# 2.06 Hausdorffness

## Video

Below the video you will find accompanying notes and some pre-class questions.

## Notes

### Definition

(0.24) A topological space $$X$$ is Hausdorff if, for all $$x,y\in X$$ with $$x\neq y$$, there exist open sets $$U,V\subset X$$ such that $$x\in U$$, $$y\in V$$ and $$U\cap V=\emptyset$$. That is, any two points can be separated by open balls.

(1.12) Any metric space is Hausdorff: if $$x\neq y$$ then $$d:=d(x,y)>0$$ and the open balls $$B_{d/2}(x)$$ and $$B_{d/2}(y)$$ are disjoint. To see this, note that if $$z\in B_{d/2}(x)$$ then $$d(z,y)+d(x,z)\geq d(x,y)=d$$ (by the triangle inequality) and $$d/2>d(x,z)$$, so $$d(z,y)>d/2$$ and $$z\not\in B_{d/2}(y)$$.

(3.57) We don't yet have a construction of topological spaces which will allow us to construct a non-Hausdorff spaces: we will see such examples when we meet the quotient topology later.

### Properties of Hausdorff spaces

(4.36) If $$X$$ is Hausdorff and $$x_n$$ is a sequence of points in $$X$$ such that $$x_n$$ converges to $$x$$ and $$x_n$$ converges to $$y$$, then $$x=y$$ (so limits are unique).

(5.27) What does convergence mean for a sequence in a topological space? It means that for any open set $$U$$ containing $$x$$, the points $$x_n$$ are in $$U$$ for all sufficiently large $$n$$ (this recovers the usual metric notion of convergence if you take $$U$$ to be a sequence of balls of smaller and smaller radius going to zero).

(6.40) Suppose that we have a sequence $$x_n$$ such that $$x_n\to x$$ and let $$y$$ be a point with $$x\neq y$$. By the Hausdorff assumption, there are disjoint open sets $$U\ni x$$ and $$V\ni y$$. Because $$x_n\to x$$ there exists an $$N$$ such that $$x_n\in U$$ for all $$n\geq N$$. Therefore $$x_n\not\in V$$ for all $$n\geq N$$, and hence $$x_n$$ does not converge to $$y$$.

(8.26) A compact subset $$K$$ of a Hausdorff space $$X$$ is closed.

We need to show that $$X\setminus K$$ is open. If $$X\setminus K=\emptyset$$ then it's open, so assume that it is nonempty. Pick a point $$y\in X\setminus K$$. For each $$x\in K$$ there is a ball $$U_x\ni x$$ and $$V_x\ni y$$ such that $$U_x\cap V_x=\emptyset$$ (by the Hausdorff assumption). We would like to say $$\bigcap_{x\in K} V_x$$ is an open neighbourhood of $$Y$$ disjoint from $$K$$. Certainly it is disjoint from $$K$$ (otherwise there is some point $$x\in V_x$$, but $$V_x\cap U_x=\emptyset$$ and $$x\in U_x$$) but it might not be open because it could be an infinite intersection.

(11.11) As $$K$$ is compact, there is a finite collection $$\{x_i\ :\ i\in I\}$$ (for a finite set $$I$$) such that $$\{U_{x_i}\ :\ i\in I\}$$ covers $$K$$. Now the intersection $$V=\bigcap_{i\in I}V_{x_i}$$ is open (as it's a finite intersection). Moreover $$V\cap K=\emptyset$$.

(13.30) Therefore $$K$$ is closed (because its complement is a union of open sets $$V$$ like we just constructed, hence open).

## Pre-class questions

1. In the proof of the final lemma of the video, I did a lousy job of explaining why $$\bigcap_{i\in I}V_{x_i}\cap K=\emptyset$$. Think about why this is true and come up with your own explanation.