Download Test Assignment for Metric Space Topology 304a

Test Assignment for Metric Space Topology 304a
Instructor: Georg Biedermann
Solutions
Exercise 1:(4) Let (X, T1 ) and (Y, T2 ) be topological spaces.
1. Show that for U1 , U2 ⊂ X and V1 , V2 ⊂ Y we have:
(U1 × V1 ) ∩ (U2 × V2 ) = (U1 ∩ U2 ) × (V1 ∩ V2 )
Solution: We have the following equivalences:
(u, v) ∈ (U1 × V1 ) ∩ (U2 × V2 )
⇔ u ∈ U1 , v ∈ V1 , u ∈ U2 , v ∈ V2
⇔ (u, v) ∈ (U1 ∩ U2 ) × (V1 ∩ V2 )
2. Show that B = {U × V |U ∈ T1 , V ∈ T2 } is a basis for a topology on X × Y .
Solution: In general, given a family B of subsets that satisfies the following
two conditions, namely
• X ×Y =
S
B∈B
B
• B1 , B2 ∈ B ⇒ B1 ∩ B2 ∈ B
we can construct a topology by defining W ⊂ X × Y to be open if and only if
there exist Bi ∈ B for some I ∈ I such that
[
W =
Bi .
i∈I
We have to check that this is indeed a topology:
a. X × Y is open by the first point above, ∅ is the union over an empty index
set, hence also open.
b. Arbitrary unions are open sets, since the union of unions is a union.
c. Finite intersections of open sets are open because of the second point above:
[
[
[
( Bi ) ∩ (
Bj ) = (Bi ∩ Bj )
i∈I
i,j
j∈J
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It remains to show that our B satisfies the two points: the first one is obvious
X × ∈ B, and the second one follows from part 1.
The topology on X × Y generated by B is called the product topology. Define
the projections prX : X × Y → X and prY : X × Y → Y by
prX (x, y) := x, prY (x, y) := y.
3. Show that prX and prY are continuous if X × Y is given the product
topology.
Solution: A map is continuous if and only if the preimage of every open set is
∈ B. This
open. Let U be an open set in X, i.e. U ∈ T1 . Then pr−1
X (U ) = U × Y
is open by the definition of the product topology. Hence the projection prX is
continuous. Analogously for prY .
4. Let f : T → X and g : T → Y be continuous maps from a topological
space (T, T3 ). Show that there exists a unique map h : T → X × Y with
prX h = f and prY h = g. Show that h is continuous.
Solution: We define a map h : T → X × Y in the following way:
h(t) := (f (t), g(t)) ∈ X × Y
We claim, that this map is the desired map. First of all, the two equations
prX h = f and prY h = g are satisfied by construction. It is also clear, that h
is determined by these equations, hence it is unique (no other map can satisfy
these two equations). We just have to prove, that h is continuous. It suffices to
check, that for every B ∈ B the preimage h−1 (B) ∈ T3 , since B is a basis for the
product topology. So let U × V ∈ B. Then we have:
h−1 (U × V ) = f −1 (U ) ∩ g −1 (V )
Both sets f −1 (U ) and g −1 (V ) are open in T , since f and g are continuous. The
intersection of two open sets is open, which proves the claim.
Property 4 is called the universal property of the product. Let C(W, Z) denote
the set of continuous maps from the space W to the space Z.
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5. Show that there is a bijection C(T, X × Y ) ∼
= C(T, X) × C(T, Y ).
Solution Let’s define a map α : C(T, X × Y ) → C(T, X) × C(T, Y ) by
h 7→ (prX h, prY h) = α(h).
The projections are continuous by part 3, hence their composition with h is
continuous. Therefore α is well-defined. Conversely let β : C(T, X) × C(T, Y ) →
C(T, X × Y ) be given by
(f, g) 7→ h = β(f, g),
where h exists by part 4. Part 4 also shows, that h is unique and continuous,
whereby β is well-defined. Obviously we have:
α ◦ β = id and β ◦ α = id
This proves, that α and β are mutually inverse bijections.
Exercise 2:(4) Let (M, d) be a metric space.
1. Show that setting d′ (x, y) :=
d(x,y)
1+d(x,y)
Solution: The map d′ : M × M →
for x, y ∈ M defines a metric on M .
R has to satisfy the axioms of a metric.
• ∀ x, y ∈ M d′ (x, y) ≥ 0
This is obvious, since d(x, y) ≥ 0.
• ∀ x, y ∈ M d′ (x, y) = 0 ⇔ x = y
We have:
d′ (x, y) = 0 ⇔ d(x, y) = 0 ⇔ x = y
Here the second equivalence follows, since d is a metric.
• ∀ x, y ∈ M d′ (x, y) = d′ (y, x)
This is immediate from the fact that d(x, y) = d(y, x).
