John Stillwell - Naive Lie Theory.pdf - Index of

170 8 Topology
Pick, say, the leftmost of the two intervals [0,1/2] and [1/2,1] not contained
in a finite union of U i and divide it into halves similarly. By the same
argument, one of the new subintervals is not contained in a finite union of
the U i , and so on.
By repeating this argument indefinitely, we get an infinite sequence of
intervals [0,1]=I 1 ⊃ I 2 ⊃ I 3 ⊃···. Each I n+1 is half the length of I n
and none of them is contained in the union of finitely many U i .Butthere
is a single point P in all the I n (namely the common limit of their left and
right endpoints), and P ∈ [0,1] so P is in some U j .
This is a contradiction, because a sufficiently small I n containing P is
contained in U j ,sinceU j is open. So in fact [0,1] is contained in the union
of finitely many U i .
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The general definition of compactness motivated by this theorem is the
following. A set K is called compact if, for any collection of open sets
O i whose union contains K, there is a finite subcollection O 1 ,O 2 ,...,O m
whose union contains K . The collection of sets O i issaidtobean“open
cover” of K , and the subcollection O 1 ,O 2 ,...,O m is said to be a “finite
subcover,” so the defining property of compactness is often expressed as
“any open cover contains a finite subcover.”
The argument used to prove the Heine–Borel theorem is known as the
“bisection argument,” and it easily generalizes to a “2 k -section argument”
in R k , proving that any closed bounded set has the finite subcover property.
For example, given a closed, bounded set K in R 2 ,wetakeasquare
that contains K and consider the subsets of K obtained by dividing the
square into four equal subsquares, then dividing the subsquares, and so on.
If K has no finite subcover, then the same is true of a nested sequence of
subsets with a single common point P, which leads to a contradiction as in
the proof for [0,1].
Exercises
The bisection argument is also effective in another classical theorem about the
unit interval: the Bolzano–Weierstrass theorem, which states that any infinite set
of points {P 1 ,P 2 ,P 3 ,...} in [0,1] has a limit point.
8.4.1 Given an infinite set of points {P 1 ,P 2 ,P 3 ,...} in [0,1], conclude that at least
one of the subintervals [0,1/2], [1/2,1] contains infinitely many of the P i .
8.4.2 (Bolzano–Weierstrass). By repeated bisection, show that there is a point P
in [0,1], every neighborhood of which contains some of the points P i .