SECTION 4.1 Basics of Set Theory 199
- Suppose thatf :R → R is a cubic function such thatf′(x) 6 = 0
for all x∈ R. Give an intuitive argument (I’m not asking for a
formal proof) thatf must be bijective. - Definef :R×R→Rby settingf(x,y) =x−y. Show thatf is
onto but is not one-to-one. - Definef :R×R → Rby setting f(x,y) =x^2 +y. Show thatf
is onto but is not one-to-one. - Let f : C → C be aquadratic function. Therefore, there are
complex constantsa, b, c∈Cwith a 6 = 0 such thatf(x) =ax^2 +
bx+c for all x ∈ C. Prove that f is onto but not one-to-one.
(Compare with Exercise 1, above.) - For the mapping given in Exercise 3, above, show that the fibre
over each point inRis a line inR×R. - What are the fibres of the mapping in Exercise 4?
- (A guided exercise) LetA be a set and let 2A be its power set.
Let’s show that there cannot exist any surjective function
f :A→ 2 A. A good way to proceed is toargue by contra-
diction, which means that we’ll assume that, in fact, a surjective
function exists and then reach a contradiction! So let’s assume that
f:A→ 2 A is surjective. Note first that for any elementa∈A, it
may or may not happen thata∈f(a) (this is important!). Now
consider the following strange subset ofA:
A 0 = {a∈A|a6∈f(a)}∈ 2 A.IsA 0 = f(a 0 ) for some element a 0 ∈A? Think about it! This
contradiction has the same flavor as Russell’s paradox!