Is every metric space a hausdorff space too?

[AdrianZ]

I've encountered the term Hausdorff space in an introductory book about Topology. I was thinking how a topological space can be non-Hausdorff because I believe every metric space must be Hausdorff and metric spaces are the only topological spaces that I'm familiar with. my argument is, take two distinct points of a topological space like p and q and choose two neighborhoods each containing one of the two points with the radius d(p,q)/2. Now I claim that these two neighborhoods must be disjoint. Suppose that they are not disjoint, therefore there must exists a z such that z lies in both neighborhoods. using the triangle inequality, I can say that d(p,q)<=d(p,z)+d(z,q). since z is in both neighborhoods we have: d(p,z)<d(p,q)/2 and d(z,q)<d(p,q)/2. using the triangle inequality again we obtain: d(p,q)<=d(p,z)+d(z,q)<d(p,q)/2 + d(p,q)/2 = d(p,q) which is a contradiction. therefore every metric space must be Hausdorff. is this a correct argument?

Now, How can we find a non-Hausdorff space? such a space must be so interesting because we can't distinguish between two points intuitively. am I right? Is there any famous examples of a non-Hausdorff space that can be visualized or understood intuitively?

 

[micromass]

I've encountered the term Hausdorff space in an introductory book about Topology. I was thinking how a topological space can be non-Hausdorff because I believe every metric space must be Hausdorff and metric spaces are the only topological spaces that I'm familiar with. my argument is, take two distinct points of a topological space like p and q and choose two neighborhoods each containing one of the two points with the radius d(p,q)/2. Now I claim that these two neighborhoods must be disjoint. Suppose that they are not disjoint, therefore there must exists a z such that z lies in both neighborhoods. using the triangle inequality, I can say that d(p,q)<=d(p,z)+d(z,q). since z is in both neighborhoods we have: d(p,z)<d(p,q)/2 and d(z,q)<d(p,q)/2. using the triangle inequality again we obtain: d(p,q)<=d(p,z)+d(z,q)<d(p,q)/2 + d(p,q)/2 = d(p,q) which is a contradiction. therefore every metric space must be Hausdorff. is this a correct argument?

Now, How can we find a non-Hausdorff space? such a space must be so interesting because we can't distinguish between two points intuitively. am I right? Is there any famous examples of a non-Hausdorff space that can be visualized or understood intuitively?

Yes, your proof is absolutely correct! A metric space is indeed a Hausdorff space (and it's even much more than that!).

Non-Hausdorff spaces can be interesting or boring. Here are some examples:
- Take an arbitrary set X and define $\emptyset$ and X as its only open sets. Then if $X\geq 2$, then it's non-Hausdorff. This is called the indiscrete or trivial topology.
- Take the set {0,1} and define $\emptyset$, {0} and {0,1} as its only open sets. This is non-Hausdorff. It is called the Sierpinski topology
- The most interesting class of non-Hausdorff spaces are the affine schemes with the Zariski topology. But these spaces are too complicated to be able to give a nice description here.

 

[micromass]

A much more interesting topology is to take an arbitrary infinite set X, and define

$$\mathcal{T}=\{A\subseteq X~\vert~X\setminus A~\text{is finite or entire}~X\}$$

this is called the cofinite topology (and it's very close to being the Zariski-topology on a scheme). As you can see, it's open sets are verry large. And this is typical for non-Hausdorff spaces. These spaces intuitively have very large open sets which fail to distinguish between points.

 

[AdrianZ]

I see. thanks for the examples.
I don't understand this one:

Then if X≥2, then it's non-Hausdorff.

what does X≥2 mean here? Can't we equip the trivial topology with the discrete metric? I mean the function that gives 0 when p=q and gives 1 otherwise.

Are non-Hausdorff spaces interesting only for pure mathematicians or there are applications for those spaces as well? Is there any geometrical application for non-Hausdorff spaces?

 

[micromass]

I see. thanks for the examples.
I don't understand this one:

what does X≥2 mean here? Can't we equip the trivial topology with the discrete metric? I mean the function that gives 0 when p=q and gives 1 otherwise.

No, that would induce the discrete topology. That is the topology in which all sets are open. Here we're talking about the indiscrete topology where only $\emptyset$ and X are open. So there are only 2 open sets!

With $X\geq 2$, I meant $|X|\geq 2$ (so the number of elements in X is 2 or more). I made a typo

Are non-Hausdorff spaces interesting only for pure mathematicians or there are applications for those spaces as well? Is there any geometrical application for non-Hausdorff spaces?

They are being studied in algebraic geometry. And they are very important in algebraic geometry! Algebraic geometry is being used in physics and string theory. So I guess that non-Hausdorff spaces have applications outside of mathematics as well.
Another famous non-Hausdorff topology is the Scott topology. This topology is used in computer science a lot!

 

[AdrianZ]

No, that would induce the discrete topology. That is the topology in which all sets are open. Here we're talking about the indiscrete topology where only $\emptyset$ and X are open. So there are only 2 open sets!

With $X\geq 2$, I meant $|X|\geq 2$ (so the number of elements in X is 2 or more). I made a typo

I see. How do you say that the discrete metric induces the topology in which all sets are open? you're using the definition of an open set that every point of it must be internal?

They are being studied in algebraic geometry. And they are very important in algebraic geometry! Algebraic geometry is being used in physics and string theory. So I guess that non-Hausdorff spaces have applications outside of mathematics as well.
Another famous non-Hausdorff topology is the Scott topology. This topology is used in computer science a lot!

Sounds quite interesting. looking forward to studying them one day soon what prerequisites do I need to know to understand algebraic geometry? I already have read baby Herstein and have watched Gilbert Strang's linear algebra lectures on MIT and Harvard's abstract algebra lectures. Is that enough to understand a bit of algebraic geometry or I still need to learn a lot before I try to read a book about algebraic geometry?

 

[micromass]

I see. How do you say that the discrete metric induces the topology in which all sets are open? you're using the definition of an open set that every point of it must be internal?

The easiest way to see this is to look at the open balls. Let x be an arbitrary point, then B(x,1/2)={x}. This means that all the singletons are open. But arbitrary unions of open sets are open, thus this means that all sets are open (since all sets are the union of its singletons).

Sounds quite interesting. looking forward to studying them one day soon what prerequisites do I need to know to understand algebraic geometry? I already have read baby Herstein and have watched Gilbert Strang's linear algebra lectures on MIT and Harvard's abstract algebra lectures. Is that enough to understand a bit of algebraic geometry or I still need to learn a lot before I try to read a book about algebraic geometry?

You need to know abstract algebra very well. Certainly ring-theory. Be acquainted with prime ideals, maximal ideal, noetherian rings, etc. Herstein is a good book, and I also recommend Artin's algebra.

If you're comfortable with ring theory (and a bit of module theory), then you can study algebraic geometry. The absolute best books for undergraduates is (according to me) "Lectures on Curves, Surfaces and Projective Varieties" by Beltrametti. It assumes a knowledge of projective geometry, ring theory and a bit of topology.

Another great resource, but a bit more advanced are Milne's lecture notes: http://www.jmilne.org/math/CourseNotes/ag.html

 

[felper]

The converse is useful to verify if some topological space is metrizable: if it's not a Hausdorff space, it can't be metrizable.