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Math Help - function between sets

  1. #1
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    function between sets

    I decided to take a Real Analysis course, and I'd appreciate some help throughout this semester on questions I have if possible. I'm alrdy having trouble w/ one of the 1st problems..

    1.) Given a function such that f : D -> R and given that a subset B "is a subset of" R, let f^(-1)(B) be the set of all the points from domain D that will get mapped into B;

    IE: f^(-1)(B) = {x "is an element of" D : f(x) "is an element of B}. Note that this is referred to as the preimage of B.

    a.) Let f(x) = x^2. Suppose A is the interval [0, 4] and B is the interval [-1, 1] (note both of these are closed), what is f^(-1)(A) and f^(-1)(B).

    So this seems elementary, and perhaps I'm overlooking it.

    We know the domain D maps to the reals..

    So for A, we'd have [0,0], [1,1], [2,4], [3,9], [4,16]
    And for B, we'd have [-1,1], [0,0], [1,1]

    I'm not quite sure what they're wanting.

    b.) Does f^(-1)(A ^ B) = f^(-1)(A) ^ f^(-1)(B) in this case? Does f^(-1)(A v B) = f^(-1)(A) v f^(-1)(B)?

    (**The v and ^ by itself means union and intersection, respectively)

    Well, A ^ B would be [0,0],[1,1] if im not mistaken, and
    A v B would be [-1,1], {A} => [-1,1], [0,0], [1,1], [2,4], [3,9], [4,16]

    But if that's f^(-1)(A ^ B) and f^(-1)(A v B), what is it by component (ex. f^(-1)(A) ^ f^(-1)(B)

    c.) The good behavior, as showed above, of these preimages is general. Show that for some arbitrary function g : R -> R, its always true that g^(-1)(A ^ B) = g^(-1)(A) ^ g^(-1)(B) and g^(-1)(A v B) = g^(-1)(A) v g^(-1)(B) for all sets A, B "is a subset of" R

    Not sure how to show this if I can't even show it for a specific function.

    Thanks for your help and clarification.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    1.) Given a function such that f : D -> R and given that a subset B "is a subset of" R, let f^(-1)(B) be the set of all the points from domain D that will get mapped into B;

    IE: f^(-1)(B) = {x "is an element of" D : f(x) "is an element of B}. Note that this is referred to as the preimage of B.

    a.) Let f(x) = x^2. Suppose A is the interval [0, 4] and B is the interval [-1, 1] (note both of these are closed), what is f^(-1)(A) and f^(-1)(B).

    So this seems elementary, and perhaps I'm overlooking it.

    We know the domain D maps to the reals..

    So for A, we'd have [0,0], [1,1], [2,4], [3,9], [4,16]
    And for B, we'd have [-1,1], [0,0], [1,1]

    I'm not quite sure what they're wanting.
    Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

    So
    f^{-1}( [0, 4] ) = [0, 2]
    and
    f^{-1}( [-1, 1] ) = [0, 1]

    Note in this last one f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] ) because the first expression is not real on the set [-1, 0).

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    a.) Let f(x) = x^2. Suppose A is the interval [0, 4] and B is the interval [-1, 1].

    b.) Does f^(-1)(A ^ B) = f^(-1)(A) ^ f^(-1)(B) in this case? Does f^(-1)(A v B) = f^(-1)(A) v f^(-1)(B)?

    (**The v and ^ by itself means union and intersection, respectively)

    Well, A ^ B would be [0,0],[1,1] if im not mistaken, and
    A v B would be [-1,1], {A} => [-1,1], [0,0], [1,1], [2,4], [3,9], [4,16]

    But if that's f^(-1)(A ^ B) and f^(-1)(A v B), what is it by component (ex. f^(-1)(A) ^ f^(-1)(B)
    A \cap B = [0, 4] \cap [-1, 1] = [0, 1]

    A \cup B = [0, 4] \cup [-1, 1] = [-1, 4]

    So
    f^{-1}(A \cap B) = f^{-1}( [0, 1] ) = [0, 1]
    and
    f^{-1}(A) \cap f^{-1}(B) = f^{-1}( [0, 4] ) \cap f^{-1}( [-1, 1] ) = [0, 2] \cap [0, 1] = [0, 1]

    So the two are, indeed equal.

    I leave it to you to do the intersection case.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    c.) The good behavior, as showed above, of these preimages is general. Show that for some arbitrary function g : R -> R, its always true that g^(-1)(A ^ B) = g^(-1)(A) ^ g^(-1)(B) and g^(-1)(A v B) = g^(-1)(A) v g^(-1)(B) for all sets A, B "is a subset of" R

    Not sure how to show this if I can't even show it for a specific function.
    I can't help you too much with this one off the top of my head. However I can remind you of one piece of information that will likely be helpful:
    "A relation f: \mathbb{R} \to \mathbb{R} is a subset of ordered pairs of the set \mathbb{R} \times \mathbb{R}."

