Winter 2016 Math 351
Homework Solutions
Problem 1.b: We first show X \ (
reverse.
S
A) ⊆
T
{X \ A | A ∈ A} and then the
S
S
Let x ∈ X \ ( A). Therefore, x ∈ X and x ∈
/ A. Therefore,
for each A ∈
T
A, x ∈
/ A. Therefore, for each A ∈ A, x ∈ X \ A. So x ∈ {X \ A | A ∈ A}.
O
T
Now let x ∈ {X \ A | A ∈ A. Therefore for each
S A ∈ A, x ∈ X \ A. So
S for
each A ∈ A, x ∈ X and x ∈
/ A. Therefore x ∈
/ A. Therefore x ∈ X \( A).
O
Having shown both inclusions, we have proven equality.
Problem 1.g: We first show A1 ∩ (
reverse.
S
A) ⊆
S
{A1 ∩ A | A ∈ A and then the
S
Let x ∈ A1 ∩ ( A). This means x ∈ A1 and x ∈ U nionA. So ∃A ∈ A with
x ∈ A. So x ∈ A1 ∩ A ⊆ {A1 ∩ A | A ∈ A} as required.
O
S
For the reverse, let x ∈ {A1 ∩ A |SA ∈ Aa. So thereSexists A ∈ A with
x ∈ A1 ∩ A. So x ∈ A1 and x ∈ A ⊆ A. So x ∈ A1 ∩ ( A) as required. O
Having shown both inclusions, we have proven equality.
Problem 1.i: We first show (
reverse.
T
A) \ C ⊆
S
{A \ C | A ∈ A} and then the
T
T
Let x ∈ ( A) \ C.TSo x ∈ A but x ∈
/ C. So, ∀A ∈ A, x ∈ A. So ∀A ∈ A,
x ∈ A \ C. So x ∈ {A \ C | A ∈ A} as required.
O
T
Now let x ∈ {A \ C | A ∈ A}. SoTfor all A ∈ A, x ∈ A \ C.
T So for all
A ∈ A, x ∈ A but x ∈
/ C. So x ∈ A but x ∈
/ C. So x ∈ ( A) \ C as
required.
O
Having shown both inclusions, we have proven equality.
—1—
Winter 2016 Math 351
Homework Solutions
Problem 1.m: Let z ∈ (A1 × B1 ) ∪ (A2 , ×B2 ). So either z ∈ A1 × B1 or
z ∈ A2 × B2 . Without loss of generality assume z ∈ A1 × B1 . So ∃a ∈ A1 and
∃b ∈ B1 with z = (a, b). However a ∈ A1 ⊆ A1 ∪ A2 and b ∈ B1 ⊆ B1 ∪ B2
means that z = (a, b) ∈ (A1 ∪ A2 ) × (B1 ∪ B2 ) as required.
Problem 1.n: Let A1 = {1}, A2 = {2}, B1 = {3}, and B2 = {4}. Then
(A1 × B1 ) ∪ (A2 × B2 ) = {(1, 3), (2, 4)}
6= {(1, 3), (1, 4), (2, 3), (2, 4)} = (A1 ∪ A2 ) × (B1 ∪ B2 ).
Problem 1.q: Let C ∈ P(A) ∪ P(B). Without loss of generality, assume
C ∈ P(A), therefore C ⊆ A ⊆ A ∪ B. So C ∈ P(A ∪ B) as required.
Problem 1.r (as written): Let A = B = {1} so A = B and
P(A) ∪ P(B) = P(A) ∪ P(A) = P(A) = P(A ∪ A) = P(A ∪ B)
Problem 1.r (as intended): Let A = {1} and B = {2}.
P(A) ∪ P(B) = {∅, {1}, {2}} =
6 {∅, {1}, {2}, {1, 2}} = P(A ∪ B)
Problem 1.s (as expected): Let a ∈ A, so {a} ∈ P(A) ⊆ P(B) so {a} ⊆ B,
so a ∈ B as required.
Problem 1.s (sneaky): A ∈ P(A) ⊆ P(B) so A ∈ P(B), so A ⊆ B.
T
T
Problem 2.d: Let y ∈ f ( A), so there exists a ∈ A with y = f (a). That
means
T that ∀A ∈ A, a ∈ A. So ∀A ∈ A, y = f (a) ∈ f (A). Therefore
y ∈ {f (A) | A ∈ A} as required.
