Download Geometry and Topology I Klausur, October 30, 2012 Name:

Geometry and Topology I
Klausur, October 30, 2012
Name:
Answer the following 12 multiple choices questions by making a cross on the
correct answers. Only one of the alternative for each question is correct! Points are
not removed in case of a wrong answer.
1. Which of the following is not a metric space
(R, d), where d(x, y) = 0 if x = y, d(x, y) = 1 if x 6= y.
(C 0 ([0, 1]), d∞ ), where d∞ (f, g) := supx∈[0,1] |f (x) − g(x)|.
({0, 1}27 , d), where d(x, y) :=number of different digits.
(R2 , d), where d(x, y) := (x1 − y1 )2 + (x2 − y2 )2 .
2. Consider the metric space ]0, +∞[ with the metric d(x, y) = |x − y|. The set (1/n)n∈N is
open,
connected,
closed,
none of the above.
3. Let A be a complete subset of a metric space X. Which of the following alternatives is
correct?
A is always open.
A is always bounded.
A is always closed.
None of the above.
4. Let X = [0, 1]2 be equipped with the euclidean distance. Let Y ⊂ X and consider a function
f : Y → R. Then
if f is continuous and Y is closed, then f is uniformly continuous,
if f is continuous and Y is connected, then f is uniformly continuous,
if f is continuous and Y is open, then f is uniformly continuous,
none of the above.
5. Let X be a metric space. A set K ⊂ X is compact if and only if
it is open and bounded,
it is closed and bounded,
every Cauchy sequence has a limit,
none of the above.
6. Let X be a topological space. Let (An )n∈N and (Bn )n∈N be collections respectively of open
and closed subset of X. Which of the following statement is correct?
S
S
n Bn \ n An is always closed,
S
T
n An \ n Bn is always open,
T
S
n An \ n Bn is always closed,
S
T
n Bn \ n An is always open.
Geometry and Topology I
Klausur, October 30, 2012
7. Let X and Y be two topological spaces. A function f : X → Y is continuous
if and only if f (U ) is open for each U ⊂ X open,
if and only if f (U ) is closed for each C ⊂ X closed,
if and only if f (K) is compact for each K ⊂ X compact.
none of the above.
8. Which of the following spaces is not separable?
R with the discrete topology,
P
(l1 , d), where d((xn ), (yn )) := ∞
n=1 |xn − yn |,
the Cantor set as a subspace of R,
R with the trivial topology.
9. Let X and Y be two topological spaces and let f : X → Y be a continuous function.
f (K) is connected if K ⊂ X is connected,
f −1 (K) is connected if K ⊂ Y is connected,
f −1 (K) is compact if K ⊂ Y is compact,
none of the above.
10. Let p ∈ N be a fixed prime number. On Z consider the p−adic metric defined by dp (a, b) = 0,
if a = b, and dp (a, b) = p−n , if a − b = pn c, with p not dividing c. The space (Z, dp ) is
T1 but not T2 ,
T2 but not T3 ,
T3 but not T4 ,
T4 .
11. In Z consider the collection of sets (Un )22
n=0 , where Un := {23j + n : j ∈ Z}. This is a basis
for a topology τ on Z. The space (Z, τ ) is
T1 but not T2 ,
T2 but not T3 ,
T3 ,
compact.
We have defined the separation axiom T3 for a topological space X in the following way:
(a) singletons {x} are closed;
(b) for every closed set C ⊂ X and any point x 6∈ C there are disjoint open sets U, V with
x ∈ U and C ⊂ V .
The space (Z, τ ) above satisfies (b) but not (a). However, several authors define the axiom
T3 requiring only (b): under this convention the third answer would also be correct.
12. A nonempty topological space X is called irreducible if X = A ∪ B with A, B closed implies
that either A or B is empty. An irreducible space is
compact,
connected,
T2 ,
none of the above.
Page 2
Geometry and Topology I
Klausur, October 30, 2012
Solve the following two exercises.
Exercise 1
Let X be a topological space. Prove that X is disconnected if and only if it can be mapped
continuously onto two distinct points of a Hausdorff space.
Is this characterization valid also for sets which are not path connected?
Solution:
Let Y be a Hausdorff space which contains at least two distinct points y and y 0 .
(⇒) Suppose X is disconnected, then there exists V, V 0 ⊂ X nonempty, open and
disjoint such that V ∪ V 0 = X. Define f : X → {y, y 0 } ⊂ Y by
f (x) = y, if x ∈ V,
and f (x) = y 0 , if x ∈ V 0 .
