Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cardinality of continuous functions f:R->R.

  1. Dec 30, 2006 #1
    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?
     
  2. jcsd
  3. Dec 30, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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
    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

    User Avatar
    Science Advisor
    Homework Helper

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    Yep. That's why I pulled out to think about it some more.
     
  10. Dec 31, 2006 #9
    so do you have any advice on this question, perhaps the cardinality isn't 2^c?
     
  11. Dec 31, 2006 #10

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  13. Dec 31, 2006 #12

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    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

    User Avatar
    Science Advisor
    Homework Helper

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

    Nice proof Hurkyl.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cardinality of continuous functions f:R->R.
Loading...