a∈Qn. Ra. Since each Ra is finite, A is a countable union of countable unions of
finite sets, and is ... Since A is countable, we conclude that T is uncountable. 1 ...
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}. 1
(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: 2
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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