Download Metric Spaces and Topology M2PM5 - Spring 2011 Solutions Sheet

Metric Spaces and Topology
M2PM5 - Spring 2011
Solutions Sheet 6 February 15th, 2011
Exercise 1 Let T = {A, T } be a topological space.
(i) Let H ⊆ A, and define
CH ≡ {E ⊆ A | E ⊇ H and E is closed (in T )}.
Prove that H = ∩E∈CH E.
(ii) Prove that
(ii.1) For any indexing set I, ∩i∈I Hi ⊂ ∩i∈I Hi , where Hi ⊆ A for each
i ∈ I.
(ii.2) H1 ∪ H2 ∪ . . . ∪ Hn = H1 ∪ H2 ∪ . . . ∪ Hn , where Hi ⊆ A for each
i = 1, 2, . . . , n.
(iii) Give an example of open sets H, K ⊆ R such that H ∩K, H ∩K, H ∩K,
and H ∩ K are all different. (Here, R is given the usual topology.)
Solution:
(i) If E is closed and H ⊆ E, then H ⊆ E = E. Therefore, H ⊆ ∩E∈CH E.
On the other hand, since H is closed, and H ⊆ H, it follows that
H ∈ CH , so ∩E∈CH E ⊆ H.
(ii.1) Since K ≡ ∩i∈I Hi is an intersection of closed sets, it is closed. Also, it
contains ∩i∈I Hi (since Hi ⊆ Hi for all i ∈ I), and so it also contains
∩i∈I Hi (since E ⊆ F ⇒ E ⊆ F , and F = F if F is closed).
(ii.2) Let K ≡ H1 ∪ H2 ∪ . . . ∪ Hn and H ≡ H1 ∪ H2 ∪ . . . ∪ Hn . For each
i = 1, 2, . . . , n, Hi ⊆ H, so Hi ⊆ H. It follows that K ⊆ H. (Note that
this argument works for any family of subsets {Hi }i∈I .) Conversely K,
being the union of a finite number of closed sets, is closed. Also, K ⊇ H.
Therefore K ⊇ H. So K = H.
(iii) Let H = (0, 2) ∪ (3, 4), K = (1, 3). Then H = [0, 2] ∪ [3, 4], K = [1, 3],
and H ∩K = (1, 2], H ∩K = [1, 2), H ∩K = [1, 2]∪{3}, H ∩ K = [1, 2].
Exercise 2 Prove that:
(i) Any subspace of a Hausdorff space is Hausdorff.
(ii) The topological product (see Exercise 2 (iii) Sheet 4) of two Hausdorff
spaces is Hausdorff.
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Metric Spaces and Topology
M2PM5 - Spring 2011
Solutions Sheet 6 February 15th, 2011
(iii) If f : A1 → A2 is injective and (T1 , T2 )-continuous and if T2 = {A2 , T2 }
is Hausdorff then so is T1 = {A1 , T1 }.
(iv) Hausdorff-ness is a topological property.
Solution:
(i) Let T = {A, T } be a topological Hausdorff space, and let H ⊆ A. Let
x, y ∈ H, x 6= y. Then x, y ∈ A, x 6= y, so there exist U ∈ T , V ∈ T
such that x ∈ U, y ∈ V , U ∩ V = ∅ (since T is Hausdorff). Let UH ≡
U ∩ H, VH ≡ V ∩ H, then x ∈ UH , y ∈ VH and UH ∩ VH = ∅, Moreover,
by the definition of the subspace topology TH in H (see Exercise 2 (iii)
Sheet 4), UH , VH ∈ TH , so {H, TH } is a topological Hausdorff space.
(ii) Let {Ai , Ti }, i = 1, 2, be two topological Hausdorff spaces, and let
T1 × T2 be endowed with the product topology (see Exercise 2 (iii)
Sheet 4). Let x = (x1 , x2 ), y = (y1 , y2 ) ∈ T1 × T2 , x 6= y. That is, either
x1 6= y1 or x2 6= y2 (or both). Assume, without loss of generality, that
x1 6= y1 . Then, since T1 is Hausdorff, there exist U1 , V1 ∈ T1 such that
x ∈ U1 , y ∈ V1 , U1 ∩ V1 = ∅. Let U ≡ U1 × A2 , V ≡ V1 × A2 . Then, by
the definition of the product topology, U and V are open in T1 × T2 .
Moreover, clearly x ∈ U, y ∈ V and U ∩ V = ∅. This proves that T1 × T2
with the product topology is Hausdorff.
