§ Lebesgue number lemma (TODO)
for a compact space X and an open cover {Uα}, there is a radius
r>0 such that any ball of such a radius will be in some open cover: For all
x∈X, for all such balls B(x,r), there exists a Ux∈{Uα} such that
B(x,r)⊆Ux. Intuitively, pick a point x. for
each open U, we have a ball B(x,r) that sits inside it since U is open.
Find the largest such radius, we can do so since {x} is the closed subset of a compact set.
This gives us a function f that maps a point x to the largest radius of ball
that can fit in some open cover around it. This function f is a continuous function (why?)
on a compact set, and thus has a minimum. So, for all points x∈X, if you give a
ball of radius minf, I can find some open cover around it.
§ Lebesgue number lemma, Version 2:
for a compact space X and an open cover {Uα}, there is a diameter d>0 such that
set of smaller radius will be in some open cover: For all x∈X, for all
opens x∈O such that diam(O)<d, there exists a Ux∈{Uα} such that O⊆Ux.
If we can find radius B(x,r) that satisfies this, then if we are given a set of diameter less than 2r,
there will be a ball B(x,r) that contains the set O of diameter at most d=2r,
and this ball will be contained in some Ux. So we will have the
containments O⊆B(x,r)⊆Ux.
§ Lebesgue number lemma, proof from Hatcher
TODO