The Basics of Proofs
7. The Completeness Axiom is sometimes called the Least Upper Bound Principle. The Completeness Axiom comes up frequently in proofs
2.6.1 Function, Domain, Codomain
Intuitively, a function is a mapping that assigns to each point of some domain A a value that resides in some codomain B. This is usually written f W A ! B. More precisely, the function f is defined as a set of ordered pairs.x; y/ where each x resides in the domain A of f and each y resides in the codomain B of f , and for each x 2 A there is exactly one y 2 B such that.x; y/ 2 f . Since there is a unique ordered pair .x; y/ 2 f for each x 2 A, f associates or links the value of y to the value of x and allows one to write f.x/ D y.
2.6.2 Surjection
The domain of the function f is exactly the set of all x that are first coordinates of the order pairs in f , that is, the domain is A D fx j.x; y/ 2 f g. The range of f is defined as the image of f , that is, the range is fy j.x; y/ 2 f g. Clearly, the codomain of f can be any set that contains the range of f . This can lead to some confusion since the codomain of f is not precisely defined. It is simply a convenience. When one defines a function f WR ! R, one means that f is defined for every real number, and that for any x 2R, the value f .x/ also lies in R. This is the case whether or not R is the range of f or if the range of f is actually some proper subset of R. It could be difficult and unnecessary to calculate exactly which subset ofR is the range of f , so it might be easier to just give the codomain asR and avoid the technicalities of figuring out just what values ofR are in the range of f . For example, the function f.x/ D 3x6 15x4C 12x3C 25x2 32x C 14 maps the real numbers into the real numbers, but to find the range of f , you would need to find the minimum value of f . This minimum exists, but it may not be possible to give its value explicitly.
2.6 Functions 41 If the range of f W A ! B is actually all of B, we say that f maps A onto B, and f is called surjective, and f is called a surjection. Thus, one can prove that a function is surjective by showing that each element of the codomain is in the range.
TEMPLATE for proving a function f is surjective
• SET THE CONTEXT: Make a statement introducing f , its domain A, and its codomain B.
• Select an arbitrary value y 2 B.
• Exhibit a value x 2 A such that y D f.x/.
• STATE THE CONCLUSION: Therefore, f is a surjection.
Note that the crucial step in proving that a function is surjective is showing the existence of an x with f.x/ D y and verifying that the x is in the domain A of the function. For example, the function f.x/ D 5x2C 1 is a surjection from the negative real numbers onto the interval.1; 1/. To prove this you would need to show that for each real number y> 1 there is a negative real number x for which f .x/ D y. But this just involves a simple algebraic manipulation. That is, if you need5x2C 1 D y, then you can solve to get5x2D y1 and x2 D y15 . Here one needs to be careful because it is easy to continue by writing x D
qy1
5 which always results in a positive value for x. The proof needs to exhibit a negative value for x, so it is important to set x D q
y1
5 . There is no need for the proof to display the steps of solving the equation for x. The goal is to produce a value of x 2 A such that f.x/ D y; how you arrived at that x is not important. It may be interesting, but it is not an essential part of the proof, and, therefore, it should not be part of the proof.
PROOF: The function f.x/ D 5x2C 1 is a surjection from the negative real numbers onto the interval.1; 1/.
• Let f.x/ D 5x2C 1.
The definition of function requires that each value x in the domain of f is found in exactly one ordered pair.x; y/ 2 f . The same does not have to hold for values in
the codomain, that is, one value y in the codomain could appear in many order pairs .x; y/ 2 f . For example, for the constant function f W R ! R given by f .x/ D 1 for all x 2 R, the value 1 appears as the second coordinate in all the ordered pairs of the function. If a function has the property that no value of y appears as the second coordinate of more than one ordered pair in f , then f is said to be injective or, less formally, that f is one-to-one. In this case the function f is called an injection. In such a case, one sees that f.x1/ D f .x2/ only if x1 D x2. This gives a procedure for proving that a function is injective.
TEMPLATE for proving a function f is injective
• SET THE CONTEXT: Make a statement introducing f , its domain A, and its codomain B.
• Assume that for two values x1and x2in A that f.x1/ D f .x2/.
• Show that x1D x2.
• STATE THE CONCLUSION: Therefore, f is an injection.
For example, the function f.x/ Dp
4x C 7 maps the positive real numbers to the positive real numbers. It is not a surjection, but it is an injection. The proof would require that you show that f.x1/ D f .x2/ implies that x1D x2. Again, this is just an algebraic manipulation.
PROOF: The function f.x/ Dp
4x C 7 is an injection from the positive real numbers to the positive real numbers.
• Let f.x/ Dp
If a function f W A ! B is both surjective and injective, that is, if f is both one-to-one and onto, then f is bijective, and f is called a bijection. In this case, f exhibits a one-to-one correspondence between the set A and the set B.
