Download Math 535 - General Topology Fall 2012 Homework 1 Solutions

Math 535 - General Topology
Fall 2012
Homework 1 Solutions
Definition. Let V be a (real or complex) vector space. A norm on V is a function k·k : V → R
satisfying:
1. Positivity: kxk ≥ 0 for all x ∈ V and moreover kxk = 0 holds if and only if x = 0.
2. Homogeneity: kαxk = |α|kxk for any scalar α and x ∈ V .
3. Triangle inequality: kx + yk ≤ kxk + kyk for all x, y ∈ V .
A normed vector space is the data (V, k·k) of a vector space V equipped with a norm k·k.
Problem 1.
Let (V, k·k) be a normed vector space. Define a function d : V × V → R by
d(x, y) := kx − yk.
Show that d is a metric on V , called the metric induced by the norm k·k.
Solution. We check the three properties of a metric.
1. Positivity:
d(x, y) = kx − yk ≥ 0 for all x, y ∈ V,
d(x, y) = 0 ⇔ kx − yk = 0
⇔x−y =0
⇔ x = y.
2. Symmetry:
d(y, x) = ky − xk
= k(−1)(x − y)k
= |−1|kx − yk
= kx − yk
= d(x, y).
3. Triangle inequality:
d(x, y) = kx − yk
= kx − z + z − yk
≤ kx − zk + kz − yk
= d(x, z) + d(z, y).
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Problem 2.
Denote by k·k2 the standard (Euclidean) norm on Rn , defined by
! 21
n
X
.
kxk2 :=
x2i
i=1
n
Now consider the function k·k1 : R → R defined by
n
X
kxk1 :=
|xi |.
i=1
a.
Show that k·k1 is a norm on Rn .
Solution. We check the three properties of a norm.
1. Positivity:
kxk1 =
n
X
|xi | ≥ 0 for all x ∈ Rn ,
i=1
n
X
kxk1 = 0 ⇔
|xi | = 0
i=1
⇔ |xi | = 0 for 1 ≤ i ≤ n
⇔ xi = 0 for 1 ≤ i ≤ n
⇔ x = 0.
2. Homogeneity:
n
X
kαxk1 =
|αxi |
i=1
n
X
=
|α||xi |
i=1
n
X
= |α|
|xi |
i=1
= |α|kxk1 .
3. Triangle inequality:
kx + yk1 =
n
X
|xi + yi |
i=1
≤
n
X
(|xi | + |yi |)
i=1
n
n
X
X
=
|xi | +
|yi |
i=1
i=1
= kxk1 + kyk1 .
2
Remark. The norms k·k1 and k·k2 are special cases of the so-called p-norm, for any real number
p ≥ 1 or p = ∞. See:
http://en.wikipedia.org/wiki/Lp_spaces#The_p-norm_in_finite_dimensions.
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b.
Find constants C, D > 0 satisfying
kxk2 ≤ Ckxk1
kxk1 ≤ Dkxk2
for all x ∈ Rn .
Solution. Claim: kxk2 ≤ kxk1 for all x ∈ Rn . In other words, we can take the constant
C = 1.
Proof 1. The claim is equivalent to
kxk22 ≤ kxk21
n
X
x2i ≤
i=1
n
X
!2
n
X
|xi |
i=1
x2i ≤
i=1
n
X
x2i +
i=1
0≤
X
|xi ||xj |
i6=j
X
|xi ||xj |
i6=j
which holds for all x ∈ Rn .
ith
z}|{
Proof 2. Write ei for the standard basis vector ei = (0, . . . , 0, 1 , 0, . . . , 0). Then we have
kxk2 = k
n
X
xi ei k2
i=1
≤
n
X
kxi ei k2
i=1
=
n
X
|xi |kei k2
i=1
=
n
X
|xi |
i=1
= kxk1 .
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For the bound kxk1 ≤ Dkxk2 , here are two solutions.
Solution 1: Crude bound. Noting |xi | ≤ kxk2 , we obtain
kxk1 =
n
X
|xi |
i=1
n
X
≤
kxk2
i=1
= nkxk2
so that we can take the constant D = n.
Solution 2: Better bound. The 1-norm can be expressed as a dot product
kxk1 =
n
X
|xi |
i=1
=
n
X
sign(xi )xi
i=1
=s·x
where s ∈ Rn is the vector with entries ±1 given by si = sign(xi ). Let’s say sign(0) = +1 by
convention.
The Cauchy-Schwarz inequality yields
kxk1 = s · x
≤ ksk2 kxk2
√
= n kxk2
so that we can take the constant D =
√
n.
