Homework 1 Solutions
Problem 2.1.
Let f : A → B, let A0 ⊂ A and B0 ⊂ B.
(a)
Proposition. A0 ⊂ f −1 (f (A0 )), with equality if f is injective.
Proof. If x ∈ A0 , then f (x) ∈ f (A0 ), so x ∈ f −1 (f (A0 )). This proves that A0 ⊂ f −1 (f (A0 )).
For the converse, suppose that f is injective, and let x ∈ f −1 (f (A0 )). Then f (x) ∈ f (A0 ),
so f (x) = f (x0 ) for some x0 ∈ A0 . Since f is injective, it follows that x = x0 , and thus
x ∈ A0 .
(b)
Proposition. f (f −1 (B0 )) ⊂ B0 , with equality if f is surjective.
Proof. If y ∈ f (f −1 (B0 )), then y = f (x) for some x ∈ f −1 (B0 ). Then f (x) ∈ B0 , which is to
say that y ∈ B0 . This proves that f (f −1 (B0 )) ⊂ B0 .
For the converse, suppose that f is surjective, and let y ∈ B0 . Since f is surjective,
y = f (x) for some x ∈ A. Then f (x) ∈ B0 , so x ∈ f −1 (B0 ), and hence y ∈ f (f −1 (B0 )).
Problem 3.5.
Let S and S 0 be the following subsets of the plane:
S = {(x, y) | y = x + 1 and 0 < x < 2},
S 0 = {(x, y) | y − x is an integer}.
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(a)
Proposition. S 0 is an equivalence relation on the real line and S 0 ⊃ S.
Proof. Note first that (x, x) ∈ S 0 for all x ∈ R, since x − x = 0 is an integer. Next, if
(x, y) ∈ S 0 , then y − x is an integer, so x − y is an integer, and hence (y, x) ∈ S 0 . Finally, if
(x, y), (y, z) ∈ S 0 , then y − x and z − y are integers, so z − x = (z − y) + (y − x) is an integer,
and hence (x, z) ∈ S 0 .
For the second part, if (x, y) ∈ S, then y = x + 1, so y − x = 1 is an integer, and hence
(x, y) ∈ S 0 .
The equivalence relation S 0 has one equivalence class for each number x ∈ [0, 1), namely
the set {x + n | n ∈ Z}.
(b)
Proposition. If C is a collection of equivalence relations on a set A, then
T
C is an equiv-
alence relation on A.
Proof. If x ∈ A, then (x, x) ∈ E for each equivalence relation E ∈ C, and hence (x, x) ∈
Next, if (x, y) ∈
T
C.
C, then it must be the case that (x, y) ∈ E for each equivalence relation
E ∈ C. Then (y, x) ∈ E for each E ∈ C, and therefore (y, x) ∈
T
T
T
C. Finally, if (x, y), (y, z) ∈
C, then (x, y), (y, z) ∈ E for each equivalence relation E ∈ C. Then (x, z) ∈ E for each
E ∈ C, so (x, z) ∈
T
C.
(c) T is the following equivalence relation on the real line
T = {(x, y) | y − x is an integer and either x = y or x, y ∈ (0, 3)}.
The equivalence classes for T are the following sets:
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1. The set {1, 2}.
2. The set {x, x + 1, x + 2} for each x ∈ (0, 1).
3. The set {x} for each x ≤ 0 or x ≥ 3.
Problem 5.4.
(a) f (x1 , . . . , xm ) = (x1 , . . . , xm , x1 , . . . , x1 ), with x1 repeated n − m times at the end.
(b) g(x1 , . . . , xm ) = ((x1 , . . . , xm ), (y1 , . . . , yn )) = (x1 , . . . , xm , y1 , . . . yn ).
(c) h(x1 , . . . , xn ) = (x1 , . . . , xn , x1 , x1 , . . .)
(d) k((x1 , . . . , xn ), (y1 , y2 , . . .)) = (x1 , . . . , xn , y1 , y2 , . . .)
(e) l((x1 , x2 , . . . , ), (y1 , y2 , . . .)) = (x1 , y1 , x2 , y2 , . . .).
(f) m((a11 , a12 , . . .), (a21 , a22 , . . .), . . . , (an1 , an2 , . . .)) = (a11 , a21 , . . . , an1 , a12 , a22 , . . . , an2 , . . .).
Problem 6.3.
Let S = {(x1 , x2 , . . .) ∈ X ω | x1 = 0}. Then S is a proper subset of X ω , and the function
f : X ω → S defined by
f (x1 , x2 , . . .) = (0, x1 , x2 , . . .)
is a bijection.
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