Download Math 535: Topology Homework 1

Math 535: Topology
Homework 1
Mueen Nawaz
Mueen Nawaz
Math 535 Topology Homework 1
Problem 1
Problem 1
Find all topologies on the set X = {0, 1, 2}.
Solution:
In the list below, a, b, c ∈ X and it is assumed that they are distinct from one another.
• P(X) the power set of X (discrete topology).
• {φ, X} (the trivial topology)
• {φ, X, {a}}, a ∈ {0, 1, 2}. There are 3 of these.
• {φ, X, {a, b}} and other combinations of single subsets of size two from X. There will
be 3 of these.
• {φ, X, {a, b}, {d}}, d ∈ X. d need not be distinct from a or b. There will be 9 of these
(3 ways of choosing the subset of size two, and three of choosing the singleton).
• {φ, X, {a}, {a, b}, {a, c}}. There will be 3 of these (the subsets of size two are determined by the element a).
• {φ, X, {a}, {b}, {a, b}}. There will be 3 of these.
Problem 2
(i) Show that if
every function f
(ii) Show that if
every function f
X is a set with the trivial topology and Y is any topological space, then
: Y → X is continuous.
X is a set with the discrete topology, and Y is any topological space, then
: X → Y is continuous.
Solution:
(i) Let U ⊆ X be open. As X has the trivial topology, this means that U is either φ
or X. If U = φ, then f −1 (U ) = φ, which is open. If U = X, then f −1 (U ) = Y , and Y is
always open, as it is a member of any topology defined on Y . Thus, if U is any open set in
X, f −1 (U ) is also an open set in Y . Hence, f is continuous.
(ii) Let U be any open set in Y . Clearly, f −1 (U ) is a subset of X. However, as X has
the discrete topology, all its subsets are open. Hence, f −1 (U ) ∈ P(X) is open in X for any
open U in Y . Thus, f is continuous.
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Mueen Nawaz
Math 535 Topology Homework 1
Problem 3
Problem 3
Use de Morgan’s Laws to prove:
(i) The union of finitely many closed subsets of a topological space is closed.
(ii) The intersection of arbitrarily many closed subsets of a topological space is closed.
Solution:
Let X be the topological space.
(i) Let A1 , A2 , . . . , An be closed sets with Ai ⊆ X. Let Bi = X\Ai for each i. By the
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definition of closed sets, all the Bi are open. Now ni=1 Ai = X\ ni=1 (X\Ai ) = X\ ni=1 Bi
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Since Bi are open, any finite number of intersections of Bi is open. So ni=1 Bi = C is open.
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Thus, ni=1 Ai = X\C. As C is open, the finte union of closed sets is closed.
(ii) Let Ad , d ∈ D be arbitrarily many closed subsets of X. Let Bd = X\Ad . As in the
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previous section, Bd is open for all d ∈ D. Now d∈D Ad = X\ d∈D (X\Ad ) = X\ d∈D Bd .
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Since the Bd are open, any union of Bd is also union. So d∈D Bd = C is open. Thus,
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d∈D Ad = X\C. As C is open, the interesection of arbitrarily many closed sets is closed.
Problem 4
Show that if X is a set and a collection σ of subsets of X satisfies:
C1 The set σ is closed under finite unions.
C2 The set σ is closed under arbitrarily many intersections.
C2 The set σ contains φ and X.
Then the collection τ = {U ⊆ X|X\U ∈ σ} is a topology on X.
Solution:
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1. Let Ad ∈ τ with d ∈ D. Note that d∈D Ad = X\ d∈D (X\Ad ). As X\Ad ∈ σ, it
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is closed under intersections (C2). Thus, d∈D (X\Ad ) ∈ σ and d∈D Ad ∈ τ by the
manner in which τ is constructed.
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2. Let A1 , A2 , . . . , An ∈ τ . Note that ni=1 Ai = X\ ni=1 (X\Ai ). As X\Ai ∈ σ, it
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is closed under finite unions (C1). Thus, i=1 (X\Ai ) ∈ σ and ni=1 Ai ∈ τ by the
manner in which τ is constructed.
3. φ is in τ as φ = X\X and X ∈ σ. Likewise, X is in τ as X = X\φ and φ ∈ σ.
Thus, τ is a topology on X.
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Mueen Nawaz
Math 535 Topology Homework 1
Problem 5
Problem 5
Give an example of a topological space and a collection {Wα }α∈A of closed subsets such that
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their union α∈A Wα is not closed.