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Winter 2016 Math 351
Homework Solutions
Problem 2.e: Let X = {1, 2} and Y = {3} and f : X → Y be defined by
f (1) = 3 and f (2) = 3. Set A1 = {1} and A2 = {2}. Then
f (A1 ∩ A2 ) = f ({1} ∩ {2}) = f (∅) = ∅ =
6 {3} = {3} ∩ {3}f (A1 ) ∩ f (A2 )
S
S
Problem 2.f: We will first show f −1 ( B) ⊆ {f −1 (B) | B ∈ B} and then
the reverse.
S
S
Let x ∈ f −1 ( B). So f (x)
∈
B. So ∃B ∈ B with f (x) ∈ B. So
S
−1
−1
x ∈ f (B). This forces x ∈ {f (B) | B ∈ B}.
O
S
Now let x ∈S {f −1 (B) | B S
∈ B}. So ∃B ∈ B with x ∈ f −1 (B). So
−1
f (x) ∈ B ⊆ B. So x ∈ f ( B) as required.
O
Having shown both inclusions, we have proven equality.
Problem 2.i: Let X = {1, 2} and Y = {3} and f : X → Y be defined by
f (1) = 3 and f (2) = 3. Set A = {1}. Then
A = {1} =
6 {1, 2} = f −1 ({3}) = f −1 (f (A)).
Problem 2.l (as corrected): Let y ∈ f (X)\f (A). So y ∈ f (X), but y ∈
/ f (A).
So there exists x ∈ X with y = f (x). However, ∀a ∈ A, we have f (a) ∈
f (A). Since f (x) = y ∈
/ f (A), we must have x ∈
/ A. So x ∈ X \ A and
y = f (x) ∈ f (X \ A).
Problem 2.m: Let X = {1, 2} and Y = {3} and f : X → Y be defined by
f (1) = 3 and f (2) = 3. Set A = {1}. Then
f (X \ A) = f ({2}) = {3} =
6 ∅ = {3} \ {3} = f (X) \ f (A)
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Winter 2016 Math 351
Homework Solutions
Problem 2.o: Let ι = ι(A,X) . We prove ι−1 (A0 ) ⊆ A∩A0 and then the reverse.
Let x ∈ ι−1 (A0 ), then x is in A because it is the domain of ι and x = ι(x) ∈ A0
because x ∈ ι−1 (A0 ). So x ∈ A ∩ A0 as required.
O
Now let x ∈ A ∩ A0 , so ι(x) = x ∈ A0 and x ∈ ι−1 (A0 ) as required.
O
Having shown both inclusions, we have proven the equality.
Problem 3.c: Let x, x0 ∈ X with f (x) = f (x0 . Because g is a function:
(g ◦ f )(x) = g(f (x)) = g(f (x0 )) = (g ◦ f )(x0 ).
Now because g ◦ f is an injection, x = x0 as required.
Problem 3.d: Let z ∈ Z. Because g ◦ f is a surjection, ∃x ∈ X with (g ◦
f )(x) = z. Set y = f (x) ∈ Y . So we have
z = (g ◦ f )(x) = g(f (x)) = g(y).
Which proves that g is a surjection.
Problem 3.ef: Let X = Z = {1}, Y = {2, 3}, f : X → Y be defined by
f (1) = 2, and g : Y → Z be defined by g(2) = g(3) = 1. Then g ◦ f = ιX
is a bijection but f is not a surjection (3 ∈
/ f (X)) and g is not an injection
(g(2) = g(3)).
Problem 4: Let ι = ι(A,B) . Let a, a0 ∈ A with ι(a) = ι(a0 . Then we have
a = ι(a) = ι(a0 ) = a0 , as required.
Problem 13a: We show that T satisfies the four properties of a topology.
∅ ∈ T because p ∈
/ ∅.
O
X ∈ T by inspection.
O
Let U ⊆ T with U 6= ∅. We proceed by cases. If X ∈ U, then
—4—
S
U = X ∈ T.
Winter 2016 Math 351
Homework Solutions
Otherwise X ∈
/ U, then
/ U and so p ∈
/
S
S for all U ∈ U we have p ∈
U ∈ T. In all cases, U ∈ T.
