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Troubled with Definition

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    For sets A and B, the Cartesianproduct A x B is
    {<a,b>: a [tex]\in[/tex] A and b [tex]\in[/tex] B}. A subset f of A x B is said to be a function from A to B (in symbols,f: A->B) just in case, for each element a of A there is exactly one element b of B with <a, b> [tex]\in[/tex] f:


    Given B be the set {a,b}

    and

    There are four functions from B to B, namely:
    {<a,a>, <b,a>}
    {<a,a>, <b,b>}
    {<a,b>, <b,a>}
    {<a,b>, <b,b>}


    2. Relevant equations



    3. The attempt at a solution

    by definition, i know how BxB

    but i don understand B->B how to get that 4 function?
    {<a,a>, <b,a>} meaning two output? or <a,a> is input, and <b,a> the output? or that is the subset of BxB?

    this bring me to next question

    "A subset f of A x B" or let make it simpler "A subset M of N" is this mean N is the subset of M?

    help me clear my confusion
     
  2. jcsd
  3. May 16, 2010 #2

    tiny-tim

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    Hi annoymage! :smile:
    Let's call those four subsets f g h and i.

    Then f = {<a,a>, <b,a>} is a subset of AxB and is the function such that f(a) = a, f(b) = a.
    Yes, every function is a subset of the Cartesian product of its range and domain spaces.
     
  4. May 16, 2010 #3

    HallsofIvy

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    Since B contains only two members, a and b, and a "function from B" must have each of those exactly once, any function from B to another set can have only two members: <a, > and <b, >. Now, fill in those blank spots with elements of the other set. Since her the "target set" is also B, we have two things, a and b, we can put into those blank spots. since we can choose either a or b for each of those two blank spots, there are [math]2^2= 4[/math] choices:
    1) a, a
    2) a, b
    3) b, a
    4) b, b.

    That gives the 4 possible functions: {<a, a>, <b, a>}, {<a, a>, <b, a>}, {<a, b>, <b, a>}, and {<a, b>, <b, b>}.
     
  5. May 16, 2010 #4
    i'm still trying to understand what HallsofIvy trying to say, sorry but im no good in english ;P.

    anyway, maybe i understand tiny-tim, ok, check my understanding

    if B={a,b}
    C={c,d}

    possible fucntion of B->C are:

    {<a,c>,<b,c>} is a subset of BxC and is the function such that f(a)=c , f(b)=c
    {<a,d>,<b,d>} is a subset of BxC and is the function such that f(a)=d , f(b)=d
    {<a,c>,<b,d>} is a subset of BxC and is the function such that f(a)=c , f(b)=d
    {<a,d>,<b,c>} is a subset of BxC and is the function such that f(a)=d , f(b)=c

    and now suddenly i get what HallsofIvy saying

    but now, this bothers me instead

    M={<a,c>,<b,c>} is a subset of BxC and is the function such that f(a)=c , f(b)=c

    is it f(B)=c, B={a,b}???

    if it is, what about

    {<a,c>,<b,d>} is a subset of BxC and is the function such that f(a)=c , f(b)=d

    is it

    f(B)={ c , B=a
    d , B=b }

    (pice wise function)

    is it correct? sorry if i over-complicating things
     
  6. May 17, 2010 #5

    HallsofIvy

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    Be careful here. f is defined on members of b, not B itself. We can give a separate definition: If f is a function from A to B and C is a subset of A, rather than a member of A, f(C) is defined as the set [itex]\{y | y= f(x)\}[/itex] for some x in A.
    Since B contains a and b and both f(a)= c and f(b)= c, f(B)= {c}. That is, f(B) is NOT "c", it is the set containing only c.

    If f(a)= c and f(b)= d then f(B)= f({a, b))= {c, d}.

    Again, you have to distinguish between a function, defined on a set, applied to elements of that set and to subsets of that set.

    If we are looking at sets of numbers, and [itex]f(x)= x^2[/itex], then f([-2, 2])= [0, 4] because any number in the interval from -1 to 1, squared, gives a number from 0 to 4.
     
  7. May 17, 2010 #6
    understood, thank you thank you very much
     
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