Math 10850: Honors Calculus I, Fall 2020 Homework 3 Solutions I want to emphasize that all your solutions should be written out in English prose using complete sentences and correct grammar. Strive for precision, and do not simply turn in your scratchwork!!! m 1. (a) Let f : Z × Z → Q be defined by the formula f(n, m) = |n|+1 . Determine whether or not f is injective and/or surjective.

Solution: The function f is surjective. Indeed, consider x ∈ Q. Our goal is to find some (n, m) ∈ Z × Z such that f(n, m) = x. We can write x = a/b with a, b ∈ Z and b 6= 0. Multiplying the numerator and denominator of x = a/b by −1 if necessary, we can assume that b is positive. Setting (n, m) = (b − 1, a), we then have a a a f(n, m) = f(b − 1, a) = = = = x, |b − 1| + 1 b − 1 + 1 b where the third equality uses the fact that b > 0, so |b − 1| = b − 1. The claim follows.

However, the function f is not injective. Indeed, (1, 0) 6= (2, 0), but f(1, 0) = f(2, 0) = 0.

(b) Recall that R[t] is the set of polynomials in a single variable t with real coefficients. 0 0 Let D : R[t] → R[t] be the function D(f) = f , where f is the derivative of the polynomial f. Is D injective? Is D surjective? Make sure to prove your answer!

Solution: The function D is not injective. Indeed, the constant polynomials 0 and 1 are different, but D(0) = D(1) = 0.

However, the function D is surjective. Indeed, consider f ∈ R[t]. We must find some g ∈ R[t] such that D(g) = f. Write n X k f(t) = akt with a0, . . . , an ∈ R. k=0 Setting n X ak g(t) = tk+1 ∈ [t], k + 1 R k=0 we then have D(g) = f, as desired.

2. (a) Given a function f : X → Y and a subset A ⊂ X, is it always true that f −1(f(A)) = A? Either prove it or give a counterexample.

Solution: This is not always true. Indeed, let f : R → R be the constant function f(x) = 1 and let A = {0, 1, 2}. We then have f(A) = {f(0), f(1), f(2)} = {1}, so −1 −1 f (f(A)) = f ({1}) = R.

1 (b) Given a function f : X → Y and a subset B ⊂ Y , is it always true that f(f −1(B)) = B? Either prove it or give a counterexample.

Solution: This is not always true. Indeed, let f : R → R be the constant function f(x) = 1 and let B = R. We then have −1 −1 f (B) = f (R) = R, so −1 f(f (B)) = f(R) = {1}.

(c) Let A be a set and let f, g : A → A be two functions such that f ◦ f = g ◦ g. Must it be the case that f = g? Either give a proof or a counterexample.

Solution: This need not be the case. Indeed, let f : R → R be the constant function f(x) = 1 and let g : R → R be the function ( 1 if x 6= 7, g(x) = 0 if x = 7.

Thus f 6= g. However, both f ◦ f and g ◦ g are the constant function 1.

(d) Construct functions f : X → Y and g : Y → Z with the following properties: i. g ◦ f is injective but g is not injective (remember in class we proved that if g ◦ f is injective then f is injective!).

Solution: Let f : {1} → R be the function defined by f(1) = 1 and let g : R → R be the constant function g(x) = 1. The function g is not injective, but g ◦ f : {1} → R is function defined by g ◦ f(1) = 1, which is injective (this is a place where the domain really matters!).

ii. g ◦ f is surjective but f is not surjective (remember in class we proved that if g ◦ f is surjective then g is surjective!).

Let f : R → R be the constant function f(x) = 1 and let g : R → {1} be the constant function g(x) = 1. Then f is not surjective, but g ◦ f : R → {1} is the constant function g ◦ f(x) = 1, which is surjective (this is a place where the codomain really matters!).

2 2 3. Let N be the set of all ordered pairs (n, m) with n, m ∈ N. Prove that N is countably infinite. Hint: think about our proof that Q is countably infinite.

Solution: Since the minimal element of N is 1, the minimal value of n + m as (n, m) 2 2 ranges over N is n + m = 1. Enumerate N by first listing all the pairs (n, m) with n + m = 2, then listing all pairs with n + m = 3, etc:

(1, 1), (1, 2), (2, 1), (3, 1), (2, 2), (1, 3), (4, 1), (3, 2), (2, 3), (1, 4), (5, 1),...

2 th The bijection f : N → N then is defined by letting f(n) be the n term in this sequence.