(iii) Let x, y ∈ A1 , x 6= y. Then f (x), f (y) ∈ A2 and f (x) 6= f (y) since
f is injective. Since T2 is Hausdorff, there exist U2 , V2 ∈ T2 such that
f (x) ∈ U2 , f (y) ∈ V2 , and U2 ∩ V2 = ∅. Since f is continuous, U1 ≡
f −1 (U2 ), V1 ≡ f −1 (V2 ) are both open (i.e., belong to T1 ) and clearly
x ∈ U1 , y ∈ V1 . Since z ∈ U1 ∩ V1 ⇒ f (z) ∈ U2 ∩ V2 = ∅, it follows that
U1 ∩ V1 = ∅. Therefore, T1 = {A1 , T1 } is Hausdorff.
(iv) Let Ti = {Ai , Ti }, i = 1, 2 be two homeomorphic topological spaces.
That is, there exists a continuous bijection f : A2 → A1 such that
f −1 is also continuous. Assume T1 is Hausdorff. Then (iii) (used on f )
gives that T2 is Hausdorff. Conversely, assume T2 is Hausdorff. Then
(iii) (used on f −1 ) gives that T1 is Hausdorff. Hence, Hausdorff-ness is
a topological property.
Exercise 3
(i) Prove that a finite subset of a metric space has no limit points.
(ii) Prove that any space having only a finite number of points is compact.
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Metric Spaces and Topology
M2PM5 - Spring 2011
Solutions Sheet 6 February 15th, 2011
(iii) Prove that a discrete space is compact only if it is finite.
(iv) Given that H, K are compact subspaces of a space T , prove that H ∪ K
is compact.
(v) Given two topologies T1 , T2 on a set A with T1 ⊂ T2 , prove that if
{A, T2 } is compact then so is {A, T1 } .
Solution:
(i) Let M = {X, d} be a metric space, and let S = {x1 , . . . , xn } ⊆
X. Assume x ∈ X is a limit point of S. If x 6∈ S, then let δ =
mini∈{1,...,n} d(x, xi ) > 0. Then the ball Bδ/2 (x) does not contain any point from S. Thus we have a contradiction with the assumption that x is a limit point of S. Similarly, if x = xm ∈ S, then let
δ = mini∈{1,...,n},i6=m d(xm , xi ) > 0. Then the ball Bδ/2 (xm ) does not
contain any point from S, except x = xm itself. This is again contradiction with the assumption that x is a limit point of S. So S does not
have any limit points.
(ii) Assume T = {A, T }, A = {x1 , x2 , . . . , xn }, n ∈ N, and let U be any
open covering of A. Choose, for i = 1, 2, . . . , n, a Ui ∈ U such that
xi ∈ Ui . Then clearly {U1 , U2 , . . . , Un } is a finite subcover of U for A,
so T is compact.
(iii) Assume T = {A, 2A } is compact. Then, since
U = Ux Ux = {x}, x ∈ A
is an open cover for A, there exists finitely many x1 , x2 , . . . , xn ∈ A
such that A ⊆ ∪ni=1 Uxi , that is, A ⊆ {x1 , x2 , . . . , xn }, so A is finite.
(iv) Assume U is an open cover for H ∪K. Then, since H ⊆ H ∪K, U is also
an open cover for H, and since H is compact, there exists U1 , . . . , UnH ∈
H
U such that H ⊆ ∪ni=1
Ui . Similarly, there exists V1 , . . . , VnK ∈ U
nK
H
K
such that K ⊆ ∪i=1 Vi . Therefore, H ∪ K ⊆ (∪ni=1
Ui ) ∪ (∪ni=1
Vi ), so
{U1 , . . . , UnH , V1 , . . . , VnK } forms a finite subcover of U for H ∪ K, so
H ∪ K is compact.
(v) Assume U is an open (in {A, T1 }) cover for A, i.e.,
A ⊆ ∪i∈I Ui ,
U = {Ui ∈ T1 | i ∈ I, I some index set } .
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Metric Spaces and Topology
M2PM5 - Spring 2011
Solutions Sheet 6 February 15th, 2011
Note that, since T1 ⊂ T2 , each Ui is also open in {A, T2 }, i.e., Ui ∈
T2 , i ∈ I. Therefore, U is an open (in {A, T2 }) cover for A, i.e.,
A ⊆ ∪i∈I Ui ,
U = {Ui ∈ T2 | i ∈ I, I some index set } .
Since {A, T2 } is compact, there exists a finite subcover of U for A. i.e.,
{U1 , . . . , Un } ⊆ U such that A ⊆ ∪ni=1 Ui . So {A, T1 } is also compact.
