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Homework Help: Abstact math help, function proof

  1. Apr 3, 2012 #1
    Abstact math help, function proof!!

    1. The problem statement, all variables and given/known data
    Let f : X to Y . For each element b element of Y , let Y_b = f^-1({b}). Prove that the Y_b's
    have the following property:
    (a) If b and c are distinct elements of Y , then Yb intersection Yc = ∅.
    (b) X =U_element of Y ( Yb).

    2. Relevant equations

    3. The attempt at a solution

    Well a) we know that b≠c, but we have to prove that Y_b n Y_c is an empty set. I know it have to be an empty set becase Y_b=f^-1({b}) and Y_c=f^-1({c}); so it Y_b n Y_c would not be empty then f is not a function. But I don't know how to prove it.
    b) X is a union of Y_b - I have no idea how to even start. I submitted pdf file.

    Attached Files:

  2. jcsd
  3. Apr 3, 2012 #2


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    Re: Abstact math help, function proof!!

    You have to show a particular y_1 \in Y and another particular y_2 \in Y, (since Y is the domain of the inverse here), show that they project uniquely to an element of the codomain (X-here), that the pairs exists and are unique. The other part, that y_1 and y_2 are not equal you already kind of have, you just left out some verbiage
  4. Apr 3, 2012 #3


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    Re: Abstact math help, function proof!!

    Part B : Have you gone over Partitions of sets yet? That definition would be immensely helpful to you.
  5. Apr 3, 2012 #4
    Re: Abstact math help, function proof!!

    No, the problem is that we did not go over anything, the teacher does not explain anything and I am trying to figure it out myself. From my previous class I know something about inverse function, but only enough to find an inverse of given function and to find out if it is one to one or onto. So that would be :
    Let y_1 be an element of Y, and y_2 be an element of Y, but y_1 does not equal y_2. So can I say that assume y_1=f^-1({b}) and y_2=f^-1({c}) ? If I can say that then what would be the next step?
  6. Apr 3, 2012 #5


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    Re: Abstact math help, function proof!!

    I understand it's frustrating. We've all been there. Proofs like this really can be quite enjoyable once you get the hang of it.

    You're on the right track. Does your instructor knock a lot of points off for technical Proof Formatting? If so, make SURE you lay your proof out correctly to start with - it will also make it easier to compose your thoughts. My guess is that you may be a bit unsure how to lay things out which may be causing some of your angst.

    It should look like this:

    Let.....(everything that's given in the INITIAL premise - after the "2." on your attachment sheet) -leave out the part that says "Prove...following property"

    Next, you'll write:
    Let b & c \in (are elements of) Y

    .....Now, what you have to SHOW (through your writing) is that Y_b not equal to Y_c--that inequality will be your ending to Part A's Proof.

    From here: (like you had) WRITE - something along the lines of "From our premise, f is a function from X to Y, such that....a in X --> b in Y (use your symbols properly though)
    ==> (a,b) in F
    Then: (can you go on here?) You are given that F^-1(b), where b is in Y, so you KNOW that f inverse is f^-1: Y-->X

    ---That implies two properties of the function F --- Give it a shot

    ---one more hint. You'll need to show not that f(a)=w and f(b)=v, but rather IF f(a)=w=f(b), then "a" MUST equal "b" by definition.

    Try to clean up what you have and make sure to stick to the definitions of functions, inverse etc.

    Keep trying!
  7. Apr 3, 2012 #6
    Re: Abstact math help, function proof!!

    For (a) you have exactly the right idea, you just need to lay out the argument.

    Suppose f_1(a) intersect f_1(b) is not empty.

    Then there exists an element x in f_1(a) intersect f_1(b).

    That means x is is an element of EACH of _____ and _______.

    But then what is f(x)? It must be both _____ and ______. But that's impossible because f is a function.
  8. Apr 3, 2012 #7
    Re: Abstact math help, function proof!!

    Thank you all, I got it now
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