• ∀ x, y, z ∈ M d′ (x, z) ≤ d′ (x, y) + d′ (y, z)
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For all a, b ∈
R≥0 and d := a + b + ab we have:
b
a + b + 2ab
d
a
+
=
≥
1+a 1+b
1 + a + b + ab
1+d
x
is monotonically increasing
We also have d ≥ a + b and that the map x 7→ 1+x
on ≥0 . This follows from calculus, since for x ∈ ≥0 the derivative
R
R
d
x
1
(
)=
dx 1 + x
(1 + x)2
has positive values. So for any c ≤ a + b it follows:
d
a
b
c
≤
≤
+
1+c
1+d
1+a 1+b
The triangle inequality now follows be setting a := d′ (x, y), b := d′ (y, z) and
c := d′ (x, z).
2. Show that (M, d′ ) is bounded, i.e. ∃C > 0 ∀ x, y ∈ M d′ (x, y) < C.
Solution: For all X, y ∈ M we have d′ (x, y) :=
d(x,y)
<1
1+d(x,y)
3. Show that idM is (d, d′ )-continuous and (d′ , d)-continuous.
This shows that every metric space is topologically equivalent to a bounded
metric space.
Solution: Let’s denote for a metric m the corresponding open ǫ-ball by Bǫm (x) :=
{y ∈ M |m(x, y) ≤ ǫ}. Since for all x, y ∈ M we have d′ (x, y) ≤ d(x, y), it follows:
′
Bǫd (x) ⊂ Bǫd (x)
This proves, that id : (M, d) → (M, d′ ) is continuous. For the other direction
observe, that we have for all 0 ≤ a ≤ 1:
a
a
≤
2
1+a
So for arbitrary 1 > ǫ > 0 we see:
′
d
(x) ⊂ Bǫd (x)
Bǫ/2
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This proves, that id is (d′ , d)-continuous.
Rn. Show the following inequalities:
Exercise 3:(4) Let x, y ∈
1
n−1 d1 (x, y) ≤ n− 2 d2 (x, y) ≤ d∞ (x, y) ≤ d2 (x, y) ≤ d1 (x, y)
This shows that the three metrics d1 , d2 and d∞ are Lipschitz equivalent.
P 2P 2 P
si ≥ ( ri si )2 . If we
Solution: The Cauchy-Schwarz inequality says
ri
put ri = 1 and si = |xi − yi | for all 1 ≤ i ≤ n we obtain:
n
n
X
i=1
n
X
|xi − yi |)2
|xi − yi |2 ≥ (
i=1
All terms are positive, so by taking the square root this is equivalent to:
n
n
X
√ X
√
n d2 (x, y) = n (
|xi − yi | = d1 (x, y)
|xi − yi |2 ≥
i=1
i=1
This proves the first inequality. Next we obviously have:
n
X
(xi − yi )2 ≤ n max |xi − yi |2
i=1,...,n
i=1
Again since all terms are positive, this is equivalent to the second inequality:
1
√
n
n
X
i=1
(xi − yi )2
!− 12
max |xi − yi | = d∞ (x, y)
≤
i=1,...,n
Furthermore we have the obvious inequalities
max |xi − yi |2 ≤
i=1,...,n
n
X
i=1
(xi − yi )2 ≤
n
X
i=1
!2
|xi − yi |
which prove the remaining inequalities of the exercise.
Exercise 4:(3) Let X be a space with the trivial topology and let Y be a space
with the discrete topology. Let T be an arbitrary topological space.
1. Describe the sets C(T, X) and C(Y, T ).
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Solution: A map is continuous if and only if the preimage of every open set
is open. Let f : T → X be an arbitrary map. The only open subsets of X are
X and ∅, their preimages are T and ∅ both of which are open in T . Hence f is
continuous. So C(T, X) is just the set of all maps from T to X.
Now let g : Y → T be an arbitrary map. The preimage of every subset of T is
open, simply because every subset of Y is open. In particular every map from
Y to T is continuous. So C(Y, T ) is the set of all maps from Y to T .
2. Let f : X → T be a continuous map. What can you say about the subspace f (X) ⊂ T ?
Solution: The subspace f (X) ⊂ T has to have the trivial topology: Let U ⊂
f (X) be open. Then its preimage f −1 (U ) is open, so has to be either X or ∅.
Now f : X → f (X) is surjective, which means:
f (f −1 (U )) = U
This implies, that U has to be either f (X) or ∅.
3. Show that a map g : T → Y is continuous if and only if it is locally
constant, i.e. for all t ∈ T there exists an open set t ∈ U ⊂ T such that g
is constant on U .
Solution: Let g be continuous. Every subset of Y is open. In particular for
every y ∈ Y the sets g −1 ({y}) are open. Let t ∈ T , then t ∈ g −1 ({g(t)}), which
is open. Of course, g is constant on g −1 ({g(t)}). This means, that g is locally
constant.
Conversely let g be locally constant. Let U ⊂ Y be an arbitrary subset. (Remember, all of them are open, since Y is discrete.) We have to show, that
g −1 (U ) is open. Since g is locally constant, there exists for every t ∈ g −1 (U ) an
open set Vt ⊂ X such that t ∈ Vt and g is constant on Vt . It follows:
[
Vt
g −1 (U ) =
t∈g−1 (U)
So g −1 (U ) is the union of open sets, hence open.
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