Remark. In fact, this is the best possible bound, because equality
is achieved by the vector
√
u = (1, 1, . . . , 1). Indeed, we have kuk1 = n whereas kuk2 = n.
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Definition. Two norms k·k1 and k·k2 on a vector space V are equivalent if they can be
compared as in Problem 2b.
Definition. Two metrics d1 and d2 on a set X are topologically equivalent if for every
x ∈ X and > 0, there is a δ > 0 satisfying
d1 (x, y) < δ ⇒ d2 (x, y) < d2 (x, y) < δ ⇒ d1 (x, y) < .
In other words, the identity function (X, d1 ) → (X, d2 ) is a homeomorphism.
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Problem 3.
metrics.
Show that equivalent norms on a vector space V induce topologically equivalent
Solution. Let k·k1 and k·k2 be two norms on V , and let C, D > 0 be constants satisfying
kxk2 ≤ Ckxk1
kxk1 ≤ Dkxk2
for all x ∈ V . Thus the induced metrics satisfy
d2 (x, y) = kx − yk2
≤ Ckx − yk1
= Cd1 (x, y)
and likewise d1 (x, y) ≤ Dd2 (x, y) for all x, y ∈ V .
Take K := max{C, D}. Let x ∈ V and > 0 be given. Pick δ :=
chosen uniformly, independently of x). Then we have
d1 (x, y) < δ ⇒ d2 (x, y) ≤ Cd1 (x, y)
< Cδ
≤ Kδ
=
and likewise
d2 (x, y) < δ ⇒ d1 (x, y) ≤ Dd2 (x, y)
< Dδ
≤ Kδ
= .
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K
(which in this case can be
Problem 4. (Bredon Prop. I.1.3) Show that topologically equivalent metrics induce the
same topology (which explains the terminology). In other words, if d1 and d2 are topologically
equivalent metrics on X, then a subset U ⊆ X is open with respect to d1 if and only if it is
open with respect to d2 .
Solution. By symmetry of the situation, it suffices to show one side of the equivalence. Let
U ⊆ X be a subset which is open with respect to d1 . We want to show that U is open with
respect to d2 .
Let x ∈ U . Because U is open with respect to d1 , there is an -ball (in the d1 metric) around
x entirely contained in U , i.e. B1 (x) ⊆ U , for some > 0.
Because d2 is topologically equivalent to d1 , there is some δ > 0 satisfying Bδ2 (x) ⊆ B1 (x). In
particular we have
Bδ2 (x) ⊆ B1 (x) ⊆ U
so that U is open with respect to d2 .
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Definition. Let (X, T ) be a topological space. A subset C ⊆ X is closed (with respect to T )
if its complement C c := X \ C is open (with respect to T ).
Problem 5.
Show that the collection of closed subsets of X satisfies the following properties.
1. The empty subset ∅ and X itself are closed.
2. An arbitrary intersection of closed subsets is closed: Cα closed for all α implies
closed.
T
α
Cα is
3. A finite union of closed subsets is closed: C, C 0 closed implies C ∪ C 0 is closed.
Solution.
1. The empty set ∅ is closed because its complement ∅c = X is open.
The entire set X is closed because its complement X c = ∅ is open.
T
2. Let Cα be closed for all α and consider the intersection α Cα . Its complement
!c
\
[
Cα =
Cαc
α
α
is a union of open sets, therefore open. Thus
T
α
Cα is closed.
3. Let C, C 0 be closed and consider the union C ∪ C 0 . Its complement
c
(C ∪ C 0 ) = C c ∩ C 0c
is a finite intersection of open sets, therefore open. Thus C ∪ C 0 is closed.
Remark. In fact, a collection of subsets satisfies these properties if and only if their complements form a topology. Moreover, open subsets and closed subsets determine each other.
Upshot: One might as well define a topology via a collection of “closed subsets” satisfying the
three properties above. Their complements then form the topology in question.
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Problem 6. Let X be a set. Consider the collection of cofinite subsets of X together with
the empty subset:
Tcofin := {U ⊆ X | X \ U is finite} ∪ {∅}.
a.
Show that Tcofin is a topology on X, called the cofinite topology.
Solution. By problem 5, it suffices to check that the collection of subsets
{F ⊆ X | F is finite} ∪ {X}.
satisfies the axioms of closed subsets.
1. The empty set ∅ is finite, and therefore belongs to the collection, whereas X belongs to
the collection by definition.
