The Bolzano-Weierstrass theorem:
The interval I = [0,1] is (sequentially) compact.

Pf:
- It is complete and totally bounded.
- Totally bounded =>
  every sequence has a Cauchy subsequence
- Complete =>
  every Cauchy sequence converges

Why is it totally bounded?
B/c it's 1-dimensional.
Given a subset A of a k-dimensional space,
of diameter d, you can cover it with n^k pieces of diameter \sqrt{k}d/n.
Pf: duh. the box [min(i-th coord), max(i-th coord)] covers it
and has sides at most d.
Chop each dimension into k pieces.
[NB: you only need to do this in 1 dimension:
 compactness is closed under products
 -- but it's geometrically easier to see.]

Again (like irrationality of sqrt(2)), this is -precisely-
what you get back analyzing the slick proof:
each step becomes an axiom.
Indeed, that's how you should see it:
- come up with a clever proof
- see what properties you needed
  -for this proof to work-
- make these properties axioms,
  as they're likely interesting)
  (at least, these kinda proofs may work)

Cor:
$A \subset \bR^n$ is compact iff it is closed and bounded
Pf:
Compact => closed \& bounded obvious
Closed -> complete, bounded -> totally bounded

