John Stillwell - Naive Lie Theory.pdf - Index of

8.4 Compact sets 169
8.3.1 Show that if P is a limit point of S and f is a continuous function defined
on S and P,then f (P) is a limit point of f (S ).
8.3.2 If f is a continuous bijection, deduce from Exercise 8.3.1 that
f (closure(S )) = closure( f (S )).
8.3.3 Give examples of continuous functions f on subsets of R such that f (open)
is not open and f (closed) is not closed.
8.3.4 Also, give an example of a continuous function f on R and a set S such
that
f (closure(S )) ≠ closure( f (S )).
8.4 Compact sets
A compact set in R k is one that is closed and bounded. Compact sets
are somewhat better behaved than unbounded closed sets; for example, on
a compact set a continuous function is uniformly continuous, and a realvalued
continuous function attains a maximum and a minimum value. One
learns these results in an introductory real analysis course, but we will
prove one version of uniform continuity below. In Lie theory, compact
groups are better behaved than noncompact ones, and fortunately most of
the classical groups are compact.
We already know from Section 8.2 that O(n) and SO(n) are closed. To
see why they are compact, recall from Section 3.1 that the columns of any
A ∈ O(n) form an orthonormal basis of R n . This implies that the sum of
the squares of the entries in any column is 1, hence the sum of the squares
of all entries is n. In other words, |A| = √ n,soO(n) is a closed subset of
R n2 bounded by radius √ n.
There are similar proofs that U(n), SU(n), andSp(n) are compact.
Compactness may also be defined in terms of open sets, and hence it is
meaningful in spaces without a concept of distance. The definition is motivated
by the following classical theorem, which expresses the compactness
of the unit interval [0,1] in terms of open sets.
Heine–Borel theorem. If [0,1] is contained in a union of open intervals
U i , then the union of finitely many U i also contains [0,1].
Proof. Suppose, on the contrary, that no finite union of the U i contains
[0,1]. Then at least one of the subintervals [0,1/2] or [1/2,1] is not contained
in a finite union of U i (because if both halves are contained in the
union of finitely many U i , so is the whole).