2. T
Let Cα be a family of subsets that are either finite or all of X. Then the intersection
α Cα is finite (if at least one Cα is finite) or all of X (if all Cα are X).
3. Let C, C 0 be subsets that are either finite or all of X. Then the union C ∪ C 0 is finite (if
both C and C 0 are finite) or all of X (if at least one of C or C 0 is X).
b. Assuming X is infinite, show that the cofinite topology on X cannot be induced by a
metric on X.
Solution. Let Br (x) be an open ball not containing all of X. If the complement Br (x)c is
infinite, then we are done: we have found an open subset which is not cofinite.
If the complement Br (x)c is finite, then pick a point y ∈ Br (x)c and choose a radius > 0
small enough so that the open ball B (y) does not intersect Br (x); any value ≤ d(x, y) − r
will do. Then B (y) is finite, and thus its complement B (y)c is infinite, since X is infinite. We
are done: we have found an open subset B (y) which is not cofinite.
Slightly different solution. If X is infinite, then:
• Any cofinite subset of X is infinite;
• Any cofinite subset and any infinite subset must intersect.
In particular, any two non-empty open subsets of X intersect.
However, in any metric space containing at least two points, we can find non-empty open subsets
that do not intersect. Indeed, pick distinct points x and y, and take small enough open balls
Br (x) and Br (y) around them; any radius r ≤ d(x,y)
will guarantee Br (x) ∩ Br (y) = ∅.
2
Therefore the cofinite topology on X cannot be induced by a metric.
Remark. In a few lectures, we will say that such a topology is not Hausdorff, hence not metrizable.
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Definition. Let X be a set.
• The discrete topology on X is the one where all subsets are open:
Tdisc = P(X) = {U ⊆ X}.
• The anti-discrete (or trivial) topology on X is the one where only the empty subset
and X itself are open:
Tanti = {∅, X}.
Problem 7.
a.
Let D be a discrete topological space and A an anti-discrete topological space.
Describe all continuous maps f : D → X, where X is an arbitrary topological space.
Solution. The condition that f −1 (V ) be open in D for any open V ⊆ X is automatically
satisfied, since every subset of D is open. Thus every function f : D → X is continuous.
Remark. We will come back to the question of mapping into a discrete space when discussing
the notion of connectedness.
b.
Describe all continuous maps f : X → A, where X is an arbitrary topological space.
Solution. For any function f : X → A, we have f −1 (∅) = ∅ and f −1 (A) = X, both of which
are open in X. Since ∅ and A are the only open subsets of A, every function f : X → A is
continuous.
c.
Describe all continuous maps f : A → X, where X is a metric space.
Solution. Let f : A → X be a continuous. We claim that f is a constant function.
Pick some a ∈ A and look at its value x := f (a) ∈ X. For any other point y ∈ X, pick an open
subset V ⊆ X satisfying x ∈ V but y ∈
/ V . Since f is continuous, the preimage f −1 (V ) is open
in A. The condition f (a) = x ∈ V yields a ∈ f −1 (V ) so that f −1 (V ) is non-empty and must
therefore be all of A, the only non-empty open subset of A.
Now the condition f (A) ⊆ V implies that f never takes the value y ∈
/ V . Since y was an
arbitrary point distinct from x, we conclude that f is the constant function with value x.
Remark. The proof still holds whenever X is a T1 space, a property that will be discussed in a
few lectures.
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Problem 8.
a.
Let f : X → Y be a function between topological spaces, and let x ∈ X.
Show that the following conditions (defining continuity of f at x) are equivalent.
1. For all neighborhood N of f (x), there is a neighborhood M of x such that f (M ) ⊆ N .
2. For all open neighborhood V of f (x), there is an open neighborhood U of x such that
f (U ) ⊆ V .
3. For all neighborhood N of f (x), the preimage f −1 (N ) is a neighborhood of x.
Solution. (1 ⇒ 2) Let V be an open neighborhood of f (x). Since V is in particular a
neighborhood of f (x), the assumption (1) guarantees that there is a neighborhood M of x
such that f (M ) ⊆ V . Let U be an open of X satisfying x ∈ U ⊆ M . Then U is an open
neighborhood of x satisfying f (U ) ⊆ f (M ) ⊆ V .
(2 ⇒ 3) Let N be a neighborhood of f (x). Let V be an open of Y satisfying f (x) ∈ V ⊆ N .
Then we have x ∈ f −1 (V ) ⊆ f −1 (N ). By the assumption (2), f −1 (V ) is an open neighborhood
of x, so that f −1 (N ) is a neighborhood of x.