Solution:
Let R be the space with the usual topology (i.e. based on the usual metric). Let
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An = (− n1 , n1 ), n ∈ N. Define Wn = R\An . As An is open, Wn is closed. Now n∈N Wn =
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R\ n∈N An . But this is just R\{0}, which is open as it is the union of two open intervals:
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(−∞, 0) (0, ∞). Therefore, this union of closed sets is not closed.
Problem 6
Let Q ⊆ R be the subset of rational numbers. Show that Q is neither open nor closed.
Solution: Note that between any two rationals, there exists an irrational. Likewise, between
any two irrationals, there exists a rational.
Let x ∈ Q. Then every open ball (i.e. interval) around x necessarily contains an irrational.
For example, if x ∈ (a, b) where a and b are rational numbers, we know that an irrational r
exists with a < r < b — regardless of the values of a and b. Therefore, since one cannot find
any open intervals about x ∈ Q that are subsets of Q, Q cannot be open.
Likewise, consider A = R\Q. This is the set of irrationals. Let x ∈ A. Again, using
the same argument above, let x ∈ (a, b) with a, b ∈ A. There exists a rational y such that
a < y < b. Therefore, there is no open interval about x that is contained in A. Hence, A is
not open, and therefore Q = R\A is not closed.
Problem 7
Let X be any set and define a nonempty subset U ⊆ X to be open if X\U is finite. Show
that this defines a topology on X.
Solution:
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(i) Let Ad , d ∈ D be open subsets of X. Thus, Bd = X\Ad is finite.
d∈D Ad =
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X\ d∈D Bd . Since d∈D Bd ⊆ Bd and since Bd is finite, therefore d∈D Bd is finite. Thus,
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A
=
X\C
where
C
=
d
d∈D
d∈D Bd is finite and is therefore an open set as well. Hence,
open sets are closed under arbitrary unions.
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(ii) Let A1 , A2 , . . . , An be open subsets of X. Thus, Bi = X\Ai is finite. Now ni=1 Ai =
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X\ ni=1 Bi . However, ni=1 Bi is the finite union of finite sets, and hence is itself finite. So
Problem 7 continued on next page. . .
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Mueen Nawaz
Tn
i=1
Ai = X\C where C =
Math 535 Topology Homework 1
Sn
i=1
Problem 7 (continued)
Bi is finite and is thus an open set.
(iii) Assume X is nonempty. If X is finite, then X = X\φ, and so X is open as well (φ is
trivially finite) and is in the topology. Likewise, φ = X\X, and since X is finite, φ is open
and in the topology.
If X is infinite, pick two distinct elements a and b from X. Now A = X\{a} and
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B = X\{b} are open sets, by definition. As X = A B, X is also open. Also, φ = A B,
so φ is also open.
So regardless of whether X is finite or infinite, φ and X will be in the topology.
Problem 8
Let B be a basis for a topology on X and define a subset U ⊆ X to be open, if for every
x ∈ U , there is a V ∈ B such that x ∈ V ⊆ U . Show that this satisfies the definition of a
topology.
Solution:
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(i) Let Ad , d ∈ D be open subsets of X. Let E = d∈D Ad . Let x ∈ E. There must
exist a Ad such that x ∈ Ad . Since Ad is open, there exists a Vx ∈ B such that x ∈ Vx ⊆ Ad .
However, this means that x ∈ Vx ⊆ E as Ad ⊆ E. Thus, by definition, E is also open.
Therefore this collection of open sets is closed under arbitrary unions.
(ii) First, note that if V1 , V2 , . . . Vn ∈ B, and there exists an x such that x ∈ Vi ∀i ∈
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{1, 2, . . . , n}, then there exists a V ∈ B such that x ∈ V ⊆ ni=1 Vi . This follows from a
straightforward induction on the definition of a basis.
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Now let A1 , A2 , . . . , An be open sets in X. Let x ∈ E = ni=1 Ai . Thus x ∈ Ai ∀ i. Thus,
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for each i, there exists a Vi such that x ∈ Vi ⊆ Ai . Therefore, x ∈ ni=1 Vi . By the note in
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the previous paragraph, there is a V ∈ B such that x ∈ V ⊆ ni=1 Vi . Since ni=1 Vi ⊆ E, we
have V ⊆ E. Thus, E is open. Therefore open sets in X are closed under finite intersections.
(iii) φ is open trivially, as there is no x in φ. By the definition of a basis, if x ∈ X, there
is a V ∈ B such that x ∈ V . As V ⊆ X, this means that X is also open.
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