Topology by James Munkres: Solution Manual

172 Pages·2019·0.952 MB·English

by  Dan Whitman

Topology
Second Edition
by James Munkres
Solutions Manual
by Dan Whitman
April 14, 2019
Chapter 1 Set Theory and Logic
§1 Fundamental Concepts
Exercise 1.1
Check the distributive laws for ∪ and ∩ and DeMorgan’s laws.
Solution:
Suppose that A, B, and C are sets. First we show that A∩(B∪C)=(A∩B)∪(A∩C).
Proof. We show this as a series of logical equivalences:
x∈A∩(B∪C)⇔x∈A∧x∈B∪C
⇔x∈A∧(x∈B∨x∈C)
⇔(x∈A∧x∈B)∨(x∈A∧x∈C)
⇔x∈A∩B∨x∈A∩C
⇔x∈(A∩B)∪(A∩C),
which of course shows the desired result.
Next we show that A∪(B∩C)=(A∪B)∩(A∪C).
Proof. We show this in the same way:
x∈A∪(B∩C)⇔x∈A∨x∈B∩C
⇔x∈A∨(x∈B∧x∈C)
⇔(x∈A∨x∈B)∧(x∈A∨x∈C)
⇔x∈A∪B∧x∈A∪C
⇔x∈(A∪B)∩(A∪C),
which of course shows the desired result.
Now we show the first DeMorgan’s law that A−(B∪C)=(A−B)∩(A−C).
Proof. We show this in the same way:
x∈A−(B∪C)⇔x∈A∧x∈/ B∪C
⇔x∈A∧¬(x∈B∨x∈C)
⇔x∈A∧(x∈/ B∧x∈/ C)
⇔(x∈A∧x∈/ B)∧(x∈A∧x∈/ C)
⇔x∈A−B∧x∈A−C
⇔x∈(A−B)∩(A−C),
which is the desired result.
Lastly we show that A−(B∩C)=(A−B)∪(A−C).
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Proof. Again we use a sequence of logical equivalences:
x∈A−(B∩C)⇔x∈A∧x∈/ B∩C
⇔x∈A∧¬(x∈B∧x∈C)
⇔x∈A∧(x∈/ B∨x∈/ C)
⇔(x∈A∧x∈/ B)∨(x∈A∧x∈/ C)
⇔x∈A−B∨x∈A−C
⇔x∈(A−B)∪(A−C),
as desired.
Exercise 1.2
DeterminewhichofthefollowingstatementsaretrueforallsetsA,B,C,andD. Ifadoubleimplication
fails,determinewhetheroneortheotherofthepossibleimplicationsholds. Ifanequalityfails,determine
whetherthestatementbecomestrueifthe“equals”symbolisreplacedbyoneortheotheroftheinclusion
symbols ⊂ or ⊃.
(a) A⊂B and A⊂C ⇔A⊂(B∪C). (j) A⊂C and B ⊂D ⇒(A×B)⊂(C×D).
(b) A⊂B or A⊂C ⇔A⊂(B∪C). (k) The converse of (j).
(c) A⊂B and A⊂C ⇔A⊂(B∩C). (l) The converse of (j), assuming that A and B
(d) A⊂B or A⊂C ⇔A⊂(B∩C). are nonempty.
(e) A−(A−B)=B. (m) (A×B)∪(C×D)=(A∪C)×(B∪D).
(f) A−(B−A)=A−B. (n) (A×B)∩(C×D)=(A∩C)×(B∩D).
(g) A∩(B−C)=(A∩B)−(A∩C). (o) A×(B−C)=(A×B)−(A×C).
(h) A∪(B−C)=(A∪B)−(A∪C). (p) (A−B)×(C−D)=(A×C−B×C)−A×D.
(i) (A∩B)∪(A−B)=A. (q) (A×B)−(C×D)=(A−C)×(B−D).
Solution:
(a) We claim that A⊂B and A⊂C ⇒A⊂(B∪C) but that the converse is not generally true.
Proof. Suppose that A ⊂ B and A ⊂ C and consider any x ∈ A. Then clearly also x ∈ B since
A⊂B so that x∈B∪C. Since x was arbitrary, this shows that A⊂(B∪C) as desired.