Now notice that for each U ⊂ Y open, f −1 (U ) can only be equal to V , V 0 , X or
∅, which are all open. Hence f is continuous.
(⇐) Suppose there exists a continuous function f : X → {y, y 0 } ⊂ Y , such that there
are x, x0 ∈ X satisfying f (x) = y and f (x0 ) = y 0 . Since Y is T2 , there exist
disjoint open neighborhood Uy , Uy0 of y and y 0 respectively. Since f is continuous
f −1 (Uy ) =: Vx and f −1 (Uy0 ) =: Vx0 are open. Moreover Vx ∩Vx0 = ∅ and Vx ∪Vx0 =
X. Since x ∈ Vx and x0 ∈ Vx0 , they are both nonempty, and X is disconnected.
From the lecture we know that path-connectedness is not equivalent to connectedness,
hence the answer to the second question is no.
Page 3
Geometry and Topology I
Klausur, October 30, 2012
Exercise 2
Let c0 be the space of real sequences converging to 0, that is
n
o
c0 := (xn )n ⊂ R : lim xn = 0 .
n→∞
Let d : c0 × c0 → R be defined by
d((xn )n , (yn )n ) := sup |xn − yn |.
n∈N
Prove that (c0 , d) is a metric space. Is it complete? Is it compact?
Solution:
d is a metric. Since every convergent sequence is bounded, d((xn )n , (yn )n ) ∈ R, for every
(xn )n , (yn )n ∈ c0 . Moreover, d((xn )n , (yn )n ) is obviously nonnegative. Furthermore:
• d((xn )n , (yn )n ) = 0 ⇐⇒ |xi − yi | ≤ supn∈N |xn − yn | = 0 ∀i ∈ N
⇐⇒ xn = yn ∀n ∈ N ⇐⇒ (xn )n = (yn )n ;
• d((xn )n , (yn )n ) = supn∈N |xn − yn | = supn∈N |yn − xn | = d((yn )n , (xn )n );
• For any (xn )n , (yn )n , (zn )n ∈ c0 ,
d((xn )n , (yn )n ) = sup |xn − yn | ≤ sup |xn − zn | + sup |zn − yn |
n∈N
n∈N
n∈N
≤ sup |xn − zn | + sup |yn − zn | ≤ d((xn )n , (zn )n ) + d((zn )n , (yn )n ).
n∈N
b∈N
(k)
(c0 , d) is complete. Let (x(k) )k = ((xn )n )k be a Cauchy sequence in c0 . Since for every n
(j)
(k) (j)
|x(k)
n − xn |n ≤ d(x , x )
∀k, j ∈ N ,
(k)
(k)
(xn )k is a Cauchy sequence in R and so there is xn ∈ R such that xn → xn as k → ∞.
Consider the sequence x := (xn )n . We first show that
lim sup |xkn − xn | = 0
(1)
k→∞ n∈N
and then that x ∈ c0 , which thus shows x(k) → x. For any ε > 0 let N ∈ N such that
(j)
k (j)
sup |x(k)
n − xn | = d(x , x ) ≤ ε
∀k, j ∈ N .
n∈N
(k)
(k)
(j)
(k)
If k ≥ N we have |xn − xn | = limj |xn − xn | ≤ ε, which implies supn |xn − xn | ≤ ε. The
arbitrariness of ε shows therefore (1). We now show x ∈ c0 . Let δ > 0 be given and, by (1)
choose K ∈ N such that
sup |x(K)
− xn | < δ .
(2)
n
n
Since
x(K)
∈ c0 , there exists N ∈ N such that
|x(K)
n | < δ,
∀n > N .
(K)
(3)
(K)
By (2) and (3) we conclude that |xn | ≤ |xn − xn | + |xn | < 2δ for any n ∈ N . The
arbitrariness of δ shows that xn → 0, i.e. that x ∈ c0 .
Page 4
Geometry and Topology I
Klausur, October 30, 2012
(c0 , d) is not compact. Indeed consider the sequence (x(k) )k defined by
1 if n = k
(k)
xn :=
0 if n 6= k
(k)
Since |xn | = 0 for each n > k, x(k) ∈ c0 for each k ∈ N. However d(x(k) , x(l) ) = 1 for any
k 6= l. No subsequence of (x(k) )k can therefore be convergent.
Page 5