S
U and
O
Assume U1 , U2 ∈ T. We proceed by cases. If p ∈ U1 and p ∈ U2 , then
U1 = U2 = X and U1 ∩ U2 = X ∈ T. Otherwise p ∈
/ U1 or p ∈
/ U2 , we have
p∈
/ U1 ∩ U2 and so U1 ∩ U2 ∈ T. In all cases, U1 ∩ U2 ∈ T.
O
Having shown T satisfies the four properties of a topology, we have shown T
is a topology.
Problem 13b:
Open: sets which exclude α,
{κ, o, ω, π, }, {δ, o, γ, ι, }, {τ, o, π, λ, γ, υ}
Closed: sets which include α,
{α, β, γ, δ, , φ}, {γ, ι, ρ, α, φ, }, {δ, α, o, ρ, υ, ν}
Clopen: ∅, X and sets which simultaneously both exclude and include α,
∅, X
Neither: everything either avoids α or its complement avoids α, so no sets
are neither in this topology.
Problem 13c: The only open set which contains 0 is R because any other
open set must exclude the particular point which is 0.
Problem 14a: We show that T satisfies the four properties of a topology.
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Winter 2016 Math 351
Homework Solutions
∅ ∈ T by inspection.
O
X \ X = ∅ is finite, so X ∈ T.
O
S
Let U ⊆ T with U 6= ∅. We proceed by cases. If U = {∅}, then U = ∅ ∈ T.
Otherwise, ∃U∗ ∈ U with U∗ 6= ∅. Therefore, X \ U∗ must be finite and
[ [
X\
U = {X \ U | U ∈ U} ⊆ X \ U∗
S
S
S
forces X \ ( U) to be finite and U ∈ T. In all cases, U ∈ T.
O
Let U1 , U2 ∈ T. We proceed by cases. If U1 = ∅ or U2 = ∅, then U1 ∩ U2 =
∅ ∈ T. Otherwise X \ U1 and X \ U2 are both finite, so
X \ (U1 ∩ U2 ) = (X \ U1 ) ∪ (X \ U2 )
forces X \ (U1 ∩ U2 ) to be finite and U1 ∩ U2 ∈ T. In all cases, U1 ∩ U2 ∈ T. O
Having shown T satisfies the four properties of a topology, we have shown T
is a topology.
Problem 14b:
Closed: finite sets other than ∅,
{0}, {1, 2}
Open: sets other than R whose complements are finite,
R∗
Clopen: ∅, R and sets which are finite with finite complements,
∅, R
—6—
Winter 2016 Math 351
Homework Solutions
Neither: infinite sets with infinite complements,
Z, Q, R \ Q, R \ Z, R \ (1, 10)
Problem 14c: Q, Q∗ , Q \ {5}
Problem 17a: We show that T satisfies the four properties of a topology.
∅ ∈ T because 0 ∈
/ ∅.
O
X ∈ T because (−1, 1) ⊆ X.
O
Let U ⊆ T with U 6= ∅. We proceed by cases. If ∃U∗ ∈ U with (−1, 1) ⊆ U∗ ,
then
[
(−1, 1) ⊆ U∗ ⊆
U,
S
S
S
so U ∈ T. SOtherwise, ∀U ∈ U, we have 0 ∈
/ U , so 0 ∈
/ U and U ∈ T.
In all cases, U ∈ T.
O
Let U1 , U2 ∈ T. We proceed by cases. If 0 ∈
/ U1 or 0 ∈
/ U2 , then 0 ∈
/ U1 ∩U2 , so
U1 ∩ U2 ∈ T. Otherwise (−1, 1) ⊆ U1 and (−1, 1) ⊆ U2 , so (−1, 1) ⊆ U1 ∩ U2
and therefore U1 ∩ U2 ∈ T. In all cases, U1 ∩ U2 ∈ T.
O
Having shown T satisfies the four properties of a topology, we have shown T
is a topology.
Problem 17b: Note, if a set is open because it contains (−1, 1) then its complement won’t contain 0, so its complement will be open, so it will be clopen.
Clopen: ∅, X
∅, [−1, 1), [−1, 1]
Open: avoids 0 but isn’t clopen
{1/2},
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Winter 2016 Math 351
Homework Solutions
Closed: is a subset of {±1} or contains 0,
{0}, (−1/2, 1/2)
Neither: ignoring ∅ which is clopen, a non-open set must contain 0, but then
its complement will not contain 0, so its complement will be open and the
original set will be closed, so the set cannot be neither.