Exercise 4
(i) Recall that a topological space T is compact if every open cover of T
has a finite subcover.
Assume S is a subspace of T . Prove that S is a compact topological
space if and only if given any cover U of S by sets open in T , there is
a finite subcover of U for S.
(ii) Let M = {X, d} be a compact metric space. Prove that there cannot
exist a sequence {xn }n∈N ⊂ X with the following property: There exists
> 0 such that d(xn , xm ) ≥ , n 6= m.
Solution:
(i) Assume S is a compact space (in the topology TS induced by the topology T of T ), and let U be a cover of S by sets open in T :
U = {Ui ⊆ T |Ui ∈ T , i ∈ I} ,
with I some index set. Let
US ≡ {Ũi | Ũi = Ui ∩ S, Ui ∈ T } .
Then by definition of TS , and since U is a cover of S, US is an open
cover of S (i.e., a cover of S by sets open in its topology). Since S is
compact, there exists a finite subcover, {Ũi1 , . . . , Ũin } (i1 , . . . , in ∈ I).
Then {Ui1 , . . . , Uin } clearly forms a finite subcover of U for S.
Conversely, assume that given any cover U of S by sets open in T , there
is a finite subcover of U for S. Let V be an open cover for S. That is,
a cover of S by sets open in the induced topology TS :
V = {Vi | Vi ∈ TS , i ∈ I} ,
with I some index set. By definition of TS , for all i ∈ I there exists
Ui ∈ T such that Vi = Ui ∩ S. With this notation,
U ≡ {Ui | Vi = Ui ∩ S, i ∈ I}
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Metric Spaces and Topology
M2PM5 - Spring 2011
Solutions Sheet 6 February 15th, 2011
forms a cover of S by sets open in T . By assumption, there exists a
finite subcover {Ui1 , . . . , Uir } of U for S, and clearly the corresponding
subcover {Vi1 , . . . , Vir } of V covers S. So S is compact (in the topology
TS ).
(ii) Assume that there exists such a sequence, and let S = {xn | n ∈ N} ⊆
X be the set of its members. Then, since S is closed, it is compact
(closed subsets of compact sets are compact). Let B/2 (xn ), n ∈ N,
be the sequence of balls of radius /2 and centres xn . Notice that, by
the assumption on the sequence, B/2 (xn ) ∩ B/2 (xm ) = ∅ for n 6= m.
Clearly, {B/2 (xn ) | n ∈ N} forms an open cover of S. We claim that
S = S (and hence, {B/2 (xn ) | n ∈ N} forms an open cover of S). In
fact, assume x is a limit point of S, but x 6∈ S. Then Bδ (x) ∩ S 6= ∅ for
every δ > 0; but since d(xn , xm ) ≥ , m 6= n, this implies the existence
of a unique xn0 ∈ S such that Bδ (x) ∩ S = {xn0 } for δ < /2. This
implies x = xn0 , which contradicts x 6∈ S. So indeed S = S. Thus we
have a countable open cover of the compact set S(= S) with no finite
subcover. Contradiction.
Exercise 5
Definition Two metrics d1 , d2 on a set A are uniformly equivalent iff the
identity map of A is uniformly (d1 , d2 )-continuous and uniformly (d2 , d1 )continuous.
(i) Let X be a set, and d1 and d2 two metrics on X. Prove that if {X, d1 }
is compact and if d1 , d2 are topologically equivalent metrics then they
are uniformly equivalent.
(ii) Prove that any two Lipschitz equivalent metrics are uniformly equivalent.
(iii) Prove that for any metric space M = {X, d} the metric d˜ given by
˜ y) = d(x, y)/(1 + d(x, y)) is uniformly equivalent to d, but not
d(x,
necessarily Lipschitz equivalent to d.
(iv) Prove that
d(x, y) ≡ |x−1 − y −1 | ,
x, y ∈ (0, 1)
defines a metric on (0, 1) ⊂ R which is topologically equivalent but not
uniformly equivalent to the Euclidean metric on (0, 1).
Solution:
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Metric Spaces and Topology
M2PM5 - Spring 2011
Solutions Sheet 6 February 15th, 2011
(i) Recall: The two metrics d1 , d2 are topologically equivalent iff for any
metric spaces {B, d}, {C, d0 }, and any maps f : B → X, g : X → C,
we have
(a) f is (d, d1 )-continuous iff f is (d, d2 )-continuous.
(a) g is (d1 , d0 )-continuous iff g is (d2 , d0 )-continuous.
