Topology – Solutions Manual by D. Whitman (for J.R. Munkres Textbook)

SOLUTIONS MANUAL




Page 1

,Topology 2nd edition Solutions Manual by Dan Whitman
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

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
equality fails, determine whether the statement becomes true if the “equals” symbol is replaced by
one or the other of the inclusion symbols ⊂ or ⊃.


(a) A ⊂ B and A ⊂ C ⇔ A ⊂ (B ∪ (j) A ⊂ C and B ⊂ D ⇒ (A × B) ⊂ (C × D).
C). (k) The converse of (j).
(b) A ⊂ B or A ⊂ C ⇔ A ⊂ (B ∪ C).
(l) The converse of (j), assuming that A and B
(c) A ⊂ B and A ⊂ C ⇔ A ⊂ (B ∩ are nonempty.
C). (m) (A × B) ∪ (C × D) = (A ∪ C) × (B ∪ D).
(d) A ⊂ B or A ⊂ C ⇔ A ⊂ (B ∩ C).
(n) (A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D).
(e) A − (A − B) = B.
(o) A × (B − C) = (A × B) − (A × C).
(f) A − (B − A) = A − B.
(p) (A −B) ×(C −D) = (A ×C −B ×C) −A ×D.
(g) A ∩ (B − C) = (A ∩ B) − (A ∩
(q) (A × B) − (C × D) = (A − C) × (B − D).
C).
(h) A ∪ (B − C) = (A ∪ B) − (A ∪
C).
(i) (A ∩ B) ∪ (A − B) = A.




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, 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 ⊄ B and A ⊄ C. Hence it is not true that A ⊂ B or A ⊂ C.




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