Page 75 - DMTH401_REAL ANALYSIS

Unit 4: Compactness

Notes
T  S =  m = yi . Finite union, each T  V can be covered by finite subfamily of U, so T  S can
j 1 T V
yj
be covered by finite subfamily of U.
4.5 Compactness and Continuity

Proposition: Cts image of compact space compact.
–1
Proof: f: T S cts, T compact. U open cover of f(T). f (U) open " U  U.
–1
–1
Cover T since " x  T f(x) in some U  U. Hence  f (U ),... , f (U ) subcover of T. " y  f(T) have
1 n
y = f(x) where x  T so x  f (U ) for some i so y  U . Hence U ,... ,U .
–1
i i 1 n
Theorem: Cts bijection of compact T onto Hausdorff S is homeomorphism.
Proof: U open in T, T\U closed so compact.
Therefore (f ) (U) = f(U) = S\f(T\U) open, so f cts.
–1
–1 –1
Corollary: Let T be compact. Cts f : T   is bdd and attains max and min.
Proof: f(T) compact so closed.

Then sup f(T)  f(T) = f(T).

–1
Alternative proof: Let c = sup T f(x). If f not attain c then f(x) < c" x so {x : f(x) < r} = f (–¥, a) where
x
n
r < c s.t. T   i 1 {x : f(x) r }< i . Then f(x) < max {r ,..., r } " x so c = sup xT f(x)  max {r ,..., r } < c
=
1
1
n
n
Contradiction.
Definition: Given cover U of metric M,  > 0 called Lebesgue number of U if " x M U U s.t.
B(x, )  U.
Proposition: Every open cover U of compact metric space has a Lebesgue number.
Proof: " x M pick r(x) > 0 s.t. B(x, r(x)) contained in some set of U. Then M  x M B ( x, r(x) ) so x ,.

2
1
min {r(x ),...,r(x )}
. . , x s.t. M  j i 1 B ( x , r( i x ) ) . Let  = 1 j . Then " x  M pick i s.t. x B( x , r( i x ) ) and
j = i 2 i 2
2
B(x, )  B (x , r(x )) subset of some set from U.
i i
Theorem: Cts map of compact metric M to metric N is uniformly cts.
e
–1
Proof: Let  > 0. Then sets U = f (B (f(z), )) z  M open cover of M. Let be Lebesgue number.
z N
2
If x, y  M, d (x, y) <  y  B(x, )  U some z so d (f(x), f(y))  d (f(x), z) + d (f(y), z) < e.
M z N N N
4.6 Compact Sets in  n
n
Theorem: (Heine-Borel). A   compact if f closed and bdd.
Proof: () Metric spaces are Hausdorff, so A closed.
n
n
() C   bdd  [a, b]   s.t. C  [a, b]  . . .  [a, b]. This compact by Tychanov. If C closed
then closed subset of compact space so compact.

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