## Counter-Examples: The Particular Point Topology.

### November 3, 2010

I just used this counter-example, so I felt like I should share it with all of you guys.

The particular point topology is defined in the following way: given some space , we let be a distinguished (or particular) point. It can be any point, really. Then we let a set be open if it is the empty set, or if it contains . Convince yourself that this is, in fact, a topology by going over the definition of a topology.

Below, I made a little drawing of a five point space. The red rings are the open sets with exactly two elements, the green dot is , and we note that the set is also open. I’ve drawn one three-element set which is open, and there are a few more. There are also a few four-element open sets, and, of course, the entire space.

A good easy exercise for the reader is, given a finite space of size , then prove there are open sets in the topology.

If we think of the real line, and the origin as our distinguished, there are a ton of open sets. Any set that contains the origin will be open! Crazy!

Let’s also note that generally we want to think of the space as having more than two points so that nothing really trivial happens. Below are some properties if the space has more than two points in it.

Is such a space with this topology going to be Hausdorff? Given are there two open sets separating them? No, every open set has in common. Is it regular? No, same reasoning. Is it normal? No, same reasoning.

There are two really cool properties about these spaces though, which I want to discuss right now. These are the two I keep in mind.

**If** ** is a space with the particular point topology and** ** has more than two points, then:**

**The interior of every closed set is empty.****Depending on the space, the closure of a compact set may not be compact (!).**

Let’s just show these things quickly. The first one is not too bad: the compliment of an open set is just going to be a set not containing , and so if a point is in the interior of this set, it has an open neighborhood around it completely contained in the closed set. Unfortunately, every open set contains and so it can never be contained in our closed set.

The second one is really the one that I love this space for. One thing that I hear a bunch is, "If is compact, then the closure of is just ." This probably stems from the idea that in Euclidean space a set is compact if and only if it is closed and bounded. Let’s suppose we’re given ${\mathbb R}$ with as the distinguished point. Given the compact set (which is compact since any one non-empty open set covers it). What is the closure of this? Note that every non-distinguished point of the space is a limit point (since every open set containing will also contain ) and so the closure is actually all of .

You might get a bit ahead of yourself and say, "Oh, nice, we’re done because is not compact." but we actually don’t know that yet. It may be compact in this topology! Luckily for this post, it is not. In particular, the open cover has no finite subcover, as you can check. Thus, we have shown the closure of a compact set is not necessarily compact.

What did we actually need in that last part? All we really needed was for our space to have more than finitely many points: any space which is countable or higher would work.