2 4. Let X be the set of all finite subsets of N. Thus some elements of X include ∅ and {1, 5, 9} and {3, 346} and {1}; however, the set of even natural numbers is not an element of X. Prove that X is countable. Hint: the best way to think about this is to try to figure out a way to systematically enumerate X like we did in class with Q.

Solution: Define σ : X → N to be the function taking S ∈ X to the sum of all elements in the finite subset S ⊂ N. Thus for instance σ(∅) = 0 and σ({1, 5, 9}) = 15. For all n ∈ N, the set of all elements S ∈ X with σ(S) = n is finite. Enumerate X by first listing all elements S with σ(S) = 0, then all elements of S with σ(S) = 1, then all elements of S with σ(S) = 2, etc:

∅, {1}, {2}, {1, 1}, {3}, {1, 2}, {1, 1, 1}, {4},...

2 th The bijection f : N → N then is defined by letting f(n) be the n term in this sequence.

5. Let A ⊂ N be an infinite subset. Prove that A is countably infinite.

Solution: Start by listing all elements of N in order:

1, 2, 3, 4,...

Delete all elements from this list that do not lie in A. The result is a listing

a1, a2, a3,...

of all elements of A. The bijection f : N → A is then defined by letting f(n) = an.

6. Prove that R has the same cardinality as an open interval (a, b). Hint: try to construct a bijection f : R → (a, b) by coming up with an appropriate formula. Think geometrically!

Solution: We first observe that the function φ: (0, 1) → (a, b) defined by φ(x) = a + 1 (b−a)x is a bijection with two-sided inverse ψ : (a, b) → (0, 1) defined by ψ(x) = b (x−a). It is thus enough to construct a bijection between (0, 1) and R. There are many ways to do this. A nice selection of some of them is at https://math.stackexchange.com/questions/200180/

1 1 2 2 2 7. Let S denote the unit circle in the (x, y)-plane, so S = {(x, y) ∈ R | x + y = 1}. 2 1 (a) Let X = {(x, y) ∈ R | (x, y) ∈ S and (x, y) 6= (1, 0)}. Prove that X has the same cardinality as R. Hint: find a nice formula for the bijection.

Solution: In the previous problem, we established that there is a bijection f : R → (0, 2π). The desired bijection g : R → X is then

g(x) = (cos(f(x)), sin(f(x))).

1 (b) Prove that S has the same cardinality as R by finding a bijection between R and S1. Though S1 only has one more point than X, this is trickier than you might expect! You shouldn’t expect a simple formula...

3 Solution: Let g : R → X be the bijection from the previous part and let G: X → 1 R be its two-sided inverse. Define h: R → S via the formula  g(x) if x∈ / ,  N h(x) = g(x − 1) if x ∈ N and x 6= 1, (1, 0) if x = 1.

In other words, we modify g precisely on the natural numbers to “shift its values up by 1 and make g(1) the missing point (1, 0).” This is a bijection whose 1 two-sided inverse is the function H : S → R defined via the formula  G(p) if p 6= (1, 0) and p∈ / g( ),  N H(p) = G(p) + 1 if p ∈ g(N), 1 if p = (1, 0).

8. (Extra credit) Say that x ∈ R is truly excellent if for all n ≥ 1, there is a sequence of n digits that appears infinitely many times in the decimal expansion of x. For instance, all rational numbers are truly excellent, which is best illustrated in the following examples: • 1/4 is truly excellent since its decimal expansion is 0.250, where the overline indicates that 0 repeats. For any n ≥ 1, you can thus find a sequence of n zeros infinitely many times in the decimal expansion of 1/4. • 5/13 is truly excellent since its decimal expansion is 0.384615. For any n ≥ 1, you can thus find a length-n sequence composed of many copies of 384615 (with possibly only a portion of this at the end) that repeats infinitely many times. 2 Problem: Prove that if x ∈ R is truly excellent, then x is truly excellent.

Solution: This is a trick problem: in fact, every element of R is truly excellent! Indeed, consider some x ∈ R. Letting n ≥ 1, we must prove that there is a sequence of n digits in the decimal expansion of x that repeats infinitely many times. Let A be the finite set of all length-n sequences of digits chosen from {0, 1,..., 9}. We then consider the decimal expansion of x. Let a1 ∈ A be the first n digits in that decimal expansion after the decimal point, let a2 ∈ A be the next n digits in that decimal expansion, let a3 ∈ A be the next n digits in that decimal expansion, etc. Thus x equals

±(nonnegative-whole-number).a1a2a3a4 ··· ,

where each ai ∈ A represents n digits. Since A is finite, there must be some a ∈ A that appears infinitely many times in this expression. The claim follows.

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