Part A

1. Let A,B and C be sets. Prove A×(B∪C)=(A×B)∪(A×C).
2. For each of the following statements, determine if it is true or false for all sets A,B,C and D. In the former case, give a proof. In the latter, give an explicit counterexample, including an explanation of how your purported counterexample shows the given statement to be false
   1. (A×B)∪(C×D)=(A∪C)×(B∪D)
   2. (A×B)∩(C×D)=(A∩C)×(B∩D)
3. Suppose that A and B are sets. Prove that there exists an injection f:A→B if and only if there exists a surjection g:B→A.

Part B

4. Here are some important facts about images and inverse images: for any sets X and Y, any subsets A,B⊆X and C,D⊆Y, and any function f:X→Y, we have
   • f(A∪B)=f(A)∪f(B).
   • f(A∩B)⊆f(A)∩f(B).
   • f−1(C∪D)=f−1(C)∪f−1(D)
   • f−1(C∩D)=f−1(C)∩f−1(D)
   • f−1(f(A))⊇A
   • f(f−1(C))⊆C
   1. Prove the first assertion above.
   2. Prove the fourth assertion above.
   3. Give an explicit example which shows that the second fact is false if we replace “⊆” with equality. (Though not stated explicitly at such, this implicitly asks you to provide specific values of A,B,C,D,X,Y and f, and then prove that the equality of set fails in this case.)
   4. Prove that equality holds in the fifth fact for all subsets A⊆X if and only if f is injective.
5. If f:X→Y and g:U→V satisfy f(X)⊆U, then we define the composition function g∘f:X→V by the following rule: for any x∈X, we let (g∘f)(x)=g(f(x)).
   1. Prove that if f and g are injections, then g∘f is an injection.
   2. Suppose f and g are surjections and that Y=U. Prove that g∘f is a surjection.
6. Prove that |(0,1)|=|[0,1]| by constructing an explicit bijection from (0,1) to [0,1].[Hint: (1) We’ll see later in the semester that it’s impossible for there to be a continuous bijection from (0,1) to [0,1], so the function you create will wind up being discontinuous. (2) Rosenlicht gives an explanation for why infinite sets are precisely the sets which have a proper subset with which they are in bijection; the idea he uses can be adapted for your construction.]