Problem 17c: There are only four such sets: [−1, 1], [−1, 1), (−1, 1], (−1, 1).
Problem 19a: We show that T satisfies the four properties of a topology.
∅ ∈ T because the condition is vacuously satisfied.
O
R ∈ T because if u ∈ R and n ∈ Z+ , then u + n ∈ R.
O
S
+
Assume U ⊆ T and U 6= ∅. Let u ∈ USand n ∈ ZS
. So ∃U ∈ U, with
u ∈ U. However, this means u + n ∈ U ⊆ U and so U ∈ T.
O
Assume U1 , U2 ∈ T. Let u ∈ U1 ∩ U2 and n ∈ Z+ . So u ∈ U1 and u ∈ U2 . So,
u + n ∈ U1 and u + n ∈ U2 . So u + n ∈ U1 ∩ U2 . So U1 ∩ U2 ∈ T.
O
Having shown T satisfies the four properties of a topology, we have shown T
is a topology.
Problem 19b: Open: N, Z+ , [3, ∞), (3, ∞)
Closed: (−∞, 0), (−∞, 2]
Clopen: ∅, Z, Q, R \ Q, R
Neither: {0}, (−1, 1), [−1, 2), [−1, 3), (3, 5]
—8—
Winter 2016 Math 351
Homework Solutions
Problem 19c: N, Z, R are three open sets containing 0.
Problem 20a: We show that T satisfies the four properties of a topology.
∅ ∈ T because the condition is vacuously satisfied.
O
R ∈ T because if u ∈ R and n ∈ Z, then u + n ∈ R.
O
S
Assume U ⊆ T and U 6= ∅. Let u ∈S U and n S
∈ Z. So ∃U ∈ U, with u ∈ U.
However, this means u + n ∈ U ⊆ U and so U ∈ T.
O
Assume U1 , U2 ∈ T. Let u ∈ U1 ∩ U2 and n ∈ Z. So u ∈ U1 and u ∈ U2 . So,
u + n ∈ U1 and u + n ∈ U2 . So u + n ∈ U1 ∩ U2 . So U1 ∩ U2 ∈ T.
O
Having shown T satisfies the four properties of a topology, we have shown T
is a topology.
Problem 19b: Open: All open sets are clopen.
Closed: All closed sets are clopen.
Clopen: ∅, Z, Q, R \ Q, R
Neither: N, Z+ , {0}, (−1, 1), [−1, 2), [−1, 3), (3, 5], (−∞, 0), (−∞, 2],
[3, ∞), (3, ∞)
Problem 19c: N, Z, R are three open sets containing 0.
Problem 22a: We show that T satisfies the four properties of a topology.
∅ ∈ T because the condition is vacuously satisfied.
O
R ∈ T because if u ∈ R, then [u, u + 1) ⊆ R.
O
Assume U ⊆ T and U 6= ∅. Let u ∈
S
U. So ∃U ∈ U, with u ∈ U. However,
—9—
Winter 2016 Math 351
Homework Solutions
this means there exists > 0 with [u, u + ) ⊆ U ⊆
S
U and so
S
U ∈ T. O
Assume U1 , U2 ∈ T. Let u ∈ U1 ∩ U2 . So u ∈ U1 and u ∈ U2 . So, there exists
1 , 2 > 0 with [u, u + 1 ) ⊆ U1 and [u, u + 2 ) ⊆ U2 . Set = min{1 , 2 } So
[u, u + ) ⊆ U1 ∩ U2 . So U1 ∩ U2 ∈ T.
O
Having shown T satisfies the four properties of a topology, we have shown T
is a topology.
Problem 22b: Open: (−1, 1), (3, ∞)
Closed: N, Z, Z+ , {0}, [−1, 2], (−∞, 2]
Clopen: ∅, R, [−1, 3), (−∞, 0), [3, ∞)
Neither: Q, R \ Q, (3, 5]
Problem 22c: [0, 1), [0, 2), [0, 3) are three open sets containing 0.
Problem 13 - Separation: This topology is T0 . Let x1 , x2 ∈ X with x1 =
6 x2 .