(3 ⇒ 1) Let N be a neighborhood of f (x) and take M := f −1 (N ). By the assumption (3),
f −1 (N ) is a neighborhood of x, and moreover it satisfies f (f −1 (N )) ⊆ N .
b. Find an example of function f : X → Y between metric spaces which is continuous at a
point x ∈ X, but there is an open neighborhood V of f (x) such that the preimage f −1 (V ) is
not an open neighborhood of x.
Upshot: The description “preimage of open is open” is really about global continuity, not
pointwise continuity (or even local continuity).
Solution. Consider the “step” function f : R → R defined by
(
0 if x ≤ 43
f (x) =
1 if x > 43.
Then f is continuous at x = 0 (in fact everywhere except at 43). However, take the open
neighborhood (− 21 , 12 ) of f (0) = 0. Its preimage under f is
1 1
−1
f
(− , ) = (−∞, 43]
2 2
which is not open in R.
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Problem 9.
Let X be a topological space and B a collection of open subsets of X.
a. Show that B is a basis for the topology of X if and only if for every open subset U ⊆ X
and x ∈ U , there is a B ∈ B satisfying x ∈ B ⊆ U .
Solution. (⇒) Assume B is a basis for the topology. Let U ⊆ X be open
S (WLOG non-empty)
and x ∈ U . Because B is a basis, U can be written as a union U = α Bα for some family of
subsets Bα ∈ B. Thus x is in at least one of those subsets Bαx , yielding x ∈ Bαx ⊆ U .
(⇐) To show that B is a basis, there are two things to check.
1) Every union of members of B is open in X. This is automatic, because each B ∈ B was
assumed to be open.
2) Let U ⊆ X be open (WLOG non-empty). We want to show that U is a union of subsets
in the collection B. By assumption
S on B, for each x ∈ U , there is some Bx ∈ B satisfying
x ∈ Bx ⊆ U . Thus we have U = x∈U Bx .
b.
Assuming X is a metric space, show that the collection of open balls
B = {B 1 (x) | x ∈ X, n ∈ N}
n
is a basis for the topology of X.
Solution. We use the criterion from part (a).
Let U ⊆ X be open and x ∈ U . Then there is some radius r > 0 such that the open ball of
radius r centered at x is contained within U , i.e. Br (x) ⊆ U .
Pick n large enough so that
1
n
≤ r. Then we have x ∈ B 1 (x) ⊆ Br (x) ⊆ U .
n
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Problem 10.
a.
Let X be a set and S a collection of subsets of X.
Show that the collection
T :=
( n
α
[\
)
Sα,i | Sα,i ∈ S
α i=1
of (arbitrary) unions of finite intersections of members of S is a topology on X.
Solution. We check the properties of a topology.
1. Because unions indexed by the empty family are allowed, the empty set ∅ =
T.
S
∅
Sα is in
T
Because intersections indexed by the empty family are allowed, the entire set X = ∅ Sα
is in T .
S T α
S Tnβ
2. Finite intersections of members of T are in T . Let U = α ni=1
Sα,i and V = β j=1
Sβ,j
be members of T . Their intersection is
!
!
nβ
nα
[\
[\
U ∩V =
Sα,i ∩
Sβ,j
α i=1
=
nα
[ \
α,β
β j=1
Sα,i ∩
i=1
nβ
\
!
Sβ,j
j=1
which is in T since each Sα,i and Sβ,j is in S.
3. Arbitrary unions of members of T are in T . Let {U γ =
members of T . Then their union is


(γ)
α
[
[ [ n\
γ 

Uγ =
Sα,i
γ
γ
α i=1

=
[
(γ)
n\
α

γ,α
which is in T .
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i=1

γ 
Sα,i
S Tn(γ)
α
α
i=1
γ
}γ be a family of
Sα,i
b. Show that T is the topology TS generated by S. In other words: T contains S and any
other topology T 0 containing S must satisfy T ≤ T 0 .
Solution.
1. T contains S: Any S ∈ S can be viewed as the union of one set which is the intersection
of one set, namely S itself, which is in S. Therefore we have S ⊆ T .
2. Let T 0 be
S. For any family of subsets Sα,i ∈ S, the finite interT αa topology containing
section ni=1
Sα,i is in T 0 , since T 0 is a topology. Moreover, the union
nα
[\
Sα,i
α i=1
is also in T 0 , since T 0 is a topology. Thus we have T ⊆ T 0 , as claimed.
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