Anonymous asked in Science & MathematicsMathematics · 9 years ago

Sets and Maps, I need some advice on the properties of these maps?

I have a few questions that I need help on, for two of them I believe I have a way to do them and in some cases an answer, however I am not certain that they are accurate.

1) For the map of phi: the set of natural numbers (starting with 0) goes to the set of integers with x mapped to x^6 (x--->x^6) Determine whether the map is bijective.

I was thinking that this map would be injective but not surjective therefore making it not bijective. Is this correct?

2) This problem I need some help starting it: Let M and N be sets with k and l elements, respectively. Determine the number of possible maps from M to N.

3) Find and example of the maps phi:M-->N and psi: N-->P such that Psi of Phi is surjective but phi is not surjective. For this problem I was thinking that I would need to find a map where phi is not surjective and a map where psi is bijective in order to get psi of phi to be bijective and phi not to be surjective. Any thoughts?



Would number 2 have an infinite number of maps between M--->N?

3 Answers

  • 9 years ago
    Best Answer

    1. You're right.

    2. Since M has k elements, let's name the elements: M = {x1, ..., xk}. A map, or function, from M to N consists of k ordered pairs, each one looking like {xi, y}, where 1 <= i <= k and y is some member of N. There are l (lower-case L) choices for y in each of those ordered pairs, so the total number of ways you can choose the y's is l times l ... times l (k factors) or l^k. Definitely not an infinite number, to answer your follow-up question.

    3. Let's make psi of phi bijective. (You said surjective the first time, but if it's bijective it'll certainly be surjective.) If M, N, and P can be any sets, this is actually pretty easy. For example, let M and P be the non-negative real numbers, and let N be all the real numbers. Let phi be the identity function (in other words, phi(x) = x for all x), and let psi be the absolute value function. Then phi is not surjective, because its range doesn't contain any negative numbers, but psi of phi of any non-negative real number x is x, so psi of phi is the identity function from M to P.

    Notice that psi isn't injective, let alone bijective, on the set of all real numbers (for example, psi(-1) = psi(1) = 1); however, its restriction to the non-negative real numbers (which is what psi of phi amounts to) is injective and in fact bijective.

  • 9 years ago

    1) you are correct. I'll denote the map with f.

    injectivenes: for any x different from y ==> f(x) different from f(y). This is obvious.

    surjectivenes: you would have to find that for any natural y there exists at least a natural x such that y=f(x). So if you find one counterexample the function is not surjective. y=8 let's see what natural number x has the property that x^6 = 8 ==> x = 8^(1/6) = (2^3)^(1/6) = 2^[3*(1/6)] = 2^(1/2) = sqrt(2). Any respectable math student knows that proof that sqrt(2) is not rational (yet alone natural!). So f is not surjective. ==> f is not bijective

    2) A map is a relation that associates each element of the domain to one element from the range.

    Imagine that the elements of M are k boxes. In each box you can put any one of the l elements of N. Multiply the possibilities and you get l*l*l*...*l*l ---> that is k times l ==> l^k functions.

    I'll give you an example:


    1) f(a)=f(b)=f(c)=1

    2) f(a)=f(b)=1 , f(c)=2

    3) f(b)=f(c)=1 , f(a)=2

    4) f(c)=f(a)=1 , f(b)=2

    5) f(a)=f(b)=2 , f(c)=1

    6) f(b)=f(c)=2 , f(a)=1

    7) f(c)=f(a)=2 , f(b)=1

    8) f(a)=f(b)=f(c)=2

    you got 2^3 = 8 functions.

    3) It's a interval's game here:

    f:R-->R , f(x)=x^2 is not surjective because -1 = x^2 has no real solution

    g:R-->R , g(x)=|x| is not surjective because -100 = |x| has no real solution

    BUT, when we define them like this: f,g:R-->[0, infinity) they are surjective because all values in the range have at least one real solution


    phi:R-->R , phi(x) = x^2 is not surjective

    psi:R-->[0; infinity) , psi(x) = |x|

    (psi_of_phi):R-->[0; infinity) , (psi_of_phi)(x) = psi(phi(x)) = |x^2| which is surjective as noticed above.

    So, here are your examples.

    Best regards!

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