At least one of these points is not p. Assume x1 6= p, set U1 = {x1 }. Then
x1 ∈ U 1 , x2 ∈
/ U1 and U1 open (because p ∈
/ U1 ).
This topology is not T1 . Choose x1 = p and x2 6= p. Assume x1 ∈ U1 and
U1 is open. This forces U1 = X and therefore x2 ∈ U2 . So any open set
containing x1 must also contain x2 .
Problem 14 - Separation: This topology is T1 . Let x1 , x2 ∈ X with x1 6= x2 .
Set U1 = X \ {x2 } and U2 = X \ {x1 }. Then we have x1 ∈ U1 , x2 ∈
/ U1 and
U1 open (because it has a finite complement).
—10—
Winter 2016 Math 351
Homework Solutions
This topology is not T2 . Let x1 , x2 ∈ X with x1 6= x2 . Let U1 , U2 be open
sets with x1 ∈ U1 , x2 ∈
/ U1 , x2 ∈ U2 , and x1 ∈
/ U2 . Then neither U1 nor U2
are ∅ and therefore X \ U1 and X \ U2 are both finite sets. This forces
X \ (U1 ∩ U2 ) = (X \ U1 ) ∪ (X \ U2 )
to be finite. Because X is infinite, there must be an infinite number of
elements in U1 ∩ U2 and so U1 ∩ U2 6= ∅.
Problem 17 - Separation: The topology is T0 . Let x1 , x2 ∈ X with x1 6= x2 .
At least one of these points is not 0. Without loss of generality, assume
x1 6= 0. Choose U1 = {x1 }. Then x1 ∈ U1 , x2 ∈
/ U1 and U1 is open (because
0∈
/ U1 ).
The topology is not T1 . Let x1 = 0 and x2 = 1/2. Assume U1 is an open set
with x1 ∈ U1 . This forces (−1, 1) ⊆ U1 and so x2 ∈ U1 . So any open set
containing x1 must also contain x2 .
Problem 19 - Separation: The topology is T0 . Let x1 , x2 ∈ X with x1 6= x2 .
Without loss of generality, assume x1 < x2 . Choose U2 = [x2 , ∞). Then
x2 ∈ U2 , x1 ∈
/ U2 and U2 is open (because for all u ∈ U2 , n ∈ Z+ , we have
u + n ∈ U2 ).
The topology is not T1 . Choose x1 = 0 and x2 = 1. Let U1 be open with
x1 ∈ U1 . Because U1 is open, x2 = x1 + 1 ∈ U1 . So any open set containing
x1 must also contain x2 .
Problem 20 - Separation: The topology is not T0 . Choose x1 = 0 and x2 = 1.
Let U1 be open with x1 ∈ U1 . Because U1 is open, x2 = x1 + 1 ∈ U1 . So
any open set containing x1 must also contain x2 . Similarly, let U2 be open
with x2 ∈ U2 . Because U2 is open, x1 = x2 + (−1) ∈ U2 . So any open set
containing x2 must also contain x1 .
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Winter 2016 Math 351
Homework Solutions
Problem 22 - Separation: The topology is T2 . Let x1 , x2 ∈ R with x1 6= x2 .
Without loss of generality, assume x1 < x2 . We set U1 = (−∞, x2 ) and
U2 = [x2 , ∞). Then we have U1 , U2 both open, x1 ∈ U1 , x2 ∈ U2 , and
U1 ∩ U2 = ∅.
Problem 13 - Connectedness: X is connected. Let A 6= ∅ be clopen. We will
show that A = X. Because A is closed, X \ A is an open set which is not
the entire space. Therefore, p ∈
/ X \ A so p ∈ A. Therefore A = X.
Problem 14 - Connectedness: X is connected. Let A 6= ∅ be clopen. We will
show A = X. Because A is open and not equal to ∅, we know that B = X \A
is finite. However, B = X \ A is also open and A = X \ B is not finite (else
X = A ∪ B would be finite) therefore B = ∅ and A = X.
Problem 17 - Connectedness: X is disconnected as X = (−1, 1) t {−1, 1}.
Problem 19 - Connectedness: X is disconnected as X = Z t (R \ Z).
Problem 20 - Connectedness: X is disconnected as X = Z t (R \ Z).
Problem 22 - Connectedness: X is disconnected as X = (−∞, 0) t [0, ∞).
—12—