To show that the converse is not true, suppose that A={1,2,3}, B ={1,2}, and C ={3,4}. Then
clearly A⊂{1,2,3,4}=B∪C but it neither true that A⊂B (since 3∈A but 3∈/ B) nor A⊂C
(since 1∈A but 1∈/ C).
(b) We claim that A⊂B or A⊂C ⇒A⊂(B∪C) but that the converse is not generally true.
Proof. Suppose that A⊂B or A⊂C and consider any x∈A. If A⊂B then clearly x∈B so that
x∈B∪C. If A⊂C then clearly x∈C so that again x∈B∪C. Since x was arbitrary, this shows
that A⊂(B∪C) as desired.
The counterexample that disproves the converse of part (a), also serves as a counterexample to the
converse here. Again this is because A⊂B∪C but neither A⊂B nor A⊂C, which is to say that
A(cid:54)⊂B and A(cid:54)⊂C. Hence it is not true that A⊂B or A⊂C.
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(c) We claim that this biconditional is true.
Proof. (⇒) Suppose that A⊂B and A⊂C and consider any x∈A. Then clearly also x∈B and
x ∈ C since both A ⊂ B and A ⊂ C. Hence x ∈ B∩C, which proves that A ⊂ B∩C since x was
arbitrary.
(⇐) Now suppose that A⊂B∩C and consider any x∈A. Then x∈B∩C as well so that x∈B
andx∈C. SincexwasanarbitraryelementofA,thisofcourseshowsthatbothA⊂B andA⊂C
as desired.
(d) We claim that only the converse is true here.
Proof. To show the converse, suppose that A ⊂ B∩C. It was shown in part (c) that this implies
that both A⊂B and A⊂C. Thus it is clearly true that A⊂B or A⊂C.
As a counterexample to the forward implication, let A = {1}, B = {1,2}, and C = {3,4} so that
clearly A ⊂ B and hence A ⊂ B or A ⊂ C is true. However we have that B and C are disjoint so
that B∩C =∅, therefore A(cid:54)⊂∅=B∩C since A(cid:54)=∅.
(e) We claim that A−(A−B)⊂B but that the other direction is not generally true.
Proof. First consider any x∈A−(A−B) so that x∈A but x∈/ A−B. Hence it is not true that
x ∈ A and x ∈/ B. So it must be that x ∈/ A or x ∈ B. However, since we know that x ∈ A, it has
to be that x∈B. Thus A−(A−B)⊂B since x was arbitrary.
NowletA={1,2}andB ={2,3}. ThenweclearlyhaveA−B ={1},andthusA−(A−B)={2}.
So clearly B is not a subset of A−(A−B) since 3∈B but 3∈/ A−(A−B).
(f) Here we claim that A−(B−A)⊃A−B but that the other direction is not generally true.
Proof. First suppose that x∈A−B so that x∈A but x∈/ B. Then it is certainly true that x∈/ B
or x∈A so that, by logical equivalence, it is not true that x∈B and x∈/ A. That is, it is not true
that x∈B−A, which is to say that x∈/ B−A. Since also x∈A, it follows that x∈A−(B−A),
which shows the desired result since x was arbitrary.
To show that the other direction does not hold consider the counterexample A = {1,2} and B =
{2,3}. Then B−A = {3} so that A−(B−A) = {1,2} = A. We also have that A−B = {1} so
that 2∈A−(B−A) but 2∈/ A−B. This suffices to show that A−(B−A)(cid:54)⊂A−B.
(g) We claim that equality holds here, i.e. that A∩(B−C)=(A∩B)−(A∩C).
Proof. (⊂) Suppose that x ∈ A∩(B−C) so that x ∈ A and x ∈ B−C. Thus x ∈ B but x ∈/ C.
Since both x ∈ A and x ∈ B we have that x ∈ A∩B. Also since x ∈/ C it clearly must be that
x∈/ A∩C. Hence x∈(A∩B)−(A∩C), which shows the forward direction since x was arbitrary.
