Download Topology Midterm Exam November 25, 2015 1. Let X be a set and let T

Topology
Midterm Exam
November 25, 2015
1. Let X be a set and let T = {Ua | a ∈ I} be a nonempty collection of subsets of X.
(a) (8 points) State the definition when T is called a topology on X.
Solution: T = {Ua | a ∈ I} is called a topology on X if it satisfies the following conditions.
(a) 0/ and X ∈ T ;
(b) For any subset J of I, we have ∪a∈JUa ∈ T ;
(c) For any finite subset {U1 , . . . ,Un } of T , we have ∩ni=1Ui ∈ T .
(d) (4 points) State the definition of a base B for the topology T .
Solution: B is called a base for the topology T if it satisfies the following conditions.
(a) B ⊆ T ;
(b) For each U ∈ T , there exists {Ua | a ∈ I} ⊆ B such that U = ∪a∈IUa .
(c) (4 points) Let X be a topological space with topology T . State the definition when X is called a
Hausdorff space.
Solution: X is called a Hausdorff space if for each pair of distinct points x 6= y ∈ X, there exist
a pair of open sets U,V ∈ T such that x ∈ U, y ∈ V and U ∩V = 0.
/
2. Let A, B be subsets of a topological space X. Prove that
(a) (10 points) A ∪ B = A ∪ B.
0
Solution: A ∪ B = A ∪ B ∪ A ∪ B = A ∪ B ∪ A0 ∪ B0 = A ∪ A0 ∪ B ∪ B0 = A ∪ B,
0
where A ∪ B , A0 and B0 denotes the set of limit points of A ∪ B , A and B, respectively.
(b) (10 points) Int(A ∩ B) = Int(A) ∩ Int(B).
Solution: Since Int(A) ∩ Int(B) is an open
set contained in A ∩ B, we have Int(A) ∩ Int(B) ⊆
Int A ∩ B . Conversely, since Int A ∩ B is an open set contained in both A and B, we have
Int A ∩ B ⊆ Int(A) ∩ Int(B). Hence, we have Int A ∩ B = Int(A) ∩ Int(B).
3. (10 points) If A is a dense subset of a space X, and if O is open in X, show that O ⊆ A ∩ O.
Solution: Since A is dense and X \ O is closed, we have
A ∩ (X \ O) ⊆ A ∩ X \ O = X ∩ X \ O = X \ O = X \ O. (∗)
On the other hand, using (∗) and the assumption that A is dense,
O ∪ (X \ O) = X = A = (A ∩ O) ∪ (A ∩ (X \ O)) = (A ∩ O) ∪ (A ∩ (X \ O)) ⊆ (A ∩ O) ∪ (X \ O).
Thus, we have
O ∪ (X \ O) ⊆ (A ∩ O) ∪ (X \ O),
Topology
Midterm Exam (Continued)
Fall Semester 2015
and
O ⊆ (A ∩ O).
4. (10 points) If A, B are disjoint closed subsets of a metric space, find disjoint open sets U, V such that
A ⊆ U and B ⊆ V.
Solution: By the Urysohn Lemma, there exists a continuous function f : X → R such that


if x ∈ A
−1
f (x) =
1
if x ∈ B


r ∈ (−1, 1)
if x ∈ X \ (A ∪ B)
Then U = f −1 ((−∞, 0)) and V = f −1 ((0, ∞)) are disjoint open subsets satisfying A ⊆ U and B ⊆ V.
5. Let A be a subset of a topological space X.
(a) (4 points) State the definition of a limit point of A.
Solution: A point x in X is called a limit point of A if every open neighborhood U of x
intersects A \ {x}, i.e. U ∩ (A \ {x}) 6= 0.
/
(b) (10 points) Prove that A is closed if and only if A contains all its limit points.
Solution: Consult the proof of Theorem (2.2) on page 29, or
(⇒) If A is closed, its complement X \ A is open. For each x 6= A, since X \ A is an open
neighborhood of x such that (X \ A) ∩ (A \ {x}) ⊆ (X \ A) ∩ A = 0,
/ x is not a limit point of A.
Therefore, A contains all its limit point.
(⇐)Conversely, suppose A contains all its limit points and let x ∈ X \ A. Since x is not a limit
point of A there is an open neighborhood U of x such that U ∩ A = U ∩ (A \ {x}) = 0/ which
implies that U ⊂ X \ A and X \ A is open. Therefore, A is closed.
Alternative Solution: Since A = A ∪ A0 ,
A is closed ⇔ A = A = A ∪ A0 ⇔ A0 ⊆ A.
6. (10 points) Let f : X → Y be an onto continuous function and assume that X is compact. Show that Y
is compact.
Solution: Consult the proof of Theorem (3.4) on page 47.
7. Let X be a topological space and let the diagonal map ∆ : X → X × X be defined by ∆(x) = (x, x) for
x ∈ X.
Topology
Midterm Exam (Continued)
Fall Semester 2015
(a) (10 points) Show that the diagonal map ∆ : X → X × X is continuous.
Solution: Let W be an open neighborhood of (x, x) ∈ X × X. Since W is a union of sets of
the form U ×V, where U,V are open neighborhood of x ∈ X, ∆−1 (W ) is a union of sets of the
form ∆−1 (U ×V ) = U ∩V which is open in X. Thus, ∆−1 (W ) is open in X and ∆ : X → X × X
is continuous.
(b) (10 points) Prove that X is Hausdorff if and only if ∆(X) = {(x, x) | x ∈ X} is closed in X × X.
Solution: X is Hausdorff
⇔ For each (x, y) ∈
/ ∆(X), i.e. x 6= y, there exist disjoint open neighborhoods U and V of x and
y, respectively.
⇔ For each (x, y) ∈ X × X \ ∆(X), there exists an open neighborhood U ×V ⊂ X × X \ ∆(X).
⇔ X × X \ ∆(X) is open.
⇔ ∆(X) is closed in X × X.