Download MATH 202A - Problem Set 9

MATH 202A - Problem Set 9
Walid Krichene (23265217)
November 3, 2012
(9.1) A collection of sets {Eα }α∈A has the countable intersection property if for every countable subset
A0 ⊆ A, ∩α∈A0 Eα is non empty.
Let X be a topological space, and suppose that X is Lindelof. Let {Eα }α∈A be a collection of closed
subsets that has the countable intersection property. Then ∩α∈A Eα 6= ∅
proof We show the contrapositive: Assume that ∩α∈A Eα = ∅. Then for all x ∈ X, there exists αx ∈ A
such that x ∈
/ Eαx , i.e. x ∈ Eαc x . This defines an open cover of X: {Eαc x }x∈X . Since X is Lindelof, this open
cover has a countable subcover: there exists a sequence (xn )n of elements of X such that X = ∪i∈N Eαc x .
i
Taking the complement, we have ∩i∈N Eαxi = ∅. Therefore {Eα }α∈A does not have the countable intersection
property.
(9.2)
Let A, B be disjoint closed subsets of a normal space X. Then the following properties are equivalent:
1. There exists continuous f : X → [0, 1] such that x ∈ B ⇒ f (x) = 1 and x ∈ A ⇔ f (x) = 0.
2. A is a countable intersection of open sets.
proof
• 1 ⇒ 2 Let f : X → [0, 1] be continuous such that x ∈ B ⇒ f (x) = 1 and x ∈ A ⇔ f (x) = 0. For all
n ∈ N∗ , let An = {x ∈ X : f (x) < 1/n} = f −1 [0, 1/n). Since f is continuous and [0, 1/n) is open in
the relative topology, An is open. And we have
A = ∩∞
n=1 An
indeed, if x ∈ A, then f (x) = 0, thus ∀n ≥ 1, f (x) < 1/n, i.e. x ∈ An . Conversely, if x ∈ ∩∞
n=1 An ,
then ∀n ≥ 1, 0 ≤ f (x) < 1/n, therefore f (x) = 0, i.e. x ∈ A.
This proves that A is the countable intersection of open sets.
• 2 ⇒ 1 Suppose A is the countable intersection of open sets An . For all n ∈ N, let Vn = Acn ∪ B.
We have Vn is closed, and A ∩ Vn = ∅ since A ∩ B = ∅ and A ∩ Acn = ∅. Thus A and Vn are closed
disjoint subsets, and by Urysohn’s lemma, there exists a continuous function fn : X → [0, 1] such that
x ∈ Vn ⇒ fn (x) = 1, and x ∈ A ⇒ fn (x) = 0.
Now consider
f : X → [0, 1]
x 7→ f (x) =
∞
X
n=1
Then we have
1
2−n fn (x)
x ∈ B ⇒ ∀n, x ∈ Vn
since B ⊆ Vn
⇒ ∀n, fn (x) = 1
X
⇒ f (x) =
2−n = 1
n≥1
And if x ∈
/ A, then there exists n0 ∈ N such that x ∈
/ An0 , i.e. x ∈ Acn0 ⊆ Vn0 . Therefore fn0 (x) = 1,
and
f (x) ≥ 2−n0 fn0 (x) = 2−n0 > 0
thus x ∈
/ A ⇒ f (x) 6= 0. Conversely, if x ∈ A, then ∀n, fn (x) = 0, thus f (x) = 0. This proves that
f (x) = 0 ⇔ x ∈ A.
Finally, f is continuous: let x ∈ X, and let > 0. Let N ∈ N such that 2−N < /2. Then we have for
all y ∈ X
N
∞
X
X
|f (y) − f (x)| ≤
2−n |fn (y) − fn (x)| +
2−n |fn (y) − fn (x)|
n=1
n=N +1
and
∞
X
2−n |fn (y) − fn (x)| ≤
n=N +1
∞
X
2−n
n=N +1
≤ 2−N
< /2
now for all n ∈ {1, . . . , N }, fn is continuous, thus there exists an open Un ⊆ X such that x ∈ Un , and
∀y ∈ Un , |fn (y) − fn (x)| ≤ /(2N ). Let U = ∩N
n=1 Un . Then U is open, contains x, and for all y ∈ U ,
|f (y) − f (x)| <
N
X
2−n |fn (y) − fn (x)| + /2
n=1
<
N
X
/(2N ) + /2
n=1
<
This proves that f is continuous at y.
