(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.