Let B = X, f = Id (Id(x) = x for all x ∈ X), and d = d1 . Then Id
is (d1 , d1 )-continuous, so also (d1 , d2 )-continuous. Then, since {X, d1 }
is compact, Id(X) = X is compact in the space {X, d2 } (the image of
a compact set under a continuous function is compact). Furthermore,
Id is uniformly continuous on X since {X, d1 } is compact. Reversing
the argument (using that {X, d2 } is now known to be compact) gives
that Id is also uniformly (d2 , d1 )-continuous. Therefore d1 and d2 are
uniformly equivalent metrics.
(ii) Assume d1 , d2 are two Lipschitz equivalent metrics on the set X, that
is, there exist constants h, k > 0 such that for any x, y ∈ X,
h d1 (x, y) ≤ d2 (x, y) ≤ k d1 (x, y) .
(1)
Let Id : X → X be the identity map. Let > 0 and choose δ ≡ /k.
Then, if d1 (x, y) < δ for some x, y ∈ X, then, by (1),
d2 (Id(x), Id(y)) = d2 (x, y) ≤ k d1 (x, y) < .
So, Id is uniformly (d1 , d2 )-continuous. Similarly (using the other inequality in (1)) one gets that Id is uniformly (d2 , d1 )-continuous. Hence,
d1 and d2 are uniformly equivalent.
˜ y) ≤ d(x, y). Hence, Id is uniformly (d, d)-continuous
˜
(iii) Note that d(x,
(take δ = ). Conversely, let > 0 be given, and let d = min{/2, 1/2}.
˜ y) < δ implies d(x,
˜ y) < 1/2. Therefore, d(x,
˜ y) < δ implies
Then d(x,
that
d(Id(x), Id(y)) = d(x, y) =
˜ y)
d(x,
˜ y) < ,
≤ 2 d(x,
˜ y)
1 − d(x,
˜ d)-continuous. So d and d˜ are uniformly equihence Id is uniformly (d,
valent.
On R,
˜ y)
d(x,
1
=
d(x, y)
1 + |x − y|
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Metric Spaces and Topology
M2PM5 - Spring 2011
Solutions Sheet 6 February 15th, 2011
has greatest lower bound 0, so there is no constant h > 0 such that
˜ y) for all x, y ∈ R. Hence, d˜ is not Lipschitz equivalent
h d(x, y) ≤ d(x,
to d, when d is the usual Euclidean metric on R.
(iv) Clearly, d is both positive definite and symmetric. Also, if x, y, z ∈ (0, 1)
then, by the triangle inequality for | · |,
d(x, y) = |x−1 − y −1 | ≤ |x−1 − z −1 | + |z −1 − y −1 | = d(x, z) + d(z, y) ,
so d is indeed a metric.
Let d2 be the usual Euclidean metric. Note that Id : {(0, 1), d2 } →
{(0, 1), d} is uniformly continuous iff the map f : (0, 1) → R defined by f (x) = 1/x is uniformly continuous on (0, 1), with both (0, 1)
and R equipped with the usual Euclidean metric d2 (since d(x, y) =
d2 (f (x), f (y)) for x, y ∈ (0, 1)).
But f is not uniformly continuous on (0, 1): Take 0 = 1, δ > 0. Choose
n > 1/δ. Then |f (1/2n)−f (n)| = n ≥ 1 = 0 although |1/2n−1/n| < δ,
1/2n, 1/n ∈ (0, 1). This shows that Id : {(0, 1), d2 } → {(0, 1), d} is not
uniformly continuous, and so d2 and d are not uniformly equivalent
metrics.
Since d2 (x, y) = |x − y| ≤ d(x, y) (check!), we have that if U ⊆ (0, 1) is
d2 -open , then it is d-open. Conversely: Given a ∈ (0, 1), and > 0, let
δ ≡ min{a2 /2, a/2}. If |x − a| < δ then
d(x, a) =
2|x − a|
|x − a|
<
< .
|xa|
a2
Hence, the identity-map Id is (d2 , d)-continuous, so if U ⊆ (0, 1) is dopen, it is d2 -open. It follows that d and d2 are topologically equivalent.
NB: Conclusions: d1 , d2 Lipschitz equivalent ⇒ d1 , d2 uniformly equivalent
⇒d1 , d2 topologically equivalent. But d1 , d2 uniformly equivalent 6⇒ d1 , d2
Lipschitz equivalent (in general), and d1 , d2 topologically equivalent 6⇒ d1 , d2
uniformly equivalent (in general). But if the space is compact (in either d1 or
d2 ), then d1 , d2 topologically equivalent ⇒ d1 , d2 uniformly equivalent.
Thomas Østergaard Sørensen
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