(⊃) Now suppose that x ∈ (A∩B)−(A∩C). Hence x ∈ A∩B but x ∈/ A∩C. From the former
of these we have that x ∈ A and x ∈ B, and from the latter it follows that either x ∈/ A or x ∈/ C.
Since we know that x ∈ A, it must therefore be that x ∈/ C. Hence x ∈ B −C since x ∈ B but
x∈/ C. Since also x∈A we have that x∈A∩(B−C), which shows the desired result since x was
arbitrary.
(h) Here we claim that A∪(B −C) ⊃ (A∪B)−(A∪C) but that the forward direction is not
generally true.
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Proof. First consider any x∈(A∪B)−(A∪C) so that x∈A∪B and x∈/ A∪C. From the latter,
it follows that x ∈/ A and x ∈/ C since otherwise we would have x ∈ A∪C. From the former, we
have that x∈A or x∈B so that it must be that x∈B since x∈/ A. Therefore we have that x∈B
and x ∈/ C so that x ∈ B−C. From this it obviously follows that x ∈ A∪(B−C), which shows
that A∪(B−C)⊃(A∪B)−(A∪C) since x was arbitrary.
To show that the forward direction does not always hold, consider the sets A = {1,2}, B = {2,3},
and C = {2}. Then we clearly have that B −C = {3}, and hence A∪(B −C) = {1,2,3}. On
the other hand, we have A∪B = {1,2,3} and A∪C = {1,2} so that (A∪B)−(A∪C) = {3}.
Hence, for example, 1 ∈ A∪(B −C) but 1 ∈/ (A∪B)−(A∪C), which suffices to show that
A∪(B−C)(cid:54)⊂(A∪B)−(A∪C) as desired.
(i) We claim that equality holds here.
Proof. We show this with a chain of logical equivalences:
x∈(A∩B)∪(A−B)⇔x∈A∩B∨x∈A−B
⇔(x∈A∧x∈B)∨(x∈A∧x∈/ B)
⇔x∈A∧(x∈B∨x∈/ B)
⇔x∈A∧True
⇔x∈A,
where we note that “True” denotes the fact that x ∈ B ∨x ∈/ B is always true by the excluded
middle property of logic.
(j) We claim that this implication is true.
Proof. Suppose that A⊂C and B ⊂D. Consider any (x,y)∈A×B so that x∈A and y ∈B by
thedefinitionofthecartesianproduct. Thenalsoclearlyx∈C andy ∈D sinceA⊂C andB ⊂D.
Hence (x,y)∈C×D, which shows the result since the ordered pair (x,y) was arbitrary.
(k) We claim that the converse of (j) is not always true.
Proof. Consider the following sets:
A=∅ C ={1}
B ={1,2} D ={2} .
ThenwehavethatA×B =∅sincetherearenoorderedpairs(x,y)suchthatx∈A(sinceA=∅).
Hence it is vacuously true that (A×B)⊂(C×D). However, clearly it is not the case that B ⊂D,
and so, even though A⊂C, it is not true that A⊂C and B ⊂D.
(l) We claim that the converse of (j) is true with the stipulation that A and B are both nonempty.
Proof. Suppose that (A×B) ⊂ (C ×D). First consider any x ∈ A. Then, since B (cid:54)= ∅, there is a
b ∈ B. Then (x,b) ∈ A×B so that clearly also (x,b) ∈ C ×D. Hence x ∈ C so that A ⊂ C since
x was arbitrary. An analogous argument shows that B ⊂ D since A is nonempty. Hence it is true
that A⊂C and B ⊂D as desired.
(m) Here we claim that (A×B)∪(C×D)⊂(A∪C)×(B∪D) but that the other direction is not
always true.
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Proof. Firstconsiderany(x,y)∈(A×B)∪(C×D)sothateither(x,y)∈A×B or(x,y)∈C×D.
In the first case x ∈ A and y ∈ B so that clearly x ∈ A∪C and y ∈ B ∪D. Hence (x,y) ∈
(A∪C)×(B ∪D). In the second case we have x ∈ C and y ∈ D so that again x ∈ A∪C and
y ∈B∪D are still both true. Hence of course (x,y)∈(A∪C)×(B∪D) here also. This shows the
result in either case since (x,y) was an arbitrary ordered pair.
