Cardinality of continuous functions f:R->R.

  • #1
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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?
 

Answers and Replies

  • #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).
 
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  • #3
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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?
 
  • #4
matt grime
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Reread the edited post above.
 
  • #5
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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. )-:
 
  • #6
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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,...
 
  • #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.
 
  • #8
matt grime
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Yep. That's why I pulled out to think about it some more.
 
  • #9
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so do you have any advice on this question, perhaps the cardinality isn't 2^c?
 
  • #10
CRGreathouse
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so do you have any advice on this question, perhaps the cardinality isn't 2^c?

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
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|
 
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  • #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.
 
  • #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.
 
  • #15
AKG
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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.
Well this is entirely obvious, just count the constant functions.

Nice proof Hurkyl.
 

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