    "A function f: \mathbb{R} \to \mathbb{R}: x \mapsto f(x) is a relation such that for each value x in its domain there exists at most one f(x) in its image."

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

    So
    f^{-1}( [0, 4] ) = [0, 2]
    and
    f^{-1}( [-1, 1] ) = [0, 1]

    Note in this last one f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] ) because the first expression is not real on the set [-1, 0).

    -Dan
    Edit: I'm a tad uncomfortable about defining the function f^{-1} on [-1, 1] in this case. Presumably this simply means f(x) = x^2 is injective (since I would presume that [-1, 1] is a subset of the image of f, perhaps range would indeed be the better word here), but I'm just not in tune with it since f^{-1} typically indicates the inverse of a bijective (not merely injective) function.

    -Dan
    Last edited by topsquark; September 6th 2007 at 05:52 AM.
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  6. #6
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    Quote Originally Posted by topsquark View Post
    Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

    So
    f^{-1}( [0, 4] ) = [0, 2]
    and
    f^{-1}( [-1, 1] ) = [0, 1]

    Note in this last one f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] ) because the first expression is not real on the set [-1, 0).

    -Dan
    How exactly is f^(-1)([0,4]) = [0,2] ?

    I don't quite see what's going on. Sorry if this is very elementary.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    How exactly is f^(-1)([0,4]) = [0,2] ?

    I don't quite see what's going on. Sorry if this is very elementary.
    You have f(x) = x^2 so formally f^{-1}(x) = \sqrt{x}

    So if we operate f^{-1} on an interval, it will take the square root of all values on that interval.

    Thus
    f^{-1}([0,4]) = [ \sqrt{0} , \sqrt{4} ] = [0, 2]

    We have to be a bit more careful about f^{-1}( [-1, 1] ) since some of the values returned by the square root are imaginary. (This is why I mentioned I was uncomfortable in the other post.) However we may ignore this problem in the sense that f^{-1}(A) is going to only give us the real part of the "interval" that is returned. So f^{-1}( [-1, 1] ) = [0 , 1] .

    -Dan
    Last edited by topsquark; September 6th 2007 at 05:55 AM. Reason: Addendum
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  8. #8
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    Quote Originally Posted by topsquark View Post
    Edit: I'm a tad uncomfortable about defining the function f^{-1} on [-1, 1] in this case.
    There is no need to worry about that.
    \begin{array}{rcl}<br />
 x \in f^{ - 1} (A \cap B)\quad  & \Leftrightarrow & \quad f(x) \in A \cap B \\ <br />
  & \Leftrightarrow & \quad f(x) \in A \wedge f(x) \in B \\ <br />
  & \Leftrightarrow & \quad x \in f^{ - 1} (A) \wedge x \in f^{ - 1} (B) \\ <br />
  & \Leftrightarrow & \quad x \in \left[ {f^{ - 1} (A) \cap f^{ - 1} (B)} \right] \\ <br />
 \end{array}
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  9. #9
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    Quote Originally Posted by topsquark View Post
    Bear in mind that [0, 4] is an interval. It is the entire real line segment from 0 to 4, not just 0, 1, 2, 3, and 4.

    So
    f^{-1}( [0, 4] ) = [0, 2]
    and
    f^{-1}( [-1, 1] ) = [0, 1]

    Note in this last one f^{-1}( [-1, 1] ) = f^{-1}( [0, 1] ) because the first expression is not real on the set [-1, 0).

    -Dan
    Apparently, according to my professor, f^{-1}( [-1,1] ) = [-1,1]

    And, f^{-1}( [0,4] ) = [-2,2]
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  10. #10
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    It really does depend of what the D in the original is. I would agree with your instructor if D includes those two intervals.
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  11. #11
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    c.) The good behavior, as showed above, of these preimages is general. Show that for some arbitrary function g : R -> R, its always true that g^(-1)(A ^ B) = g^(-1)(A) ^ g^(-1)(B) and g^(-1)(A v B) = g^(-1)(A) v g^(-1)(B) for all sets A, B "is a subset of" R
    x\in g^{-1}(A\cap B)\Leftrightarrow g(x)\in A\cap B\Leftrightarrow (g(x)\in A)\wedge (g(x)\in B)\Leftrightarrow
    \Leftrightarrow (x\in g^{-1}(A))\wedge (x\in g^{-1}(B))\Leftrightarrow x\in g^{-1}(A)\cap g^{-1}(B)

    Prove the second part in the same way.
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