To show that the other direction does not always hold, consider A = B = {1} and C = D = {2}.
ThenweclearlyhaveA×B ={(1,1)}andC×D ={(2,2)}sothat(A×B)∪(C×D)={(1,1),(2,2)}.
We also have A∪C =B∪D ={1,2} so that (A∪C)×(B∪D)={(1,1),(1,2),(2,1),(2,2)}. This
clearly shows that (A×B)∪(C×D)(cid:54)⊃(A∪C)×(B∪D) as desired.
(n) We claim that the equality holds here.
Proof. We can show this by a series of logical equivalences:
(x,y)∈(A×B)∩(C×D)⇔(x,y)∈A×B∧(x,y)∈C×D
⇔(x∈A∧y ∈B)∧(x∈C∧y ∈D)
⇔(x∈A∧x∈C)∧(y ∈B∧y ∈D)
⇔x∈A∩C∧y ∈B∩D
⇔(x,y)∈(A∩C)×(B∩D)
as desired.
(o) We claim that equivalence holds here as well.
Proof. (⊂) First consider any (x,y)∈A×(B−C) so that x∈A and y ∈B−C. From the latter
of these we have that y ∈B but y ∈/ C. We clearly then have that (x,y)∈A×B since x∈A and
y ∈B. Italsohastobethat(x,y)∈/ A×C sincey ∈/ C eventhoughitistruethatxinA. Therefore
(x,y)∈(A×B)−(A×C) as desired.
(⊃)Nowsupposethat(x,y)∈(A×B)−(A×C)sothat(x,y)∈A×B but(x,y)∈/ (A×C). From
the former we have that x ∈ A and y ∈ B. It then must be that y ∈/ C since (x,y) ∈/ (A×C) but
we know that x∈A. Then we have y ∈B−C since y ∈B but y ∈/ C. Since also x∈A, it follows
that (x,y)∈A×(B−C) as desired.
(p) We claim the equivalence hold for this statement.
Proof. (⊂) Suppose that (x,y) ∈ (A−B)×(C −D) so that x ∈ A−B and y ∈ C −D. Then we
have that x ∈ A, x ∈/ B, y ∈ C, and y ∈/ D. So first, clearly (x,y) ∈ A×C. Then, since x ∈/ B,
we have that (x,y) ∈/ B×C, and hence (x,y) ∈ A×C −B×C. Since y ∈/ D, we also have that
(x,y) ∈/ A×D, and thus (x,y) ∈ (A×C −B×C)−A×D. This clearly shows the desired result
since (x,y) was arbitrary.
(⊃) Now suppose that (x,y) ∈ (A×C −B ×C)−A×D so that (x,y) ∈ A×C −B ×C but
(x,y)∈/ A×D. From the former we have that (x,y)∈A×C and (x,y)∈/ B×C. Thus x∈A and
y ∈C so that it has to be that x∈/ B since (x,y)∈/ B×C but we know that y ∈C. It also must be
thaty ∈/ D since(x,y)∈/ A×D butx∈A. Thereforewehavethatx∈A,x∈/ B,y ∈C,andy ∈/ D,
fromwhichitreadilyfollowsthatx∈A−B andy ∈C−D. Thusclearly(x,y)∈(A−B)×(C−D),
which shows the desired result since (x,y) was arbitrary.
(q) Here we claim that (A×B)−(C ×D)⊃(A−C)×(B−D) but that the forward direction is
not true in general.
Page 5
Proof. First consider any (x,y) ∈ (A−C)×(B−D) so that x ∈ A−C and y ∈ B−D. Thus we
have x∈A, x∈/ C, y ∈B, and y ∈/ D. From this clearly (x,y)∈A×B but (x,y)∈/ C×D. Hence
(x,y)∈(A×B)−(C×D), which clearly shows the desired result since (x,y) was arbitrary.
