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Math 310 - Summer 2006 Some Homework 2 Solutions
1.51 When f : A → B and S ⊆ B, we define If (S) = {x ∈ A : f (x) ∈ S}. Let X and Y be subsets of B.
a) Determine whether If (X ∪ Y ) = If (X) ∪ If (Y ).
Proof. These are in fact equal to each other as follows:
x ∈ If (X ∪ Y ) ⇐⇒ f (x) ∈ X ∪ Y ⇐⇒ f (x) ∈ X ∨ f (x) ∈ Y ⇐⇒ x ∈ If (X) ∨ x ∈ If (Y ) ⇐⇒ x ∈ If (X) ∪ If (Y ).
This shows that every element of If (X ∪ Y ) is an element of If (X) ∪ If (Y ) and vice versa. Thus since the two sets have the same elements, they are equal. Note that it is not enough to say If (X ∪ Y ) “distributes” across the union. That is what you are trying to prove happens!
b) Determine whether If (X ∩ Y ) = If (X) ∩ If (Y ).
Proof. Essentially the same proof works as above. Just replace “∪” with “∩” and “∨” with “∧”.
2.4 Let A and B be sets of real numbers, let f be a function from R to R, and let P be the set of positive real numbers. Without using works of negation, for each statement below write a sentence that expresses its negation.
a) For all x ∈ A, there is a b ∈ B such that b > x. There is an x ∈ A such that for all b ∈ B, b ≤ x.
b) There is an x ∈ A such that for all b ∈ B, b > x. For all x ∈ A, there is a b ∈ B such that b ≤ x.
c) For all x, y ∈ R, f (x) = f (y) =⇒ x = y. There are x, y ∈ R such that f (x) = f (y) and (but) x 6 = y.
d) For all b ∈ R, there is an x ∈ R such that f (x) = b. There is a b ∈ R such that for all x ∈ R, f (x) 6 = b.
e) For all x, y ∈ R and all ε ∈ P , there is a δ ∈ P such that |x − y| < δ implies |f (x) − f (y)| < ε. There are x, y ∈ R and ε ∈ P such that, for all δ ∈ P , |x − y| < δ and |f (x) − f (y)| ≥ ε.
2.16 Let f be a function from R to R.
a) Prove that f can be expressed in a unique way as the sum of two functions g and h such that g(x) = g(−x) and h(−x) = −h(x) for all x ∈ R.
Proof. Let g(x) = (f (x) + f (−x))/2 and h(x) = (f (x) − f (−x))/2. It is simple to check that f (x) = g(x) + h(x). Notice also that g(−x) = (f (−x) + f (x))/2 = g(x) and h(−x) = (f (−x) − f (x))/2 = −h(x)as desired. Further suppose that f (x) = g 1 (x) + h 1 (x) where g 1 and h 1 satisfy the same properties as g and h. Then
0 = f (x) − f (x) = g(x) + h(x) − g 1 (x) − h 1 (x)
so that g(x) − g 1 (x) = h 1 (x) − h(x).
Then
g(x) − g 1 (x) = g(−x) − g 1 (−x) = h 1 (−x) − h(−x) = −h 1 (x) + h(x) = −(g(x) − g 1 (x)).
The only way this can happen is if g(x) − g 1 (x) = 0 or g(x) = g 1 (x). Thus g is unique. A similar argument shows h is unique.
b) When f is a polynomial, express g and h in terms of the coefficients of f.
Proof. For a polynomial, g is the sum of the even powers with their coeffi- cients. Similarly h is the sum of the odd powers times their coefficients.
2.19 There is more than one possible solution to this problem.
