguise. And it has the same weak homotopy type as the geometric realization of its nerve, which can be any finite simplicial complex. So these spaces look quite common, using homotopy glasses. – BS Oct 29 2010 at 14:22
See also this post and comments from the Secret Blogging Seminar: sbseminar.wordpress.com/2007/08/11/… – Todd Trimble Oct 29 2010 at 14:45
Just to drop a keyword for those who want to read more about such things: Alexandroff spaces. – Pete L. Clark Feb 3 2011 at 13:09

My question is: how should one think about (and work with) these topologies? I find it very difficult to think of such topological spaces as geometric objects, due to the lack of the intuitive Hausdorff axiom (and its natural consequences)

A geometric object is not just a set together with a topology. It also consists (or sometimes, a priori only consists) of a set or rather sheaf of admissible or regular functions on it. I think that these are more important than the topology. Polynomial functions often cannot separate "points", whereas continuous functions in most application can. There you get the hausdorff property, it's already in the sheaf of regular functions.

I never had any trouble concerning non-hausdorff spaces. For me, the cited little lemmas about hausdorff spaces are not natural at all. They are useful, of course, but what is so natural about the condition that all compact subsets of a topological space are closed?

Every geometry has its characteristic models and methods. When you try a translation between two geometries, you have to make sure that all its 'partial translations' are compatible with each other. In the case of manifolds vs. varieties, the translation hausdorff <-> separated has been fruitful.

-
Geometry may be hard to define, but I know it when I see it. – Tom Goodwillie Oct 29 2010 at 16:59
@Tom Goodwillie Awesome Kenneth Clarke reference!!! – B. Bischof Oct 29 2010 at 20:56
I thought it was Potter Stewart? – Todd Trimble Oct 29 2010 at 22:11
Todd, yes. And if I mean anything serious by this joke is that I would shy away from trying to insist that in order for topology to qualify as a kind of geometry it should conform to some general pattern that fits a number of other kinds of geometry. – Tom Goodwillie Oct 29 2010 at 23:53
Conversely, the fact that polynomial functions cannot separate points is one reason it took me a long time to accept that "algebraic geometry" is really a kind of geometry---so what's "intuitive" depends on where you're coming from. On the other hand, topology is arguably just the special case of geometry where functions take values in the poset $0\le 1$. – Mike Shulman Nov 1 2010 at 20:27
show 1 more comment

I'll expand upon my comment in Andre's answer. In some sense (which I am about to make precise), non-Hausdorff spaces occur when trying to "naturally" close Hausdorff spaces under colimits. Let's say the only spaces you think are "real" are compact Hausdorff spaces (this is somewhat reasonable, from certain viewpoints). But then, you might want to consider an infinite disjoint union of such spaces as still being a space, so you arrive at having to consider locally compact Hausdorff spaces. In fact, EVERY compactly generated space (not assuming any separation axioms) is the quotient of a (possibly) infinite disjoint union of compact Hausdorff spaces.

To see this: Any compactly generated space $X$ is a (possibly large) colimit of compact Hausdorff spaces. Consider the set $P(X)\O(X)$ of non-open subsets of $X$. Then for element $V$, there exists a map $p_V:T_V \to X$ from a compact Hausdorff space such that $p^{-1}\left(V\right)$ is not open. Now, the colimit of the diagram $\left(p_V:T_V \to X\right)$ is $X$. The colimit is ALMOST formed by taking a quotient of the disjoint union of each of these $T_V$s- this is true once we know that all points of $X$ are hit, but, we can fix this by adding in a bunch of constant maps to this family.

The converse, that the quotient of a sum of compact Hausdorff spaces is compactly generated is clear.

So, Andre's answer is the "total" answer, in that it includes all (compactly generated) spaces. So yes, (almost) every example of a non-Hausdorff space is really just considering points to actually be subsets of a particular Hausdorff one.

-
 Very nice! This is quite intuitive. – Qiaochu Yuan Oct 29 2010 at 20:23

Some of the non-Hausdorff topologies that turn up are actually not that hard to get an intuition for. For example, you can think of the Zariski topology on a classical algebraic variety $V$ as just being a collection of information describing all the subvarieties of $V$ (e.g., the Zariski topology on $\mathbb{A}^3_k$ describes all the algebraic curves, surfaces, and points in 3-space).

It might seem at first glance that the topologies involved get hard to understand when we move from varieties to schemes, but really the topology of a scheme is not hard to get a handle on either. The key to understanding the topologies of schemes is to understand the generic points and to understand these you just need to get some intuition about the concept of specialization and generalization.

Given two points $x,y$ in a topological space $X$, we say that $x$ is a specialization of $y$ (or that $y$ is a generalization of $x$) if $x$ is contained in the closure of $y$. What this means is that $y$ is contained in every open neighbourhood of $x$. I like to think of this as meaning that $y$ is infinitesimally close to $x$. Similarly, given a subset $F\subseteq X$ we say that a point $x\in F$ is a generic point of $F$ if $F$ is the closure of $x$. Evidently a necessary condition for such an $F$ to possess a generic point is that $F$ be a (non-empty) irreducible closed subset of $X$. It is not hard to show that in a $T_0$-space every irreducible closed subset has at most one generic point. But in fact the topology of a scheme is nicer than this: the topology of a scheme has the nice property that every (non-empty) irreducible closed subset has a unique generic point. (Such as space is called a sober space).

