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Übungsblatt 6
MAT701 Geometrie-Topologie Herbstsemester 2012
Prof. Dr. Camillo De Lellis
Abgabe: Dienstag 06.11.2012 bis 13 Uhr
Die Übungsblätter werden am jeden Montag auf der Homepage
http://www.math.unizh.ch/hs12/mat701
publiziert.
Für eine Wertung sind mindestens die ersten beiden Aufgaben zu lösen. Generell soll der Herleitungsweg
von Resultaten übersichtlich und vollständig sein, und es wird gebeten, leserlich zu schreiben.
Bitte die Lösungen am Abgabetag bis spätestens 13 Uhr in den Fächern auf den Briefkästen im K–Stock
des Instituts für Mathematik, Bau 27 deponieren.
Exercise 1 (5 points)
Let X be an infinite set with the cofinite topology. Show that the product topology on X × X is
strictly finer than the cofinite topology on X × X.
Solution:
×
Let us denote by τcof the cofinite topology on X × X and by τcof
the product topology. First
×
×
we prove that τcof is finer than τcof , that is τcof ⊂ τcof . Indeed let U ∈ τcof , then U =
Sm
X × X \ j=1 {(zj , yj )}, for some m ∈ N. Notice that
X ×X \
m
[
{(zj , yj )} =
j=1
m n
m
[
[
o
(X \ {zj }) × X ∪ X × (X \ {yj }) =:
Uj
j=1
j=1
and Uj is the product of two open sets in X with the cofinite topology.
Now to prove that the inclusion is strict just notice that Uy := X × (X \ {y}), for some y ∈ X,
×
is an element of τcof
. However
|(X × X) \ Uy | = |X| = ∞
so that Uy ∈
/ τcof .
Exercise 2 (5 points)
Consider X := R × {0, 1} with the topology inherited by the euclidean one in R2 . Consider the
following equivalence relation:
(x, y) ∼ (x0 , y 0 )
⇐⇒ x = x0 , x, x0 6= 0
and (0, 1) ∼ (0, 1), (0, 0) ∼ (0, 0). Prove that the quotient space X/ ∼ with the quotient topology is
not Hausdorff.
1
Solution:
Let π : X → X/ ∼ be the projection onto the quotient, we are going to prove that π(0, 0) and
π(0, 1) do not have disjoint open neighborhoods, so that the quotient space cannot be T2 . Indeed
let U, V ⊂ X/ ∼ be open neighborhood of π(0, 0) and π(0, 1) respectively. Then π −1 (U ) and
π −1 (V ) are open in X. Therefore there exist ε0 , ε1 > 0, such that
(−ε0 , ε0 ) × {0} ⊂ π −1 (U )
and
(−ε1 , ε1 ) × {1} ⊂ π −1 (V ).
Notice that if π(x, 0) ∈ U , then π(x, 1) ∈ U , and viceversa. This implies that, if we set
ε := min{ε0 , ε1 }, then
(−ε, ε) × {0, 1} \ {(0, 1)} ⊂ π −1 (U )
and
(−ε, ε) × {0, 1} \ {(0, 0)} ⊂ π −1 (V ).
Since ε > 0, this implies that π −1 (U ) ∩ π −1 (V ) 6= ∅. Therefore U ∩ V 6= ∅, as desired.
Exercise 3 (5 points)
Prove that a topological space X is Hausdorff if and only if the diagonal ∆ := {(x, x) : x ∈ X} is
closed in X × X with the product topology.
Solution:
if Let x and y be two distinct points of X. Then (x, y) ∈ (X × X) \ ∆. Since ∆ is closed in
the product topology, (x, y) has an open neighborhood Ux × Vy ⊂ (X × X) \ ∆. Therfore
Ux × Vy ∩ ∆ = ∅ and so Ux ∩ Vy = ∅. Since x ∈ Ux and y ∈ Vy , we have proved that X is
Hausdorff.
only if Let (x, y) ∈
/ ∆. Then the points x and y are distinct. Since X is T2 , there exist open
neighborhoods Ux and Vy of x and y respectively, such that Ux ∩ Vy = ∅. Therefore
(Ux × Vy ) ∩ ∆ = ∅. It follows that (x, y) ∈ Ux × Vy ⊂ (X × X) \ ∆, so that ∆ is closed.
