What application is there for a non-Hausdorff topological space?

I'm learning basic topology and as I understand it, a good way to intuit what an open set is, is that it determines which elements are near each other. However, in a non-Hausdorff space, it would be possible for one point to be "near" another, without the reverse being true. For example, if $X=\{a,b\}$ and the typology $T=\{\emptyset, \{a\},X\}$ then "getting close to $b$" implies getting close to $a$, but not the reverse. You can get so close to $a$ that you are no longer close to $b$.

I know that these imprecise English words like "close" and "neighborhood" shouldn't be regarded as too strongly related to topological definitions, but I'm still wondering if it's possible to give an intuitive conception of what open sets "mean" or what they are, in a non-Hausdorff space.

[Edit: Giving examples of applications, particularly ones accessible to a person who has an beginner's acquaintance with topology, would be appreciated since they can illustrate the meaning of the concept.]

• Is your question the one in the title (applications of spaces which are not Hausdorff) or the one in the body (intuition about what an open set really is) ? – Dan Rust Jun 17 '14 at 14:12
• @Daniel Rust, Either, really. If one can say what the concept is in an intuitively meaningful way, that would be great. Also, knowing applications would help me to know what this concept is. – Addem Jun 17 '14 at 14:14
• I think those two are very different questions. If I told you that the Zariski topology is useful in algebraic geometry, would that give you an intuition of what an open set is? – Dan Rust Jun 17 '14 at 14:18
• I think that Zariski topology in algebraic geometry is non-Hausdorf and is quite important: closed sets are zeros of polynomials. In that context, "closed" is more intuitive than "open"; you only have very few closed sets. – Peter Franek Jun 17 '14 at 15:26
• There is a great explanation of why the Zariski topology is useful and intuitive despite being non-Hausdorff at math.stackexchange.com/questions/161884/why-zariski-topology. – Gyu Eun Lee Jun 17 '14 at 16:04

I don't think looking for an intuition is the way to understand this. The point of abstract topological spaces is not that they are intuitive; it is the opposite: it is to understand certain aspects of the behavior of spaces even in spaces that are so strange that your intuition does not work.

Your intuition for topological spaces comes from (or should come from) $\Bbb R^n$, which is the canonical example. The axioms of topological spaces are intended to abstract certain very general properties of open sets in $\Bbb R^n$ so that we can handle those properties in more general settings. So for example we can consider the space of all distance-preserving origin-fixing linear transformations of the plane and observe that this space has two connected components, and suddenly that pulls in a whole pile of related results that we have proved about disconnected spaces in general, without our having to have a mental picture or an intuition about it, and without having to prove them all over again for this particular space.

Here is another example. You mentioned the Sierpiński space $S = \{\top,\bot\}$ whose topology is $\{S, \{\top\}, \emptyset\}$. There is an important notion in computability theory of “recursive enumerability”: a recursively enumerable set (“RE set”) is (roughly) one whose values can be listed by some automatic process; perhaps you can imagine why this might be important.

It transpires that if $X$ is some space of values then a subset $Y\subset X$ is RE if and only if its characteristic function $\chi_Y:X\to S$ is continuous, where the characteristic function is $$\chi_Y(x) = \begin{cases} \top\quad\text{if x\in Y} \\ \bot\quad\text{otherwise} \end{cases}$$

If you formulate the property of recursive enumerability in this way, you instantly get a huge amount of information about RE sets, essentially for free, all imported from the huge body of knowledge that already exists about continuous functions. For example, a finite intersection of RE sets is also RE, but an infinite intersection need not be; this is exactly analogous to the topological theorem that finite intersections of open subsets of $X$ are open, but infinite intersections need not be. The purpose here isn't to try to develop intuition about how $\{\top\}$ could be open while $\{\bot\}$ is closed. There is no intuition for that; it's just a fact. Instead, the purpose is to apply existing theory to a new set of cases.

Von Neumann is famously supposed to have said that “in mathematics, you don't understand things, you just get used to them”. I'm not sure exactly what he meant, but I think this might be a good example. You can understand the topology of $\Bbb R^2$. But I think the topology of the Sorgenfrey plane is something you get used to, not something you understand. (Even though it is Hausdorff!)

• Right, I understand that just following the Math can be important. I was just hoping to gain greater understanding, if possible, from some intuitions. But perhaps here there are no intuitions to have. – Addem Jun 17 '14 at 17:07
• The intuition should be gained from the motivation - Metric spaces. One of course then needs to be careful when working with non-metrisable topological spaces. – Dan Rust Jun 17 '14 at 17:15
• To put this all another way: to gain intuition, study examples. – Lee Mosher Jun 17 '14 at 17:19
• Great answer. But..."The point of abstract topological spaces is not that they are intuitive" I beg to differ. There is very good intuition for these things. Topologies can be seen as characterising sets of states of a system compatible with a measurement. I explain a bit about it here: mathoverflow.net/a/19156/1233 IMO, in the very place where you say "There is no intuition for that" you've just made a great start on explaining the intuition :-) – Dan Piponi Jun 17 '14 at 20:39
• @Dan, tou have no firmer admirer than I, and I remember reading your explanation of this idea when it first appeared on your blog. But I still don't think it provides much useful insight in understanding, say, the Sorgenfrey plane. – MJD Jun 17 '14 at 22:23

Consider the closed unit sphere in 3-dim space and the function $f$ from the sphere to the 2-point space $\{a,b\}$ sending the interior of the sphere to $a$ and the boundary of the sphere into $b$. Under which topology on the target space is this function continuous? Well, since $f^{-1}(a)$ is an open subset of the sphere then we want $\{a\}$ to be open and this is the only requirement. The non Hausdorff topology allows you to treat this identification map as a continuous function between two topological spaces, allowing for the whole machinery to apply. Or, in other terms, allows you to think at the sphere, for all purposes in which inner points may be identified one with the other and boundary points may be identified one with the other, as a 2-point space with non Hausdorff topology.

A typical application of this arises each time you have a topological space $X$ and an equivalence relation ${\cal R}$ on it. Then you have a canonical way of endowing the set of equivalence classes with a topology. So you can say when two equivalence classes are "close to each other"... Take the most typical example in undergraduate geometry: classification of real conics under affine transformations. Since affine classes of conics are classified by rank and signature of the corresponding symmetric matrix you end up with a finite non Hausdorff topological space. Its topology tells you exactly which kind of transition from one conic to another may happen when you have a continous family of conics, so for example governs transitions of conic type inside linear families of conics.

As was said above group actions falls into this situation, and non Hausdorff spaces typically arises as spaces of orbits under group action.