Download Math 641 Lecture #9 ¶2.3,2.4,2.5,2.6,2.7,2.8 Topological Preliminaries

Math 641 Lecture #9
¶2.3,2.4,2.5,2.6,2.7,2.8
Topological Preliminaries
Definitions (2.3). Let X be a topological space.
a) If, for any two distinct points p, q ∈ X, there are open sets U, V such that p ∈ U, q ∈ V ,
and U ∩ V = ∅, then X is Hausdorff.
b) If, for every p ∈ X, there is an open U such that p ∈ U and U (the closure of U :
the smallest closed set containing U ), is compact, (every open cover contains a finite
subcover), then X is locally compact.
Theorem (2.4). If K is compact in X, F is closed in X, and F ⊂ K, then F is compact.
Proof: Let {Vα } be an open cover of F .
S
Then K ⊂ F C ∪ ( α Vα ) = X.
Hence, there is a finite subcover {Vαi } such that
K ⊂ F C ∪ Vα1 ∪ · · · ∪ Vαn .
Thus F ⊂ Vα1 ∪ · · · ∪ Vαn since F ⊂ K.
Note. In this proof, we used the fact that F was closed when we included F C in the
open cover of K.
Corollary: If A ⊂ B and B has compact closure (i.e., B is compact), then A has
compact closure.
Proof: Since B is compact, and A ⊂ B, it follows that A ⊂ B.
Take F = A and K = B in Theorem (2.4).
Theorem (2.5). If X is Hausdorff, K is compact in X, and p ∈ K c , then there are open
sets U, W such that p ∈ U , K ⊂ W , and U ∩ W = ∅ (separation of compact subsets from
points “outside”).
Proof: For each q ∈ K there exist disjoint open sets Uq and Vq such that p ∈ Uq and
q ∈ Vq , i.e., {Vq } is an open cover of K.
Since K is compact, there are finitely many points q1 , . . . , qn ∈ K such that
K ⊂ Vq1 ∪ · · · ∪ Vqn = W.
Then p ∈ Uq1 ∩ · · · ∩ Uqn = U and U ∩ W = ∅.
Corollaries. (a) Every compact subset of a Hausdorff space is closed.
(b) If X is Hausdorff, F closed in X, and K compact in X, then F ∩ K is compact.
Proof: (a) By Theorem 2.5, for each p ∈ K c there is an open Up such that p ∈ Up and
Up ∩ K = ∅.
S
We then have that K c = p∈K c Up is open, and so K is closed.
(b) The compactness of K compact implies that K is closed by (a).
Since F closed, then F ∩ K is closed.
Thus F ∩ K ⊂ K implies that F ∩ K is compact by Theorem 2.4.
Theorem (2.6). If X is Hausdorff
T and {Kα } is a collection (possibly uncountable) of
compact subsets of X such that
empty intersection.
α
Kα = ∅, then some finite subcollection of {Kα } has
Proof: Set Vα = Kαc .
Each Vα is open because Kα is compact and X is Hausdorff (see Corollary (a) above).
Fix a member K1 of {Kα }.
No point of K1 belongs to every Kα because
T
α
Kα = ∅.
So each q ∈ K1 belongs to at least one Vα .
Thus {Vα } is an open cover of K1 .
By the compactness of K1 , there is a finite collection Vα1 , . . . , Vαn such that
K1 ⊂ Vα1 ∪ · · · ∪ Vαn .
This implies that
∅ = K1 ∩ (Vα1 ∪ · · · ∪ Vαn )c = K1 ∩ Kα1 ∩ · · · ∩ Kαn .
This completes the proof.
Theorem (2.7): If X is a locally compact Hausdorff space (LCH for short), U open
in X, and K a compact subset of U , then there is an open V such that V is compact
(i.e., V has compact closure) and K ⊂ V ⊂ V ⊂ U .
Proof: By local compactness, each point q ∈ K is contained in an open Vq for which Vq
is compact.
The collection {Vq : q ∈ K} is an open covering of K, so there is a finite subcover
{Vqi : i = 1, . . . , n} of K.
Let G = Vq1 ∪ · · · ∪ Vqn ; so K ⊂ G.
Since each Vqi is compact, G is compact (i.e., every open cover of G is an open cover of
each Vqi ; so there is a finite subcover for each Vqi ; the finite union of these finite subcovers
is a finite subcover of G; therefore G is compact).
If U = X, then choose V = G.
It follows that K ⊂ V ⊂ V ⊂ U .
Otherwise, if U 6= X, then set C = U c (6= ∅).
Since K c ⊃ C (recall K ⊂ U ), there are for each p ∈ C open sets Sp , Wp such that
p ∈ Sp , K ⊂ Wp and Sp ∩ Wp = ∅ by Theorem 2.5.
If p ∈ Wp , then p ∈ ∂Wp ≡ Wp ∩ Wpc , (Wpc is closed), while p ∈ Sp ⊂ int(Wpc ) since
Sp ∩ Wp = ∅ (Sp open, inside Wpc , so p ∈ int(Wpc )), which contradicts (intWpc ) ∩ ∂Wp = ∅.
Thus p ∈
/ Wp for every p ∈ C.
This implies that {C ∩ G ∩ Wp : p ∈ C} has empty intersection, so that by Theorem 2.6,
there is a finite subcollection {C ∩ G ∩ Wpi : i = 1, . . . , n} with empty intersection.
Choose V = G ∩ Wp1 ∩ · · · ∩ Wpn .
Then K ⊂ V because K ⊂ G and K ⊂ Wpi .
And V ⊂ U because
V ∩ (U c ) = G ∩ Wp1 ∩ · · · ∩ Wpn ∩ C
⊂ G ∩ Wp1 ∩ · · · ∩ Wpn ∩ C
= ∅.
This completes the proof.
Question. Why ⊂ and not = in the box in the proof above?
Definitions (2.8). Let f : X → [−∞, ∞] for a topological space X.
If {x : f (x) > α} is open for all α ∈ R, then f is lower semicontinuous (or lsc).
If {x : f (x) < α} is open for all α ∈ R, then f is upper semicontinuous (or usc).
Examples.
a) χE is lsc if E is open:


X
{x : χE (x) > α} = E


∅
if α < 0,
if 0 ≤ α < 1,
if α ≥ 1.
b) χE is usc if E is closed:


∅
{x : χE (x) < α} = E C


X
if α ≤ 0,
if 0 < α ≤ 1,
if α > 1.
Elementary Properties.
a) An f : X → [−∞, ∞] is continuous if and only if f is lsc and usc.
If f continuous, then f −1 ([−∞, α)) and f −1 ((α, ∞]) are open for all α ∈ R.
If f is lsc and usc, then f −1 ([−∞, α)) and f −1 ((α, ∞]) are open for all α, hence f −1 ((α, β)) =
f −1 ([−∞, β) ∩ (α, ∞]) is open too.
b) If fn : X → [−∞, ∞], n = 1, 2, 3, . . ., are lsc, then supn fn is lsc.
Recall that if g = supn fn , then
−1
g ((α, ∞]) =
∞
[
fn−1 ((α, ∞]).
n=1
(We saw this in proof that supn fn measurable if each fn is.)
c) If fn : X → [−∞, ∞], n = 1, 2, 3, . . ., are usc, then inf n fn is usc.
If g = inf n fn , then
g −1 ([−∞, α)) = {x : g(x) < α}
= {x : inf fn < α}
n
= {x : fn (x) < α for some n}
∞
[
=
fn−1 ([−∞, α)).
n=1
Remark. Actually b) and c) hold for arbitrary collections, even uncountable ones.