Download Solutions to selected exercises

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University
of Bristol [2014]
SOLUTIONS TO HOMEWORK FOR TOPICS IN MODERN GEOMETRY
Solution (Exercise 2.1).
a. Let Fα (α ∈ A) be closed sets. Then for each α ∈ A there exists an open set Uα such that
Fα = X \ Uα . By de Morgan’s laws, we have
!
\
[
(X \ Uα ) = X \
Uα .
α∈A
α∈A
This is the complement of an open set, since arbitrary unions of open sets are open. Hence
the intersection is closed.
Given a finite number of closed sets Fi = X \ Ui , a similar argument works using the
identity
F1 ∪ · · · ∪ Fn = X \ (U1 ∩ · · · ∩ Un ).
T
b. Let {Fi }i∈I denote the collection of all closed sets containing A. Then F := i∈I Fi is closed
by the previous part of the exercise.
To see T
that A ⊂ F , let x ∈ A. Then for any i ∈ I we have A ⊂ Fi , hence x ∈ Fi . It follows
that x ∈ i∈I Fi , as required.
c. For the ‘only if’ or =⇒ direction, let x ∈ A and let N be a neighbourhood of x. Then
there exists an open set U such that x ∈ U ⊂ N . If U ∩ A = ∅, then X \ U is a closed set
containing A, therefore x ∈ A ⊂ X \ U . But this contradicts the fact that x ∈ U . Hence we
must have U ∩ A 6= ∅, and so N ∩ A 6= ∅.
Conversely, for the ‘if’ or ⇐= direction, suppose that for any neighbourhood N of x we
have A ∩ N 6= ∅. To show that x ∈ A we must show that x lies in every closed set F with
A ⊂ F . If x ∈
/ F , then X \ F is an open neighbourhood of x, hence A ∩ (X \ F ) 6= ∅, which
contradicts A ⊂ F .
Solution (Exercise 2.6). Let Uα (α ∈ A) be an arbitrary collection of open sets in τ . Then
for all α ∈ A and i ∈ I we have Uα ∈ τi . Since each τi is closed under arbitrary unions
[
Uα ∈ τi .
α∈A
As this holds for all i ∈ I, we deduce that
[
\
Uα ∈
τi = τ.
α∈A
i∈I
The other two topology axioms are checked similarly.
Solution (Exercise 2.9).
a. Let (X, d) be a metric space. A set U ⊂ X is open if and only if for any x ∈ U there exists
ε(x) > 0 such that Bε(x) (x) ⊂ U . Hence
[
U=
Bε(x) (x).
x∈U
Thus every open set in X can be written as a union of open balls. The open balls in R are
precisely open intervals (x − ε, x + ε).
Date: Autumn 2014.
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c
University
of Bristol [2014]
b. Consider the subfamily of open balls Bε (x) ⊂ Rn whose radii are rational and whose centre
x = (x1 , . . . , xn ) has rational coordinates. Notice that the set of such open balls is in bijective
correspondence with Qn+1 , which is a Cartesian product of countable sets, and hence is itself
countable. It remains to show these balls form a base for the topology.
It suffices (why?) to show that for any x ∈ Rn and any ε > 0 there exists a rational vector
q ∈ Qn and m ∈ N such that
x ∈ B1/m (q) ⊂ Bε (x).
(1)
Let m ∈ N satisfy 1/m < ε/2. Since every real number can be approximated arbitrarily
closely by a rational, for each xi there exists qi ∈ Q such that
1
|xi − qi | < √ .
m n
It follows that |x − q| < 1/m < ε/2. Then by the triangle inequality, if z ∈ B1/m (q) we have
|z − x| ≤ |z − q| + |q − x| < ε/2 + ε/2 = ε.
Hence z ∈ Bε (x). This establishes (1).
Prove that a subset E of a topological space X is dense in X if and only if the closure of E
is equal to the whole space X, i.e.
E = X.
Solution (Exercise 2.9). E is dense in X if and only if for any non-empty open set U we
have E ∩ U 6= ∅. Equivalently, for any proper closed set F 6= X we have E 6⊂ F , i.e. the only
closed subset of X which contains E is X itself. This is true if and only if the intersection of
all closed sets in X containing E is equal to X itself.
Solution (Exercise 3.6). The map f is continuous if and only if for any open U ⊂ Y the
inverse image f −1 (U ) is open in the product topology. Since the open rectangles form a base
for the product topology, the set f −1 (U ) is open if and only if for any (x1 , x2 ) ∈ f −1 (U ) there
exist open sets Ui ⊂ Xi with xi ∈ Ui and such that U1 × U2 ⊂ f −1 (U ). The equivalence follows.
Solution (Exercise 3.7).
a. Let us prove the ‘only if’ or =⇒ direction. We first show that (x1 , x2 ) 7→ (x1 , x−1
2 ) is
continuous as a map G × G → G × G. Let U ⊂ X1 × X2 be open and (x1 , x−1
)
∈
U
.
By
the
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previous exercise, it suffices to show that there exist open sets Ui ⊂ G such that xi ∈ Ui and
U1 × U2−1 ⊂ U . Sine the open rectangles form a base for the product topology, there exist
open sets Vi such that (x1 , x−1
2 ) ∈ V1 × V2 ⊂ U . Since the inverse map is continuous, the set
−1
V2 is open. Taking U1 := V1 and U2 := V2−1 gives the required Ui .
The continuity of the division map now follows, since it equals the composition of the
map (x1 , x2 ) 7→ (x1 , x−1
2 ) with the product map (x1 , x2 ) 7→ x1 x2 , and the composition of two
continuous maps is continuous.
Conversely, let us suppose that the division map is continuous. We must show that the
inverse map and the product map are both continuous. If the inverse map is continuous,
then the argument given above shows that (x1 , x2 ) 7→ (x1 , x−1
2 ) is continuous. Composing
this with the division map yields the product map. Hence it suffices to prove continuity of
the inverse map.
We claim that if f : X1 × X2 → Y is continuous with X1 × X2 under the product topology,
then for any fixed x1 ∈ X1 , the function g : x2 7→ f (x1 , x2 ) is continuous as a map X2 → Y .
To see this, let U be an open subset of Y with g(x2 ) ∈ U . Then by the previous exercise,
there exist open neighbourhoods Ui = Ui (xi ) of xi such that f (U1 × U2 ) ⊂ U , in particular
g(U2 ) = f ({x1 } × U2 ) ⊂ U . It follows that g −1 (U ) is a union of the sets U2 (x2 ) as x2 ranges
over g −1 (U ).
Since the division map is continuous, we conclude that the function x2 7→ ex−1
2 is continuous as a map G → G, where e denotes the identity element. This is precisely the division
map.
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University
of Bristol [2014]
b. By part (a), G is a topological group if and only if the division map is continuous. By the
previous exercise, this is equivalent to the statement in (b).
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