To show that the forward direction does not hold, consider A={1,2}, B ={a,b}, C ={2,3}, and
D ={b,c}. We then clearly have the following sets:
A×B ={(1,a),(1,b),(2,a),(2,b)} A−C ={1}
C×D ={(2,b),(2,c),(3,b),(3,c)} B−D ={a}
(A×B)−(C×D)={(1,a),(1,b),(2,a)} (A−C)×(B−D)={(1,a)} .
This clearly shows that (A×B)−(C×D) is not a subset of (A−C)×(B−D).
Exercise 1.3
(a) Write the contrapositive and converse of the following statement: “If x<0, then x2−x>0,” and
determine which (if any) of the three statements are true.
(b) Do the same for the statement “If x>0, then x2−x>0.”
Solution:
(a) First we claim that the original statement is true.
Proof. Since x < 0 we clearly have that x−1 < x < 0 as well. Then, since the product of two
negative numbers is positive, we have that x2−x=x(x−1)>0 as desired.
The contrapositive of this is, “If x2−x ≤ 0, then x ≥ 0.” This is of course also true by virtue of
the fact that the contrapositive is logically equivalent to the original implication.
Lastly, the converse of this statement is, “If x2 −x > 0, then x < 0.” We claim that this is not
generally true.
Proof. A simple counterexample of x=2 shows this. We have x2−x=22−2=4−2=2>0, but
also clearly x=2>0 as well so that x<0 is clearly false.
(b) First we claim that this statement is false.
Proof. As a counterexample, let x = 1/2. Then clearly x > 0, but we also have x2−x = (1/2)2−
1/2=1/4−1/2=−1/4<0 so that x2−x>0 is obviously not true.
The contrapositive is then “If x2−x≤0, then x≤0,” which is false since it is logically equivalent
to the original statement.
The converse is “If x2−x>0, then x>0,” which we claim is false.
Proof. As a counterexample, consider x = −1 so that x2 −x = (−1)2 −(−1) = 1+1 = 2 > 0.
However, we also clearly have x=−1<0 so that x>0 is not true.
Exercise 1.4
Let A and B be sets of real numbers. Write the negation of each of the following statements:
(a) For every a∈A, it is true that a2 ∈B.
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(b) For at least one a∈A, it is true that a2 ∈B.
(c) For every a∈A, it is true that a2 ∈/ B.
(d) For at least one a∈/ A, it is true that a2 ∈B.
Solution:
These are all basic logical negations using existential quantifiers:
(a) There is an a∈A where a2 ∈/ B.
(b) For every a∈A, a2 ∈/ B.
(c) There is an a∈A where a2 ∈B.
(d) For every a∈/ A, a2 ∈/ B.
Exercise 1.5
Let A be a nonempty collection of sets. Determine the truth of each of the following statements and of
their converses:
(cid:83)
(a) x∈ A⇒x∈A for at least one A∈A.
A∈A
(cid:83)
(b) x∈ A⇒x∈A for every A∈A.
A∈A
(cid:84)
(c) x∈ A⇒x∈A for at least one A∈A.
A∈A
(cid:84)
(d) x∈ A⇒x∈A for every A∈A.
A∈A
Solution:
(a)Thestatementontherightisthedefinitionofthestatementontheleftsoofcoursetheimplication
and its converse are true.
(b) The implication is generally false.
(cid:83)
Proof. As a counterexample, consider A = {{1},{2}}. Then clearly A = {1,2} so that
A∈A
(cid:83)
1∈ A, but 1 is not in A for every A∈A since 1∈/ {2}.
A∈A
However, the converse is true.
Proof. Suppose that x∈A for every A∈A. Since A is nonempty there is an A ∈A. Then x∈A
0 0
(cid:83)
since A ∈A. Hence by definition x∈ A since x∈A and A ∈A.
0 A∈A 0 0
(c) The implication here is true.
(cid:84)
Proof. Supposethatx∈ Asothatbydefinitionx∈AforeveryA∈A. SinceAisnonempty
A∈A
there is an A ∈A so that in particular x∈A . This shows the desired result since A ∈A.
0 0 0
The converse is not generally true.
Proof. As a counterexample consider A = {{1,2},{2,3}}. Then 1 ∈ {1,2} and {1,2} ∈ A, but
(cid:84) (cid:84)
1∈/ A since clearly A={2}.
