1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Function on sets and subsets

  1. Oct 26, 2008 #1
    Alright, 2+1/2 problems here:


    36. Let F be a function from the set A to the set B. Let S and T be subsets of A. Show that:

    a) f(S[tex]\cup[/tex]T)=f(S)[tex]\cup[/tex]f(T)
    b)f(S[tex]\cap[/tex]T)[tex]\subseteq[/tex] f(S)[tex]\cap[/tex]f(T)

    Note: This must be done using a membership proof. i.e. prove that f(S[tex]\cup[/tex]T)[tex]\subseteq[/tex]f(S)[tex]\cup[/tex]f(T) AND f(S)[tex]\cup[/tex]f(T)[tex]\subseteq[/tex]f(S[tex]\cup[/tex]T)

    40. Let f be a function from A to B. Let S be a subset of B. We define the inverse image of S to be the subset of A whose elements are precisely all pre images of all elements of S. We denote the inverse image of S by f[tex]^{-1}[/tex](S) so f[tex]^{-1}[/tex](S)={a[tex]\in[/tex]A| f(a)[tex]\in[/tex]S}

    Prove parts a and b of 36 substituting the inverse image for f, and an equality in both parts (rather than subset)

    1) (not a question from the text). This was not covered in the lecture or in the book, but how do most people go about proving that a function is surjective from the definition?

    2. Relevant equations
    f[tex]^{-1}[/tex](S)={a[tex]\in[/tex]A| f(a)[tex]\in[/tex]S}

    3. The attempt at a solution
    Alright, I only have an idea of how to do 36, 40 I am lost. The professor said to use proof by case for 36, so here is how I thought it would go:

    a) prove:f(S[tex]\cup[/tex]T)=f(S)[tex]\cup[/tex]f(T)


    1) assume x[tex]\in[/tex] f(SUT), then, by definition x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]SUT}

    2) If x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]SUT}, then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S or T} by definition of union.

    3) If x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}, then x[tex]\in[/tex]f(S), and since f(S)[tex]\subseteq[/tex]f(S)Uf(T), thus if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cup[/tex]{f(s)|s[tex]\in[/tex]T}.

    4)If x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T}, then x[tex]\in[/tex]f(T), and since f(T)[tex]\subseteq[/tex]f(S)Uf(T), thus if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cup[/tex]{f(s)|s[tex]\in[/tex]T}.

    4)Therefore if x[tex]\in[/tex]f(SUT) then x[tex]\in[/tex]f(S)Uf(T)

    5 Now we must prove that f(S)[tex]\cup[/tex]f(T)[tex]\subseteq[/tex]f(S[tex]\cup[/tex]T, to do this, just invert the steps of this proof (obviously I wouldn't put this on a test but to cut length, this is reasonable)

    b)f(S[tex]\cap[/tex]T)[tex]\subseteq[/tex] f(S)[tex]\cap[/tex]f(T)


    1) assume x[tex]\in[/tex] f(S[tex]\cap[/tex]T), then, by definition x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S[tex]\cap[/tex]T}

    2) Then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S and T}.

    3) Thus s[tex]\in[/tex]S and s[tex]\in[/tex]T and so, through simplification, we can say:

    4) s[tex]\in[/tex]S, if s[tex]\in[/tex]S then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}

    5) We can also say s[tex]\in[/tex]T, and if s[tex]\in[/tex]T then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T}.

    6) Therefore, if s[tex]\in[/tex]S and s[tex]\in[/tex]T, which are true, we have x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S} and x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T} (we can say through addition).

    7) Note that by the definition of intersection, x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cap[/tex]{f(s)|s[tex]\in[/tex]T}.

    8) Therefore, if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S[tex]\cap[/tex]T} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cap[/tex]{f(s)|s[tex]\in[/tex]T}.

    9)if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S[tex]\cap[/tex]T} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cap[/tex]{f(s)|s[tex]\in[/tex]T}.

    10)This means f(S[tex]\cap[/tex]T)[tex]\subseteq[/tex]f(S)[tex]\cap[/tex]f(T).

    For Problem 40, I am wondering whether the proofs follow the same steps, or is there something else I have to worry about?
    Last edited: Oct 27, 2008
  2. jcsd
  3. Oct 27, 2008 #2
    Re: Sets

    To make this easier, forget problem 36, and 40, I can go in for help for that. Now my main question is on proving surjectivity: How is this done. I understand, you must pick a random member y0 of the codomain and prove that there is some value x0 from the domain such that f(x0)=y0, in practice however, this gives me trouble. For example, f(x)=x^2 isn't surjective on R-->R, however, it is surjective on R-->R+, how would one guarantee that the y0 value picked from the codomain reflects the restrictions put on it?
  4. Oct 27, 2008 #3


    User Avatar
    Science Advisor

    Re: Sets

    By saying so? If you want to prove that f(x)= x2:R-->R+ is surjective, you would start by saying "Let y be in R+." Then you are looking for x in R such that x2= y. Since y is in R+, x= y1/2 exists.
  5. Oct 27, 2008 #4
    Re: Sets

    Aha, thank you, why it couldn't be explained as clearly as that in my text is beyond me.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook