Download 5 Solution of Homework

Math 4181
Name:
Dr. Franz Rothe
December 4, 2014
14FALL\4181_fall14h50s.tex
5
Solution of 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)”. Let C ⊆ Y be any set. Hence C is closed and by assumption (3),
the preimage f −1 (C) is closed.
C ⊆ C by closure property (CL3).
f −1 (C) ⊆ f −1 (C)
f −1 (C) ⊆ f −1 (C) = f −1 (C)
by closure property (CL4’), and since the set f −1 (C) is closed.
Proof of ”(b) ⇒ (3)”. Let B ⊆ Y be any closed set. Using the assumption (b), we get
f −1 (B) ⊆ f −1 (B) ⊆ f −1 (B) = f −1 (B)
Hence f −1 (B) = f −1 (B), and f −1 (B) is closed, as claimed.
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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) = ∅.
Answer. Equivalent are
A is dense in X ⇔ A = X ⇔ int(X \ A) = X \ A = ∅
⇔ [O ⊆ X \ A and O is open in X ⇒ O = ∅] ⇔ [O ∩ A = ∅ and O is open in X ⇒ O = ∅]
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 A ∩ O 6= ∅.
Moreover, A ∩ O 6= ∅ is an open set. Using item (i) once more, now for the dense
set B, we conclude that the intersection B ∩ (A ∩ O) 6= ∅ is nonempty.
Obviously, B ∩ (A ∩ O) = (A ∩ B) ∩ O. 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.
Answer. For two open dense sets A and B the assertion follows from the previous
problem: the intersection A ∩ B is dense by the previous problem, and the intersection
of two open sets is always open.
For all natural n ≥ 1, we assume the sets An are open dense. The general case is
proved by induction. Here is the induction step ”n → n + 1”. We have already shown
that the intersection ∩1≤i≤n Ai is open dense. Too, the set An+1 is open dense. Hence
the first paragraph implies that
\
\
Ai
Ai ∩ An+1 =
1≤i≤n+1
1≤i≤n
is open and dense.
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 .
Answer. We know from proposition ?? that f (A) ⊆ f (A). Since A is dense and the
function f maps onto Y , we know A = X and f (X) = Y . Hence
Y = f (X) = f (A) ⊆ f (A) ⊆ f (X) = Y
implies f (A) = Y and hence f (A) is dense in Y .
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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))))
Answer. Let B := cl(A). Since int(B) ⊆ B, the closure property (CL4’) implies
cl(int(B)) ⊆ cl(B) = B, which yields the inclusion cl(int(cl(A))) ⊆ cl(A). Taking
complements yields int(A) ⊆ int(cl(int(A))). The remaining parts are even easier to
check.
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.
Answer.
cl(int(cl(B))) ⊆ cl(B) with B:=int(cl(A)) implies
cl(int(cl(int(cl(A))))) ⊆ cl(int(cl(A)))
int(C) ⊆ int(cl(int(C))) with C:=cl(A) implies
int(cl(A)) ⊆ int(cl(int(cl(A))))
cl(int(cl(A))) ⊆ cl(int(cl(int(cl(A))))) both inclusions imply
cl(int(cl(A))) = cl(int(cl(int(cl(A)))))
The remaining parts are even easier to check.
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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))) int(cl(int(cl(S))))
(0, 1)
(0, 1)
(2, 3) ∩ Q
∅
∅
(4 + A)
(4 + A) (4, 5)
(4, 5)
[6, 7] \ (6 + A)
∅
∅
S
[0, 1]
cl(int(cl(int(S)))) cl(int(cl(S))) cl(S)
[0, 1]
[0, 1]
[0, 1]
(2, 3) ∩ Q
∅
[2, 3]
[2, 3]
(4 + A)
[4, 5]
[4, 5]
[4, 5]
[6, 7] \ (6 + A)
∅
∅
[6, 7] \ (6 + A)
(2, 3)
∅
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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
A 6= ∅ , B 6= ∅ , A ∪ B = Y , A ∩ B = ∅ , A ∩ B = ∅
(5.1)
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.
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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 equivalent to the original conditions (5.1).
The subspace Y is connected if and only if the subspace Y has no separation 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 A ∩ B = ∅ and similarly get A ∩ B = ∅. Hence the pair A, B is a
separation and item (ii) does not hold, as claimed.
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