Hausdorff Topological Space

Wanted to explain what I think a Hausdorff is in my own words because maybe that is the root of the problem.

A Hausdorff Space is one in which for every x and y in X with x does not equal y, there exists an open set containing x and an open set containing y within x and the intersections of that is the empty set.

My first problem is I need to find a topology that isn't a Hausdorff, and that is the Cofinite topology I know from class, but I don't quite understand why.

Also, my second question is if I am given (X,T) is a metrizable space, how to I prove the topology (X,T) is a Hausdorff? I know the properties of a metrizable space, but not how they would apply to prove something is a topology, or even a Hausdorff topology.

Thanks!

• What do you mean by an open set containing y within x? How are you going to get any kind of set ‘within’ a point? – Brian M. Scott Oct 15 '14 at 2:41
• Brian I think that they mean "an open set containing $y$ within $X$" – user171177 Oct 15 '14 at 2:43

Firstly, if you don't know what is the induced topology related to a metric space is, then you can't answer your second question so you should first review that.

As for you definition of what a Hausdorff space is, which seems basically correct, try to visualize what is going on. Being Hausdorff means that any two distinct points can be separated by disjoint open sets. Try to draw a picture that illustrates that. As for an example of a non-Hausdorff space, you can start off by showing some very simple example of such spaces. For instance, the space $X={1,2}$ with the topology $\{X,\emptyset \}$ is not Hausdorff.

• "Being Hausdorff means that any two distinct points can be separated by disjoint open sets." I wanted to emphasize this clean definition. Many students hinder their understanding by storing mathematical definitions symbol by symbol in their memory (e.g. "for every $x$ and $y$ in $X$ with $x$ does not equal $y$"). Seek every opportunity to phrase definitions naturally in your native tongue. – Austin Mohr Oct 15 '14 at 17:13

Hints:

1) Let $\Bbb N=\{1,2,\cdots,n,\cdots\}$ with cofinite topology,i.e., $\tau=\{U \subset \Bbb N: N\setminus U \text{ is finite }\}$. Then this space $(\Bbb N, \tau)$ is not Hausdorff.

2) Suppose that $x\not=y$. Let $|x-y|=2\epsilon$. Then two open sets, $B(x,\epsilon)$ and $B(y,\epsilon)$, witnesses that the mrtric space is Hausdorff.

Part 1:

If $x,y$ are points in $X$ where $X$ has the finite complement (cofinite) topology (and is infinite) then pick $O_x$ as an open set that contains $x$ and $O_y$ for $y$. The complement $X \backslash O_x$ is finite and $O_y$ contains infinitely many points (because if it didn't then its complement wouldn't be finite) so $O_y$ can not be contained in $X \backslash O_x$ and $O_y \cap O_x$ is not empty.

A more trivial example of a space that isn't Hausdorff is the indiscrete topology on a space.

Part 2:

If $X$ is metrizable then $B_x = \{ y : T(x,y) < \epsilon\}$ is an open set (in the metric topology). So if you have two points $x,y$ such that $T(x,y) = d$ for some $d \not = 0$ then you can define open sets around each point such that no third point can be in both (using the triangle inequality).

For a very simple example of a topological space that isn't Hausdorff, consider the two-point set $X=\{x,y\}$ with the trivial topology, i.e. only $X$ and $\emptyset$ are open. Then there are no open sets that contain $x$ except $X$, so certainly the definition of Hausdorff is not satisfied.

(Note that this example actually fails all of the separation axioms, if you've seen these; Hausdorff is one of them.)

Now, if $X$ is an infinite set, then the cofinite topology fails to be Hausdorff for the following reason. Pick $x, y\in X$ with $x\neq y$. For any open set $U_x$ containing $x$, the complement of $U_x$ is finite by the definition of the cofinite topology, and similarly for any open set $U_y$ containing $y$. It follows that the complement of $U_x \cap U_y$ must also be finite, as it is the union of two finite sets. In particular, $U_x\cap U_y$ must be finite and cannot be empty.

For your final question, given two points $x,y\in X$ where the topology on $X$ is given by a metric, if $x\neq y$, then the distance $d(x,y) \neq 0$. Then the ball of radius $d(x,y)/2$ about $x$ and the ball of radius $d(x,y)/2$ are both open and do not intersect. Therefore, the topology is Hausdorff.