Two functions f and g whose ranges are in the real numbers can be combined arithmetically. Specifically, one can define f C g, f g, fg, and gf in natural ways:
.f C g/.x/ D f .x/ C g.x/, .f g/.x/ D f .x/ g.x/, .fg/.x/ D f .x/ g.x/, and, for x such that g.x/ ¤ 0,
f g
.x/ D fg.x/.x/. When functions f and g are combined in this way, the domain of the sum, difference, product, or quotient is assumed to be the intersection of the domain of f and the domain of g with the exception that the domain of fg also excludes values of x for which g.x/ D 0. Thus, the function f.x/ Dp
x 4 is defined for all x 4, and the function g.x/ Dp
5 x is defined for all x 5. It follows that the functionp
x 4 Cp
5 x is defined only for those x satisfying4 x 5. Similarly, the function ff.x/.x/ is only defined for x > 4 even though it is identically 1 for those x. That function has a natural extension to all real numbers.
2.6 Functions 43
A
B
C x
y
z g(x)
f(x)
f g◦ Fig. 2.5 Composition.f ı g/.x/ D z
2.6.4 Composition
If g is a function assigning values in its domain A to values in its range contained in the set B, and if f is a function assigning values in its domain B to values in its range contained in the set C, then the composition of f with g is the function .f ı g/.x/ D f
g.x/
which assigns to values in its domain A values in its range contained in set C (Fig.2.5). The main reason for considering compositions is that it is often easiest to represent complicated functions as compositions of simpler functions. For example, the function f.x/ D psinxC42x is clearly a quotient where the numerator is the composition of the function x2 with the function sin x, and the denominator is the composition of the functionp
x with the function x C4.
It is easily shown that if g W A ! B and f W B ! C are both surjective functions, then their composition, f ı g W A ! C, is also surjective. To prove this, you would follow the template for proving that a function is surjective. That requires that you select an arbitrary z 2 C and show that there is an x 2 A such that.f ı g/.x/ D z.
Why might you use the variable z here rather than the variable y? Well, that allows you to think of g as mapping x to y, and f , in turn, mapping y to z. Faced with the statement f
g.x/
D z, there is little you can do except to apply what you know about the function f , that is, that f is surjective. Because f is surjective, and z is in the codomain of f , you know that there is a y in the domain of f such that f.y/ D z.
Can you find an x such that g.x/ D y? Of course y is in the domain of f which is the codomain of g. The function g is surjective, so there must be an x in the domain of g that maps onto y. These ideas give the following proof.
PROOF: If gW A ! B and f W B ! C are both surjective functions, then their composition f ı g W A ! C is also surjective.
• Let g W A ! B and f W B ! C be two surjective functions.
• Let z 2 C.
• Then since f is a surjection from B to C, there is a y 2 B such that f.y/ D z.
• Since g is a surjection from A to B, there is an x 2 A such that g.x/ D y.
• Therefore,.f ı g/.x/ D f g.x/
D f .y/ D z.
• It follows that f ı g W A ! C is surjective.
It is also true that if g W A ! B and f W B ! C are both injective functions, then their composition, f ı g W A ! C, is also injective. You would prove this by following the template for proving that a function is injective. That is, you would assume that.f ı g/.x1/ D .f ı g/.x2/ for some x1and x2in A. Again, what can you say if you know f
g.x1/ D f
g.x2/
? All that you can do is apply what you know about the function f , that is, that f is injective. Since f is injective, you can conclude that g.x1/ D g.x2/. Then because g is injective, you can conclude x1D x2, and you are done.
PROOF: If gW A ! B and f W B ! C are both injective functions, then their composition f ı g W A ! C is also injective.
• Let g W A ! B and f W B ! C be two injective functions.
• Assume that for some x1and x2in A,.f ı g/.x1/ D .f ı g/.x2/.
• By the definition of composition f g.x1/
D f g.x2/
.
• Then since f is an injection, it follows that g.x1/ D g.x2/.
• Since g is an injection, it follows that x1 D x2.
• It follows that f ı g W A ! C is injective.
2.6.5 Exercises
Write a proof for each of the following statements.
1. For each real number r there is a real number x such that x3D r.
2. For each real number r 0 there is a real number x 0 such that x4D r.
3. If n is an odd positive integer, then for each real number r there is a real number x such that xnD r.
4. If n is an even positive integer, then for each real number r 0 there is a real number x 0 such that xnD r.
5. If h W A ! B, g W B ! C, and f W C ! D are three functions, then.f ı g/ ı h D fı .g ı h/. In other words, function composition is associative. (Hint: Show that both functions.f ı g/ ı h and f ı .g ı h/ give the same result when applied to an x2 A.)
2.6 Functions 45 6. If h W A ! B, g W B ! C, and f W C ! D are three surjective functions, then
their composition f ı g ı h is surjective.
7. If h W A ! B, g W B ! C, and f W C ! D are three injective functions, then their composition f ı g ı h is injective.