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Uncountable domain

  1. Sep 7, 2005 #1
    hi,
    the question I am working on is this:
    If f:A-->B and the range of f is uncountable, prove that the domain of f is uncountable.
    Intuitively this seems to be true. If the range is uncountable, then function has to map an uncountable number of elements from the domain to the range. I don't know how to make this a rigorous proof. Does the function have to be one to one? Can I say that the inverse exists? I know that this is an easy problem, but I am stuck. Please point me down the right path.
    Thanks,
    CC
     
  2. jcsd
  3. Sep 7, 2005 #2
    is f: said to be a bijection or 1-1 or onto?....cuz i can't see a inverse trig function satisfying that without more info.

    You can't say the inverse exists without knowing 1-1 or onto or both...or can you...its been so long.
     
  4. Sep 7, 2005 #3
    off the top of my head i would try to do it by contradiction. suppose the range of uncountable and the domain is countable. then f must map each element of A to more than one element of the range. contradiction since f is a function. (something like that anyway)

    or like this, spose f:A-->B is a function, B is uncountable and A is countable. then make a list of all the elements of the range: f(a_1), f(a_2),... etc then all the elements of the range have been listed since f can't map one thing to more than one thing, so the range is countable. contradiction.
     
  5. Sep 8, 2005 #4

    matt grime

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    Let f be a function from A to B, then if the domain of f is countable it follows that f is a surjection from a countable set onto its image, thus what can you say about the image? (this isn't a proof by contradiction but a proof by the contrpositive, ie rather than show A implies B we are proving the equivalent statement not(B) implies not(A))
     
  6. Sep 8, 2005 #5
    Hi guys,
    That's all of the info that was given for the problem. My quandry was that I need it to be bijective and I don't know how I can say that it is. I'll try the contrapositive suggestion. I tried contradiction, but I'm not confident with my results. My teacher's hint was "read the book", which I've done....over and over.
    Thanks for the input,
    CC
     
  7. Sep 8, 2005 #6
    Hi all,
    I have been working on this danged problem some more, and I am even more confused...here's why: f(x)=1. Isn't this a function that maps an uncountable domian into a countable range? If this is true then uncountable domain does not necessarily imply uncountable range. Then I am right back where I started...uncountable range, and I need to get at the domain. My head hurts.
    CC
     
    Last edited: Sep 8, 2005
  8. Sep 8, 2005 #7

    HallsofIvy

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    Yes, that's true but not relevant to the original question which was whether a countable domain could be mapped to an uncountable range. If f is not invertible, as f(x)= 1 is not, those are not the same question.

    The definition of function requires that two different y values cannot be given by the same y value. Your reverse example has many different x values giving the same y.

    Try this. Suppose the domain is countable. Then we can "list" all members of the domain:x1, x2, ..., xi,...
    For any y in the range, label it yi where y= f(xi).
    What does that tell you?
     
    Last edited: Sep 8, 2005
  9. Sep 8, 2005 #8
    does it say if the function is continous.
     
  10. Sep 8, 2005 #9
    Hi,
    Thanks for the clarification. I think I have been looking at this particular problem so long that it has stopped making sense to me. I have done the exact thing that Hallsof Ivy suggested. I just kept thinking about 1-1 and onto...but that doesn't matter. I am unproving what I have written by throwing myself some red herrings.

    No, it doesn't say that the function is continuous, just that it's a function.

    I think I am convinced that this is true. I will write up my proof and go submit it to my professor, who will hopefully agree with this reasoning.

    Thanks so much for your time and attention.
    CC
     
  11. Sep 9, 2005 #10

    matt grime

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    continuity has nothing to do with it, neurocomp.
     
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