Download 5 Homework

Math 4181
Name:
Dr. Franz Rothe
December 4, 2014
14FALL\4181_fall14h50.tex
Homework has to be turned in this handout.
For extra space, use the back pages, or blank pages between.
The homework can be done in groups up to four students
due November 31th
5
Homework
10 Problem 5.1. Let (X, I) and (Y, J ) be topological spaces and
f : X 7→ Y be any mapping. We know that a function is continuous if and only if
(3) the preimage f −1 (B) ⊆ X is closed for each closed set B ⊆ Y .
Prove that the function is continuous if and only if
(b) f −1 (C) ⊆ f −1 (C) holds for each set C ⊆ Y .
Proof of ”(3) ⇒ (b)”. The answer is left to the student.
Proof of ”(b) ⇒ (3)”. The answer is left to the student.
1
10
10
10
10
10
10
10
10
90
Definition 1. A subset A ⊆ X of a topological space X is called dense iff A = X.
10 Problem 5.2. Convince yourself that a set A is dense if and only if either
(i) Any nonempty open set contains some point of A: O 6= ∅ ⇒ A ∩ O 6= ∅; or
(ii) the complement has empty interior: int(X \ A) = ∅.
2
10 Problem 5.3. Let the subset A ⊆ X of a topological space X be open and
dense, and the subset B ⊆ X be dense. Prove that the intersection A ∩ B is dense.
Proof. Let O 6= ∅ be any open nonempty set. Since the set A is dense, by item (i) of
the last problem we conclude ?
.
Moreover, this is an open set. Using item (i) once more, now for the dense set B,
we conclude that the intersection ?
is nonempty.
The entire paragraph implies, again via item (i) that the set A ∩ B is dense.
10 Problem 5.4. Prove that the finite intersection of open dense sets is open
and dense.
3
10 Problem 5.5. Let X, Y be topological spaces, A ⊆ X be a dense subset, and
f : X 7→ Y a continuous function onto Y . Prove that f (A) is dense in Y .
4
10 Problem 5.6. Let X be a topological space and A ⊆ X be any subset. For
this problem, I find it convenient to denote the interior by int(A) and the closure by
cl(A). Prove the inclusions
int(A) ⊆ int(cl(int(A))) ⊆
int(cl(int(cl(A))))
⊆ cl(int(cl(A))) ⊆ cl(A)
cl(int(cl(int(A))))
10 Problem 5.7. Show that by applying the operations of interior and inclusion
successively to one set, no more than the six sets as in problem 5.6 can be obtained.
Especially
cl(int(cl(int(cl(A))))) = cl(int(cl(A))) and
int(cl(int(cl(int(A))))) = int(cl(int(A)))
hold for all subsets A ⊆ X.
5
10 Problem 5.8. Show that by applying the operations of interior and inclusion
successively to an appropriate set S, indeed the six sets as in problem 5.6 turn out to be
all distinct.
Let
o
n
A := ( pq − 2−2q , pq + 2−2q : 1 ≤ p < q and q ≥ 2 ⊂ (0, 1)
be an open dense set in the space [0, 1]. The Lebesgues measure of A is less than
X
X
|A| <
(q − 1)21−2q <
21−q = 1
q≥2
q≥2
since q < 2q . Hence the complement [0, 1] \ A is a closed, nonempty set. The set does
not contain any rational numbers and hence int([0, 1] \ A) = ∅. Let S be the union of
the four sets in the first column. The disjoint unions of the remaining columns give six
different sets.
S
[0, 1]
int(S)
(0, 1)
int(cl(int(S)))
(0, 1)
int(cl(int(cl(S))))
?
(2, 3) ∩ Q
?
?
?
(4 + A)
(4 + A)
?
?
[6, 7] \ (6 + A)
?
?
∅
S
[0, 1]
cl(int(cl(int(S)))) cl(int(cl(S)))
[0, 1]
?
cl(S)
[0, 1]
(2, 3) ∩ Q
?
?
?
(4 + A)
?
?
?
[6, 7] \ (6 + A)
∅
?
[6, 7] \ (6 + A)
6
Definition 2 (Connectedness). A topological space X is called connected if and only
if the only clopen sets are ∅ and X. A subset Y ⊆ X is called connected if and only if
the space Y is connected in the relative topology.
Definition 3 (Separation). A pair A, B is called a separation of a subset Y ⊆ X of
the topological space X iff
(5.1)
A 6= ∅ , B 6= ∅ , A ∪ B = Y , A ∩ B = ∅ , A ∩ B = ∅
Proposition 1. Equivalent are
(i) The topological space X is connected.
(ii) The space X has no separation.
(iii) The space X is not the union of two disjoint nonempty open sets.
(iv) The space X is not the union of two disjoint nonempty closed sets.
10 Problem 5.9. Convince yourself that this proposition is valid.
Proof. The answer is left to the student.
7
Proposition 2. Equivalent are
(i) The subset Y ⊆ X of the topological space X is connected.
(ii) The subset Y has no separation.
(iii) It is impossible that the subset Y is the union of two sets A and B for which there
exist open sets P and Q such that
∅=
6 A⊆P, ∅=
6 B ⊆ Q, A ∪ B = Y , P ∩ Q ∩ Y = ∅
(iv) It is impossible that the subset Y is the union of two sets A and B for which there
exist closed sets F and G such that
∅=
6 A⊆F, ∅=
6 B ⊆ G, A ∪ B = Y , F ∩ G ∩ Y = ∅
Proof of ”(i) ⇔ (ii)”. The closure operation for the subspace topology is given by clY (A) =
A ∩ Y . The subspace Y has a separation A, B in the relative topology iff
A 6= ∅ , B 6= ∅ , A ∪ B = Y , clY (A) ∩ B = ∅ , A ∩ clY (B) = ∅
but these conditions are ?
the original conditions (5.1).
The subspace Y is connected if and only if the subspace Y has ?
A, B in the relative topology, this happens if and only if the subset Y has no separation.
Proof of ”not (ii) ⇒ not (iii)”. Suppose the subset Y has a separation and Y = A ∪ B.
We define the open sets P := X \ B and Q := X \ A . Since
P ∩ Q = (X \ B) ∩ (X \ A) = X \ (B ∪ A) = X \ (A ∪ B) = X \ Y ⊆ X \ Y
we conclude P ∩ Q ∩ Y = ∅ as required.
Proof of ”not (iii) ⇒ not (ii)”. Suppose the subset Y is the union of two sets A and B
for which there exist open sets P and Q such that
∅=
6 A⊆P, ∅=
6 B ⊆ Q, A ∪ B = Y , P ∩ Q ∩ Y = ∅
From A ⊆ P and A ⊆ Y and P ∩ Q ∩ Y = ∅, we get
A⊆P ∩Y ⊆X \Q
and since Q is open, we conclude even A ⊆ X \ Q. Hence
A ∩ B ⊆ (X \ Q) ∩ Q = ∅
We conclude that ?
and similarly get ?
the pair A, B is a separation and item (ii) does not hold, as claimed.
8
Hence