2.06 Hausdorffness

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



(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.


Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.