A∈A A∈A
(d) The statement on the right is the definition of the statement on the left so of course the impli-
cation and its converse are true.
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Exercise 1.6
Write the contrapositive of each of the statements of Exercise 5.
Solution:
Again these involve simple logical negations of both sides of the implications:
(cid:83)
(a) x∈/ A for every A∈A⇒x∈/ A.
A∈A
(cid:83)
(b) x∈/ A for at least one A∈A⇒x∈/ A.
A∈A
(cid:84)
(c) x∈/ A for every A∈A⇒x∈/ A.
A∈A
(cid:84)
(d) x∈/ A for at least one A∈A⇒x∈/ A.
A∈A
Exercise 1.7
Given sets A, B, and C, express each of the following sets in terms of A, B, and C, using the symbols
∪, ∩ and −.
D ={x|x∈A and (x∈B or x∈C} ,
E ={x|(x∈A and x∈B) or x∈C} ,
F ={x∈A and (x∈B ⇒x∈C)} .
Solution:
First, we obviously have
D =A∩(B∪C)
E =(A∩B)∪C,
noting that D (cid:54)= E generally though they appear similar. Regarding F we have the following
sequence of logical equivalences:
x∈F ⇔x∈A∧(x∈B ⇒x∈C)
⇔x∈A∧(x∈/ B∨x∈C)
⇔x∈A∧¬(x∈B∧x∈/ C)
⇔x∈A∧¬(x∈B−C)
⇔x∈A∧x∈/ B−C
⇔x∈A−(B−C)
so that of course F =A−(B−C).
Exercise 1.8
If a set A has two elements, show that P(A) has four elements. How many elements does P(A) have if
A has one element? Three elements? No elements? Why is P(A) called the power set of A.
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Solution:
We claim that if a finite set has n elements, then its power set has 2n elements, which is why it is
called the power set.
Proof. We show this by induction on the size of the set. For the base case start with the the empty
set in which n=0. Clearly the only subset of ∅ is the trivial subset ∅ itself so that P(∅)={∅}.
This has 1 = 20 = 2n element obviously, which shows the base case. Now suppose that the power
set of any set with n elements has 2n elements. Let A be a set with n+1 elements, noting that this
is nonempty since n+1 ≥ 1 since n ≥ 0. Hence there is an x ∈ A. For any subset B ⊂ A, either
x∈/ B or x∈B. In the first case B is a subset of A−{x} and in the latter B ={x}∪C for some
C ⊂A−{x}. Therefore P(A) has twice the number of elements of P(A−{x}), one half being just
the elements of A−{x} and the other being those elements with x added in. But A−{x} has n
elements since A has n+1, and hence P(A−{x}) has 2n elements by the induction hypothesis.
Thus P(A) has 2·2n =2n+1 elements, which completes the induction.
Using this, we can answer all of the specific questions. If a set has two elements, than its power
set has 22 = 4 elements. If it has one element, then its power set has 21 = 2 elements, namely
P({x})={∅,{x}}. If a set has three elements then its power set has 23 =8 elements. Lastly, if a
set has no elements (i.e. it is the empty set), then its power set has 20 = 1 elements. As noted in
the proof we have P(∅)={∅}.
Exercise 1.9
Formulate and prove DeMorgan’s laws for arbitrary unions and intersections.
Solution:
In following suppose that A is a set and B is a nonempty collection of sets. For arbitrary unions,
we claim that
(cid:91) (cid:92)
A− B = (A−B).
B∈B B∈B
Proof. The simplest way to show this is with a series of logically equivalent statements. For any x
we have that
(cid:91) (cid:91)
x∈A− B ⇔x∈A∧x∈/ B
B∈B B∈B
⇔x∈A∧¬∃B ∈B(x∈B)
⇔x∈A∧∀B ∈B(x∈/ B)
⇔∀B ∈B(x∈A∧x∈/ B)
⇔∀B ∈B(x∈A−B)
(cid:92)
⇔x∈ (A−B),
B∈B
which of course shows the desired result.
For intersections, we claim that
(cid:92) (cid:91)
A− B = (A−B).
B∈B B∈B
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