Download Math 131: Midterm Solutions

Math 131: Midterm Solutions
(1) Let X be a completely regular topological space. Let A and B be closed
subsets of X with A ∩ B = ∅, and suppose that A is compact. Show that
there exists a continuous function f : X → [0, 1] such that f (a) = 0 for
a ∈ A and f (b) = 1 for b ∈ B.
Since X is completely regular, we can choose for each a ∈ A a continuous
function fa : X → [0, 1] such that fa (a) = 0 and fa (b) = 1 for b ∈ B. Let
Ua is an open subset of X. This open
Ua = fa−1 [0, 21 ). Since fa is continuous,
S
subset contains a, so that A ⊆ a∈A Ua . Since A is assumed to be compact,
there exists a finite subset {a1 , . . . , an } ⊆ A such that A ⊆ Ua1 ∪ · · · ∪ Uan . We
define a function f 0 : X → [0, 1] by the formula f 0 (x) = min{fai (x)}1≤i≤n , and
f : X → [0, 1] by the formula f (x) = max{0, 2f 0 (x) − 1}. If b ∈ B, then each
fai (b) is equal to 1, so that f (b) = f 0 (b) = 1. If a ∈ A, then a ∈ Uai for some
index i, so that f 0 (a) ≤ fai (a) < 21 and therefore f (a) = 0.
(2) Let X be a normal topological space and let Y ⊆ X be a closed subspace. Show that every continuous map f0 : Y → R can be extended to a
continuous map f : X → R.
Choose a homeomorphism h : R → (−1, 1), and regard h ◦ f0 as a continuous
map from Y into [−1, 1]. We proved in class that h◦f0 can be extended to a map
g : X → [−1, 1]. Then g −1 {−1, 1} is a closed subset A ⊆ X which is disjoint
from Y . By the Urysohn Lemma, we can choose a continuous function g 0 : X →
[0, 1] such that g(y) = 1 for y ∈ Y and f (a) = 0 for a ∈ A. Let g 00 : X → [−1, 1]
be defined by the formula g 00 (x) = g(x)g 0 (x). Then g 00 (y) = g(y) = (h ◦ f0 )(y)
for y ∈ Y . Moreover, g 00 (x) ∈ (−1, 1) for all x ∈ X: if x ∈ A, then g 00 (x) = 0,
and if x ∈
/ A then |g 00 (x)| = |g(x)|g 0 (x) < g 0 (x) ≤ 1. Set f = h−1 ◦ g 00 . Then f
is a continuous extension of f0 .
Let X be a Hausdorff topological space. We let K(X) denote the collection
of all compact subsets of X. We regard K(X) as a topological space, taking as
a subbasis all sets of the form
{K ∈ K(X) : K ⊆ U }
{K ∈ K(X) : K ∩ U 6= ∅}
where U is an open subset of X.
(3) Prove that that K(X) is a Hausdorff space and that K(X) is compact if
and only if X is compact.
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We first show that K(X) is Hausdorff. Let K, K 0 ∈ K(X) be distinct
compact sets. Without loss of generality there exists a point x ∈ K such that
x∈
/ K 0 . Choose disjoint open sets U, V ⊆ X such that x ∈ U , K 0 ⊆ V . Then
{Y ∈ K(X) : Y ∩ U 6= ∅}
{Y ∈ K(X) : Y ⊆ V }
are disjoint open subsets of K(X) containing K and K 0 , respectively.
Now suppose that K(X) is compact. Let {Uα }α∈A be an open covering of
X; we will show
that it admits a finite subcover. For every finite subset A0 ⊆ A,
S
let UA0 = α∈A0 Uα , and let VA0 = {K ∈ K(X) : K ⊆ UA0 }. If K ∈ K(X),
then {Uα ∩ K} is an open covering of K; the compactness of K then guarantees
the existence of a finite subcover which shows
S that K ⊆ UA0 for some finite
subset A0 ⊆ A. It follows that K(X) = A0 VA0 . Since K(X) is compact,
there exists a finite subcover: that is, there is a finite list of finite subsets
A1 , A2 , A3 , . . . , An ⊆ A S
such that K(X) = VA1 ∪· · ·∪VAn . Let A0 = A1 ∪· · ·∪An ;
we will show that X = α∈A0 Uα . To see this, we note that for each x ∈ X the
subset {x} ⊆ X is compact, so that {x} ∈ VAi for some i and therefore x ∈ Uα
for some α ∈ Ai .
Finally, suppose that X is compact. We will show that K(X) is compact.
It suffices to show that every open covering of K(X) by subbasic open sets has
a finite subcovering. To this end, suppose that we are given collections of open
sets {Uα }α∈A , {Vβ }β∈B in X such that
[
[
K(X) = (
{K ∈ K(X) : K ⊆ Uα }) ∪ (
{K ∈ K(X) : K ∩ Vβ 6= ∅}).
