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Cardinality of continuous functions f:R->R.

  1. Dec 30, 2006 #1

    MathematicalPhysicist

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    i need to find the cardinality of set of continuous functions f:R->R.
    well i know that this cardinality is samaller or equal than 2^c, where c is the continuum cardinal.
    but to show that it's bigger or equals i find a bit nontrivial.
    i mean if R^R is the set of all functions f:R->R, i need to find a 1-1 function from it to the set of continuous functions (lets call it A).
    i tried this function g:R^R->A:
    g(f)=f if f is continuous function.
    g(f)=h if f isnt continuous on R.
    where h is a continuous function on Q.
    i feel that again i did something wrong my problem is how to correct it, obviously i didnt defined h as i should be, and what's the connection of it to f, but i dont see a way to pass it.
    any help?
     
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  3. Dec 30, 2006 #2

    matt grime

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    You do not have to find a injection from R^R, the set of all functions. A map from any set with that cardinailty will do.

    About the best way of doing this (again), is to construct some subset of all continuous functions with cardinality 2^c. It might take some ingenuity, but it isn't that hard (just think of Taylor series).
     
    Last edited: Dec 30, 2006
  4. Dec 30, 2006 #3

    MathematicalPhysicist

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    any suggestions?
    i mean possible candidates are:
    {0,1}^[0,1] P([0,1]) P(R)
    i think the third set is the easiest, but gain finding the function is hard task here.
    i need to find a function from P(R)->A
    perhaps we can mapp from a subset of P(R), say B to a function in A, perhaps f|B (which is a continuous function f:B->R), will that suffice or i need something else?
     
  5. Dec 30, 2006 #4

    matt grime

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    Reread the edited post above.
     
  6. Dec 30, 2006 #5

    MathematicalPhysicist

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    i dont understand, we need to show that the cardinality of A is 2^c, iv'e showed that it's smaller or equals 2^c, now i need to show that it's bigger or equals 2^c.
    and you suggest me to:"About the best way of doing this (again), is to construct some subset of all continuous functions with cardinality 2^c"
    the only subset i can think of is the set of all functions f:R\Q->R\Q, it's cardinality is 2^c, but is it a subset, i mean every function in this subset is a continuous function cause f(R\Q) is a subset of R\Q.

    p.s
    i feel im writing jibberish. )-:
     
  7. Dec 30, 2006 #6

    MathematicalPhysicist

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    well i dont see the connection to here.
    i mean f(x)=f(x0)+(x-x0)f'(x0)+(x-x0)^2f''(x0)/2+...
    you mean the set of functions which can be written as a taylor expansion as a subset to the set of all continuous functions from R to R.
    or perhaps the set of functions of different estimations to f, such as:
    f(x0)+(x-x0)f'(x0),f(x0)+(x-x0)f'(x0)+(x-x0)^2f''(x0)/2,...
     
  8. Dec 30, 2006 #7

    AKG

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    I don't see how Taylor series will help. The most obvious interpretation of that hint shows that there are c|N| analytic functions, and although analytic functions are continuous, c|N| < 2c.
     
  9. Dec 30, 2006 #8

    matt grime

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    Yep. That's why I pulled out to think about it some more.
     
  10. Dec 31, 2006 #9

    MathematicalPhysicist

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    so do you have any advice on this question, perhaps the cardinality isn't 2^c?
     
  11. Dec 31, 2006 #10

    CRGreathouse

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    I didn't think it was -- I thought the restriction of continuity made it c -- but I don't have a proof of this offhand. Let me think about it.
     
  12. Dec 31, 2006 #11

    Hurkyl

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  13. Dec 31, 2006 #12

    Hurkyl

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    Oh, it turns out the proof's easy -- any continuous function is completely determined by its values on Q. So,

    |C(R, R)| <= |RQ| = |R||Q| = (2|N|)|Q| = 2|N||Q| = 2|NxQ| = 2|N| = |R|
     
    Last edited: Dec 31, 2006
  14. Dec 31, 2006 #13

    matt grime

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    Duh.... And there was me lamenting that analysis isn't much help 'cos I couldn't think of a way of making (different) density arguments work.
     
  15. Dec 31, 2006 #14

    mathwonk

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    hurkyls argument gives only an upper bound for the cardinality. you have to show it achieves this bound too. but maybe i came in late and missed this part of the discussion.
     
  16. Jan 1, 2007 #15

    AKG

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    Well this is entirely obvious, just count the constant functions.

    Nice proof Hurkyl.
     
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