Exercise 4 (8 points)
Let X be a topological space. For any x ∈ X let C(x) be the connected component of X containing
x. Recall from the lecture that C(x) is the biggest connected subset of X containing x, i.e. it is the
union of all connected subsets of X which contain x. Moreover, it was proved in the lecture that X
is the disjoint union of its connected components.
(a) Prove that the connected components are closed.
(b) Prove that if X has finitely many connected components, then they are open.
(c) What happens in (b) if the components are not finitely many?
Solution:
(a) It is enough to prove that
E ⊂ X connected
⇒
E connected.
(1)
Indeed by (1) and the fact that C(x) is the biggest connected set containing x, it follows
that C(x) ⊂ C(x), and so, since C(x) ⊂ C(x), we have C(x) = C(x), that is the claim.
Let us prove (1). Let U, V be two open sets such that U ∩ V = ∅ and E ⊂ U ∪ V . We are
going to prove that E is contained in only one of them. Indeed clearly E ⊂ U ∩ V , and
since E is connected, it is contained in only one of them, say E ⊂ U . Notice that X \ V
is closed, and E ⊂ X \ V , therefore E ⊂ X \ V , that is E ⊂ U .
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(b) Recall that X is the disjoint union of its connected components. Therefore, if we assume
that the connected components Ci are finitely many, say n, we can write
X=
n
[
Ci ,
the union beeing disjoint.
j=1
Therefore
Ci = X \ (C1 ∪ · · · ∪ Ci−1 ∪ Ci+1 ∪ · · · ∪ Cn ).
Since by (c) each Cj is closed, C1 ∪ · · · ∪ Ci−1 ∪ Ci+1 ∪ · · · ∪ Cn is closed, and so Ci is open
for each i = 1, . . . , n.
(c) If the components are not finitely many, claim (c) might be false. To see this just consider
Q as a subset of R with the euclidean topology. Then its connected components are all
its infinitely many points, which are closed but not open.
Exercise 5 (8 points)
Let (Ai )i be a decreasing sequence of closed connected subsets of R2 , that is
Ai ⊂ R2 and Ai+1 ⊂ Ai for every i ∈ N.
(a) Is
T∞
i=1
Ai always a connected set?
(b) What if in addition we assume A1 to be bounded?
Solution:
(a) The answer is negative. As a counterexample consider the sequence of sets (An )n defined
for each n ∈ N by
An := (x, y) ∈ R2 : x ≥ 0, y ∈ {0, 1} ∪ (x, y) ∈ R2 : x ∈ N, x ≥ n, y ∈ [0, 1] .
Each one of them is path connected, and so connected. Moreover they are closed and
Ai+1 ⊂ Ai for every i ∈ N. However
∞
\
Ai = (x, y) ∈ R2 : x ≥ 0, y = 0 ∪ (x, y) ∈ R2 : x ≥ 0, y = 1
i=1
which is disconnected.
(b) The answer is positive. Indeed notice that Ai T
⊂ A1 for each i, and so by Heine-Borel
∞
theorem each Ai is compact. Now suppose A := i=1 Ai is not connected, then there exist
2
U, V nonempty, open and disjoint subset of R , such that A = U ∪ V and A ∩ V, A ∩ U 6= ∅.
We claim that
there exists N ∈ N such that AN ⊂ U ∪ V.
(2)
Indeed suppose not, set A0n := An \ (U ∪ V ), for each n ∈ N. Since An is compact and
U, V are open, A0n is compact; also A0n+1 ⊂ A0n . Since we are assuming that (2) does not
hold, we have A0n 6= ∅, for each n ∈ N. Since R2 is a complete metric space, we have
∅=
6
∞
\
n=1
A0n =
∞
\
An \ (U ∪ V ) = ∅
n=1
which is a contradiction.
Now observe that by (2), A0n ⊂ U ∪ V , with U, V nonempty, open and disjoint. Moreover
AN ∩ U ⊃ A ∩ U 6= ∅ and AN ∩ V ⊃ A ∩ V 6= ∅, so that U and V disconnect AN , which
is a contradiction to the assumption that A is not connected.
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