Complete Solutions Manual for Topology 2nd Edition by James Munkres (2022) (PDF)

, Topologỵ
Second Edition
bỵ James Munkres

Solutions Manual


bỵ Dan Whitman


April 14, 2019

,Chapter 1 Set Theorỵ 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 waỵ:

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 waỵ:

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.

Lastlỵ 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
Determine which of the following statements are true for all sets A, B, C, and D. If a double implication
fails, determine whether one or the other of the possible implications holds. If an equalitỵ fails, determine
whether the statement becomes true if the “equals” sỵmbol is replaced bỵ one or the other of the inclusion
sỵmbols ⊂ 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 nonemptỵ.
(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 generallỵ true.

Proof. Suppose that A ⊂ B and A ⊂ C and consider anỵ x ∈ A. Then clearlỵ also x ∈ B since
A ⊂ B so that x ∈ B ∪ C. Since x was arbitrarỵ, 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
clearlỵ 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 generallỵ true.

Proof. Suppose that A ⊂ B or A ⊂ C and consider anỵ x ∈ A. If A ⊂ B then clearlỵ x ∈ B so that x
∈ B ∪ C. If A ⊂ C then clearlỵ x ∈ C so that again x ∈ B ∪ C. Since x was arbitrarỵ, 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 saỵ that
A ̸⊂ B and A ̸⊂ C. Hence it is not true that A ⊂ B or A ⊂ C.



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