α∈A
β∈B
S
Let V = β Vβ and S
let K = X − V . Then K is closed in X and therefore
compact. Since K ∈
/ β∈B {K ∈ K(X) : K ∩ Vβ 6= ∅}, we must have K ⊆ Uα
for some α ∈ A. Let K 0 = X − Uα . Then K 0 ⊆ V . Since K 0 is a closed subset
of X, S
it is compact. It follows that there exists a finite subset B0 ⊆ B such that
K 0 ⊆ β∈B0 Vβ . Then every subset of X is either contained in Uα or intersects
Vβ for some β ∈ B0 , so that
[
K(X) = {K ∈ K(X) : K ⊆ Uα } ∪ (
{K ∈ K(X) : K ∩ Vβ 6= ∅})
β∈B0
(4) Let X be a Hausdorff space. Show that the map f : K(X) × K(X) →
K(X) given by (K, K 0 ) 7→ K ∪ K 0 is continuous. Give an example to show
that the map g : K(X) × K(X) → K(X) given by (K, K 0 ) 7→ K ∩ K 0 need
not be continuous.
We first show that f is continuous. It will suffice to show that for every
subbasic open set W ⊆ K(X), the inverse image f −1 (W ) is an open subset of
K(X) × K(X). If W = {K ∈ K(X) : K ⊆ U }, then f −1 (W ) = W × W . If
W = {K ∈ K(X) : K ∩ V 6= ∅}, then f −1 (W ) = (K(X) × W ) ∪ (W × K(X)).
We now show that g is not continuous in the case X = R. Consider the
sequence of sets { n1 }. We claim that this sequence converges to {0} in K(R).
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Let W be an open subset of K(R) containing {0}; we claim that W contains
{ n1 } for all but finitely many n. Without loss of generality, we may assume that
W is a finite intersection of subbasic open sets, and is therefore of the form
{K ∈ K(X) : (K ⊆ U ) ∧ (K ∩ V1 6= ∅) ∧ · · · ∧ (K ∩ Vn 6= ∅)}
for some open sets U, V1 , . . . , Vn ⊆ X. Since {0} ∈ W , we have 0 ∈ U ∩ V1 ∩ · · · ∩
Vn . Since the intersection is open, it must contain the interval (−, ) for some
> 0. It follows that n1 ∈ U ∩ V1 ∩ · · · ∩ Vn and therefore { n1 } ∈ W whenever
1
n < .
The sequence of points {({ n1 }, {0})}n>0 converges to ({0}, {0}) in K(X) ×
K(X). If g were continuous, it would follow that the sequence {g({ n1 }, {0})}n>0
converges to g({0}, {0}) = {0} ∈ K(X). But { n1 } ∩ {0} = ∅, so the sequence
6 {0}, we obtain a contradiction (note
{g({ n1 }, {0})}n>0 converges to ∅. Since ∅ =
that K(X) is Hausdorff, so convergent sequences in K(X) can have at most one
limit).
(5) Let Z be the set of integers, endowed with the discrete topology. Let Z∨
denote the Stone-Cech compactification of Z. Prove that Z∨ is uncountable.
Let S be the set of rational numbers in the interval [0, 1]. Since S is countable, we can choose a surjective map f : Z → S. We can regard f as a continuous
map from Z to [0, 1]. Since [0, 1] is a compact Hausdorff space, f extends to
a continuous map f ∨ : Z∨ → [0, 1]. Then f ∨ (Z∨ ) is continuous image of a
compact set, and therefore a compact subset K ⊆ [0, 1]. Then K is a closed
subset of [0, 1] containing f (Z) = S. Since S is dense, we have K = [0, 1]. Since
[0, 1] is uncountable, the existence of a surjection f ∨ : Z∨ → [0, 1] guarantees
that Z∨ is uncountable.
(6) Let X be a compact Hausdorff space, and let x, y ∈ X. Assume that
for any subset U ⊆ X which is both closed and open, either x, y ∈ U or
x, y ∈
/ U . Show that there exists a connected subset of X which contains
both x and y. (For some hints, see problem 4 in §5-37 of the textbook).
Let {Uα }α∈A be the collection of T
all subsets U of X which are both closed
and open such that x ∈ U . Let S = α Uα . Then x ∈ S by construction, and
our hypothesis guarantees also that y ∈ S. We will show that S is connected.
Suppose otherwise. Then we can write S as a disjoint union of nonempty closed
subsets S 0 , S 00 ⊆ S. The sets S 0 and S 00 are closed in X and therefore compact.
It follows that we can find disjoint open sets V, W ⊆ X such that S 0 ⊆ V and
S 00 ⊆ W . Then
[
X =V ∪W ∪(
(X − Uα )).
α∈A
Since
S X is compact, there exists
T a finite subset A0 ⊆ A such that X = V ∪ W ∪
( α∈A0 (X − Uα )). Let U = α∈A0 Uα . Then U is a closed and open subset of
X containing x and y, and U ⊆ V ∪ W . It follows that U is a disjoint union of
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open sets (U ∩ V ) and (U ∩ W ), so that the sets U ∩ V and U ∩ W are also both
closed and open. Without loss of generality we may assume that x ∈ U ∩ V .
Then U ∩ V = Uα for some α ∈ A, so that S ⊆ U ∩ V ⊆ V and therefore
S 0 = S ∩ V = S. Then S 00 = S − S 0 is empty, contrary to our assumption.
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