(9.3) A topological space X is said to be completely regular if X is T1 and every closed set E ⊂ X and
x∈
/ E, there Q
exists continuous f : X → [0, 1] such that f (x) = 0 and f (y) = 1 for every y ∈ E.
Let X = s∈S Xs , where ∀s ∈ S, Xs is completely regular. Then X is completely regular.
proof X is T1 : let x, y be distinct elements in E. Then there exists s ∈ S such that xs 6= ys . Since Xs is
T1 , there exists an open subset Us ⊆ Xs such that xs ∈ Us and ys ∈
/ Us . Now consider U = ×t∈S Ut where
Ut = Xt for all t 6= s. Then U is open, x ∈ U , and y ∈
/ U . This proves that X is T1 .
To prove that X is completely regular, let E ⊆ X be a closed subset, and let x ∈
/ E.
The complement E c is open, therefore it is the union of basic open sets, E c = ∪α∈A Uα . And since
x ∈ E c , there exists α0 ∈ A such that x ∈ Uα0 . Uα0 is a basic set, thus it is of the form
Uα0 = × Us
s∈S
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where Us is open for all s, and Us = Xs for all but finitely many s. Let {s1 , . . . , sn } be the set of elements
of S such that Us 6= Xs . Then we have
x ∈ Uα0 ⇔ ∀i ∈ {1, . . . , n}, xsi ∈ Usi
/ Usci . Thus since Xsi is completely regular, there exists
For all i ∈ {1, . . . , n}, we have Usci is closed, and xi ∈
a continuous function fi : Usi → [0, 1] such that
fi (xsi ) = 0
∀ysi ∈
Now define
Usci ,
fi (ysi ) = 1
f : X → [0, 1]
x 7→
max
i∈{1,...,n}
fi (xsi )
then f is continuous as the pointwise maximum of finitely many continuous functions x 7→ fi (xsi ) (composition of fi with the projection on coordinate si , which is continuous). And we have
• For all i ∈ {1, . . . , n}, by definition of fi , fi (xsi ) = 0, thus f (x) = maxi fi (xsi ) = 0
• For all y ∈ E, we have y ∈
/ E c = ∪α∈A Uα , thus in particular y ∈
/ Uα0 , i.e. there exists i0 ∈ {1, . . . , n}
such that ysi0 ∈
/ Usi0 . Therefore by definition of fi0 , fi0 (ysi0 ) = 1, and we have
f (y) = max fi (ysi ) ≥ fi0 (ysi0 ) = 1
i
which proves f (y) = 1.
Hence f has the desired properties.
(9.4)
Any separable metrizable space has a metrizable compactification.
proof Let (X, T X ) be a separable metrizable space, and let d be a distance such that T X is the topology
induced by d. Without loss of generality, assume that for all x, y ∈ X, d(x, y) ≤ 1.
Q Since (X, d) is a separable metric space, there exists aY countable dense set D = {yn }n∈N . Let Y =
n∈N [0, 1], equipped with the product topology, denoted T . Consider the function
φ:X→Y
x 7→ φ(x)
such that ∀n ∈ N, φn (x) = d(x, yn ) (by assumption on d, d(x, yn ) ∈ [0, 1]). Then we have
• φ is injective: if φ(x) = φ(x0 ), then ∀n ∈ N, φn (x) = φn (x0 ), i.e. d(x, yn ) = d(x0 , yn ). Now since
D = {yn }n∈N is a dense subset of X, there exists a sequence of elements of D that converges to x. Let
(xn )n be such a sequence. Then we have for all n ∈ N, d(x, xn ) = d(x0 , xn ), thus limn→∞ d(x0 , xn ) =
limn→∞ d(x, xn ) = 0, therefore (xn ) converges to x0 , and by uniqueness of the limit, x = x0 .
