# 2.06 Hausdorffness

## Video

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

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## 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

- 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.

## Navigation

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**Compactness**. - Next video:
**Homeomorphisms**. - Index of all lectures.