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KOÇ UNIVERSITY, Spring 2013, MATH 402/571, QUIZ-1, February 21, 60 Minutes
SOLUTIONS BY KARATUG OZAN BIRCAN
Problem 1(10 points): Let X be an infinite set. Show that T = {U ⊆ X : U = ∅ or X \ U is finite} is
a topology on X, called the finite complement topology.
The empty set ∅ and X are in T . Let {Uα }α∈I be any family of nonempty sets belonging to T . We
observe that
[
\
X\
Uα =
(X \ Uα )
α∈I
α∈I
is finite since each X \ Uα is finite. Hence we conclude that the union
let U1 , U2 , ..., Un be nonempty sets belonging to T . Note that
X\
n
\
Uk =
k=1
n
[
S
α∈I
Uα is an element of T . Now
(X \ Uk )
k=1
is finite since each X \ Uk is finite. Hence we conclude that
topology on X.
Tn
k=1 Uk
is an element of T . Thus, T is a
PROBLEM 2 (10 points): Show that R with finite complement topology is separable.
The subset Z in R is countable. We observe that
\
Z = {C ⊆ R : Z ⊆ C, C is closed in R with the given topology} = R,
since the only closed set in R containing Z is R. Thus, R with finite complement topology is separable
because it contains a countable dense subset.
PROBLEM 3 (10 points): Suppose f, g : X → Y are continuous maps and Y is Hausdorff. Show that
the set {f (x) = g(x)} is closed in X.
Let E = {f (x) = g(x)}. Instead of proving that E is closed, we will prove that X \ E = {f (x) 6= g(x)}
is open. Let a ∈ X \ E. Then f (a) 6= g(a). Since Y is Hausdorff, there exist disjoint neighborhoods
U and V such that f (a) ∈ U and g(a) ∈ V . Also the preimages f −1 (U ) and g −1 (V ) are open in X
because f and g are continuous functions. Since finite intersection of open sets is open, their intersection
f −1 (U )∩g −1 (V ) is open in X. Note that a ∈ f −1 (U )∩g −1 (V ). Therefore the intersection f −1 (U )∩g −1 (V )
is a neighborhood around the point a. Moreover we have (f −1 (U ) ∩ g −1 (V )) ⊆ (X \ E) which implies
that a ∈ X \ E has a neighborhood contained in X \ E. Hence X \ E is open and thus E is closed.
1
2
PROBLEM 4 (10 points): Prove or disprove: If A and B are subsets of a topological space X with
A = B, then A = B.
The statement is false. Let X = R with the usual topology. Also, let A = [0, 1] and B = (0, 1). Then
A = B = [0, 1] although A 6= B.
PROBLEM 5 (10 points): Let X and Y be topological spaces and let B be a basis for Y . Show that a
map f : X → Y is continuous if and only if for every B ∈ B, f −1 (B) is open in X.
( =⇒ ) Assume that f : X → Y is continuous. Then, by definition, the preimage of every open subset
of Y is open in X. Since every B ∈ B is open, the preimage f −1 (B) is open in X.
(⇐=) Assume for every B ∈ B, the preimage f −1 (B) is open in X. Let U be an open subset of Y , and
let x ∈ f −1 (U ). Recall the basis criterion: The subset U is open in X if and only if for every p ∈ U , there
exists a basis set B ∈ B such that p ∈ B ⊆ U . Therefore by the criterion, there exists a basis set B ∈ B
such that f (x) ∈ B ⊆ U . Then since f (x) ∈ B ⊆ U , we obtain x ∈ f −1 (B) ⊆ f −1 (U ). By assumption,
the preimage f −1 (B) is open in X. Therefore f −1 (B) is a neighborhood of x that is contained in f −1 (U ).
Then by the characterization of open sets, we conclude that f −1 (U ) is open in X. Thus we proved that
f : X → Y is continuous.