• φ : X → φ(X) is surjective.
• φ is continuous: let V ⊂ Y be an open subset of Y , and let U = φ−1 (V ). Since V is open in the
product topology, it is the union of basic open sets: V = ∪α∈A Vα , where for each α, Vα is a basic open
set. Then we have
U = φ−1 (V )
= φ−1 (∪α∈A Vα )
= ∪α∈A φ−1 (Vα )
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Therefore, to show that U is open, it suffices to show that the inverse image of any basic open is open.
Let Vα be a basic open set. Then there exist open subsets of Y , W1 , . . . , Wk , and indices n1 , . . . , nk
such that
y ∈ Vα ⇔ ∀i ∈ {1, . . . , k}, yni ∈ Wi
and we have
x ∈ φ−1 (Vα ) ⇔ φ(x) ∈ Vα
⇔ ∀i ∈ {1, . . . , k}, φik (x) ∈ Wi
⇔ ∀i ∈ {1, . . . , k}, x ∈ φ−1
ni (Wi )
⇔ x ∈ ∩ki=1 φ−1
ni (Wi )
therefore φ−1 (Vα ) = ∩ki=1 φ−1
ni (Wi ) is the finite intersection of open subsets (Wi is open and φni : x 7→
d(x, yni ) is continuous), thus is open. This proves that U = φ−1 (V ) is open.
Now any open subset of φ(X) (equipped with the relative topology), is of the form V ∩ φ(X) where V
is open in Y , and we have φ−1 (V ∩ φ(X)) = φ−1 (V ) ∩ φ−1 (φ(X)) = φ−1 (V ) ∩ X = φ−1 (V ), which we
proved is open. This proves that φ is continuous.
• the inverse of φ
φ−1 : φ(X) → X
y 7→ φ−1 (y) = x
is continuous. Indeed, let x0 ∈ X, and let U ⊆ X be an open neighborhood of x0 . We want to find
an open set V ⊆ Y (containing φ(X)) such that φ−1 (V ) ⊆ U . Since {yn }n∈N is dense subset of X
and U is open, U contains yn0 for some n0 . And since yn0 ∈ U open, there exists > 0 such that
B(, yn0 ) ⊆ U . Now let V = φ(X) ∩ ×∞
n=1 Vn where
∀n 6= n0 , Vn = [0, 1]
Vn0 = [0, )
Then V is open in φ(X) (×n Vn is a basic open set in Y ), and ∀y ∈ V , let x = φ−1 (y), then we
have φ(x) ∈ V , thus φn0 (x) < , i.e. d(x, yn0 ) < , i.e. x ∈ B(, yn0 ), thus x ∈ U . This proves that
φ−1 (V ) ⊆ U , and hence φ−1 is continuous.
Therefore φ : X → φ(X) is a homeomorphism.
Next, let X ∗ = cl(φ(X)). We have (Y, T Y ) is metrizable (countable product of compact sets). Therefore
Y
(X ∗ , T ∗ ) is also metrizable, where T ∗ is the relative topology induced by T Y on X ∗ , i.e. T ∗ = T|X
∗ . We
also have X ∗ is a closed subset of the compact space Y , therefore it is compact.
φ
Y
(X, T ) → (φ(X), T|φ(X)
) ⊆ (X ∗ , T ∗ )
Therefore X is homeomorphic to φ(X), which is a dense subset of the compact metrizable X ∗ . To show
Y
that (X ∗ , T ∗ ) is a compactification of X, we need to check that the topology of φ(X), T|φ(X)
, is the relative
∗
Y
∗
topology induced by T on φ(X). In other words, we need to check that T|φ(X) = T|φ(X) . This is true since
∗
T|φ(X)
= {U ∩ φ(X), U ∈ T ∗ }
= {(V ∩ X ∗ ) ∩ φ(X), V ∈ T Y }
Y
since T ∗ = T|X
∗
since φ(X) ⊆ X ∗
= {V ∩ φ(X), V ∈ T Y }
Y
= T|φ(X)
Therefore (X ∗ , T ∗ ) is a metrizable compactification of X.
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