TEST BANK
All Chapters Included
, AN ILLUSTRATED
INTRODUCTION TO
TOPOLOGY
and
HOMOTOPY
S0LUTIONS MANUAL
FOR PART 1
TOPOLOGY
SASHO KALAJDZIEVSKI
IN COLLABORATION WITH
DEREK KREPSKI
DAMJAN KALAJDZIEVSKI
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
A C H A P M A N & HA L L BOOK
,0.1 Sets and Numbers 1
Chapter 1: Sets, Numbers and Cardinals
1.1 Sets and Numbers.
Solutions of some exercises
2. Given a set X, show that the relation is an order of the set of all subsets of X. For
which sets X is this order linear?
Solution. If A, B X are such that A B and A B , then there is b B such that
b A . Consequently B is not a subset of A, and hence is antisymmetric. If A B C
then obviously A C , and so is transitive.
If X has at least two elements, say a and b, then neither {a} {b} nor {b} {a} ,
so the order is not linear. On the other hand if X has at most one element, then the only
subsets of X are X and , and we then readily see that the order is linear.
3. Describe a linear order over (a) the set ℕ2 , and (b) the set ℝ2 .
Solution for (a). Define (n, m) ( p, q) if n p or ( n p and m q ). The parentheses in
the preceding sentence are to guarantee there is unique interpretation of the statement that
defines <. It is left to the reader to prove this relation is antisymmetric and transitive.
4. Show that if ~ is an equivalence relation over a set X, then every two equivalence
classes are either disjoint or equal.
Solution. Suppose [x] and [y] are two equivalence classes, and suppose [x] [y] .
Then there is a [x] [y] . Take any z [x] . Then a ~ x ~ z , and hence a ~ z . On the
other hand, a [y] implies that y ~ a . The transitivity of ~ applied to y ~ a and a ~ z
yields y ~ z . Hence z [y] . We proved that [x] [y] . By the symmetry of the argument,
it follows that [y] [x] . Hence [x] [y] .
7. Let X be a non-empty set and let f : X Y be any mapping. Show that “ u ~ v if and
only if f (u) f (v) ” defines an equivalence relation over X.
Solution. (i) Reflexivity: u ~ u for every u, since f (u) f (u) for every u. (ii) Symmetry:
Suppose u ~ v . Then f (u) f (v) , hence f (v) f (u) , hence v ~ u . (iii) Transitivity:
, 0.1 Sets and Numbers 2
Suppose u ~ v and v ~ w . Then f (u) f (v) and f (v) f (w) . Hence f (u) f (w), and
we conclude that u ~ w .
All Chapters Included
, AN ILLUSTRATED
INTRODUCTION TO
TOPOLOGY
and
HOMOTOPY
S0LUTIONS MANUAL
FOR PART 1
TOPOLOGY
SASHO KALAJDZIEVSKI
IN COLLABORATION WITH
DEREK KREPSKI
DAMJAN KALAJDZIEVSKI
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
A C H A P M A N & HA L L BOOK
,0.1 Sets and Numbers 1
Chapter 1: Sets, Numbers and Cardinals
1.1 Sets and Numbers.
Solutions of some exercises
2. Given a set X, show that the relation is an order of the set of all subsets of X. For
which sets X is this order linear?
Solution. If A, B X are such that A B and A B , then there is b B such that
b A . Consequently B is not a subset of A, and hence is antisymmetric. If A B C
then obviously A C , and so is transitive.
If X has at least two elements, say a and b, then neither {a} {b} nor {b} {a} ,
so the order is not linear. On the other hand if X has at most one element, then the only
subsets of X are X and , and we then readily see that the order is linear.
3. Describe a linear order over (a) the set ℕ2 , and (b) the set ℝ2 .
Solution for (a). Define (n, m) ( p, q) if n p or ( n p and m q ). The parentheses in
the preceding sentence are to guarantee there is unique interpretation of the statement that
defines <. It is left to the reader to prove this relation is antisymmetric and transitive.
4. Show that if ~ is an equivalence relation over a set X, then every two equivalence
classes are either disjoint or equal.
Solution. Suppose [x] and [y] are two equivalence classes, and suppose [x] [y] .
Then there is a [x] [y] . Take any z [x] . Then a ~ x ~ z , and hence a ~ z . On the
other hand, a [y] implies that y ~ a . The transitivity of ~ applied to y ~ a and a ~ z
yields y ~ z . Hence z [y] . We proved that [x] [y] . By the symmetry of the argument,
it follows that [y] [x] . Hence [x] [y] .
7. Let X be a non-empty set and let f : X Y be any mapping. Show that “ u ~ v if and
only if f (u) f (v) ” defines an equivalence relation over X.
Solution. (i) Reflexivity: u ~ u for every u, since f (u) f (u) for every u. (ii) Symmetry:
Suppose u ~ v . Then f (u) f (v) , hence f (v) f (u) , hence v ~ u . (iii) Transitivity:
, 0.1 Sets and Numbers 2
Suppose u ~ v and v ~ w . Then f (u) f (v) and f (v) f (w) . Hence f (u) f (w), and
we conclude that u ~ w .