How should we think about this? Well, if $F$ is a closed irreducible subset of $X$ and $\xi$ is a generic point of $F$ then this means that every point of $F$ is a specialization of $\xi$; in other words $\xi$ is contained in every open neighbourhood of every point in $F$. So this generic point is infinitesimally close to all of the points in $F$. Now, in a sober space the map sending a point to its closure provides a bijection between the set of points of the space and the set of non-empty irreducible closed subsets of the space. So if you take any scheme $X$, the closed points are the points that you should think of as being the points forming a "geometric space", and all the other points are simply generic points of the various irreducible closed subsets of this space--each non-closed point describes a unique irreducible closed subset.

For example, consider the scheme version of the affine plane: $\mathbb{A}^2_k=Spec(k[X,Y])$. The subspace of closed points (i.e., the maximal ideals) is homeomorphic to the usual variety affine plane with the Zariski topology; all the other points of the scheme are just generic points describing all the subvarieties of the affine plane.

Some of this may be a bit vague or imprecise, but the point is that it isn't too hard to develop some intuition for the (non-Hausdorff) topologies arising in algebraic geometry.

-
This is very nice! – David Carchedi Oct 29 2010 at 22:27

The fundamental category of Topological spaces (without the Hausdorff property) is the "Sober" spaces category, this is strictly related to Zariski topology, and to locales and Topos (growing in generalization). See EGA1 (Grothendieck, Dieudonne, Springer), or "Stone Spaces" of P. Johnstone.

Of course Hausdorff Topological spaces are related (roughly) to a our usual way of see the geometrical spaces, in a non Hausdorff space points are related for other intrinsic (logical, geometrical, algebraic, orders) criteria, then is right that our usually intuitive point of view lack to represent them. But when we escape from Hausdorff propriety we are near to escape from "space as set of points" concept, see the concept of locales or frames ("Stone Spaces" Johnstone).

-

One can think of a topology on a space $X$ as abstracting all the "stable" information (or "physical measurements") one can say about a state $x$ in $X$ (i.e. the open neighbourhoods of $x$ in $X$).

For instance, consider the real number $\pi$ in ${\bf R}$ (with the usual topology). We can't specify $\pi$ exactly in a stable manner, because we can perturb $\pi$ a little bit and it won't be $\pi$. (In other words, $\{\pi\}$ is not open.) But we can say, for instance, that $3.14 < \pi < 3.15$, and this is a stable piece of information (it is true even if we perturb $\pi$ a little bit). The Hausdorff nature of the real line then lets us demonstrate that two quantities are distinct even if we are only allowed to access them in a stable manner. For instance, $\pi$ and $e$ can be stably shown to be distinct, because we have a stable measurement $3 < \pi < 4$ of $\pi$ and a stable measurement $2 < e < 3$ that are disjoint from each other.

Now we work instead with the Zariski topology. Here, we are not allowed to use the $<$ sign to make stable measurements (we are now in the algebraic world rather than the semi-algebraic world). The only way to make stable measurements, then, is to use the $\neq$ sign (in conjunction with the usual arithmetic operations). For instance, one can say that $\pi$ is not equal to $3$, that $\pi^2$ is not equal to $10$, and so forth. This is of course a much weaker topology. In particular, it is no longer possible to use stable measurements to stably separate $\pi$ from $e$. ($\pi$, of course, does obey the stable measurement $\pi \neq e$, and $e$ obeys the stable measurement $e \neq \pi$, but this does not help, because the stable (i.e. open) sets $\{ x: x \neq e \}$ and $\{ x: x \neq \pi\}$ are not disjoint, and so these stable measurements do not force distinctness. In more standard notation, the Zariski topology is $T_0$ but not Hausdorff.) [This has nothing to do with the transcendental nature of $\pi$ or $e$; one also fails to separate, say, $0$ and $1$, in the Zariski topology.]

[One can also take a measurement-oriented perspective to other aspects of the Zariski topology. Thus, for instance, a set $E$ is Zariski-dense if there is no way to exclude an arbitrary point $x$ from lying in $E$ using only stable measurements of $x$. As the Zariski topology is so weak, this is a fairly weak property; there are a lot of Zariski-dense sets.]

In general, non-Hausdorff topologies are usually extremely weak topologies, in which there are very few stable measurements available and so it is hard to stably separate distinct points from each other. The most extreme case is the trivial topology, in which no non-trivial measurements are available at all.

-
 In a related direction, there is a connection between modal logic and topology, first developed by Tarski and McKinsey. To give a flavor of this: imagine that the points of the real line are continuously assigned colors in the spectrum, and you ask someone to identify which points are definitely green. There may be "grue" points on the boundary between green and blue points, but the 'definitely green' points tend to form an open set. This is related to the "necessarily true" operator used in modal logic. (This comment isn't directed at you particularly, Terry; it's more a FWIW comment.) – Todd Trimble Jun 10 2011 at 16:48