Cardinality of continuous functions f:R->R.

[MathematicalPhysicist]

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 isn't 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 don't see a way to pass it.
any help?

 

[matt grime]

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).

 

[MathematicalPhysicist]

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?

 

[matt grime]

Reread the edited post above.

 

[MathematicalPhysicist]

i don't 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 I am writing jibberish. )-:

 

[MathematicalPhysicist]

well i don't 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,...

 

[AKG]

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| < 2^c.

 

[matt grime]

Yep. That's why I pulled out to think about it some more.

 

[MathematicalPhysicist]

so do you have any advice on this question, perhaps the cardinality isn't 2^c?

 

[CRGreathouse]

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.

 

[Hurkyl]

http://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Sets_with_cardinality_c

 

[Hurkyl]

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

|C(R, R)| <= |R^Q| = |R|^|Q| = (2^|N|)^|Q| = 2^|N||Q| = 2^|NxQ| = 2^|N| = |R|

 

[matt grime]

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.

 

[mathwonk]

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.

 

[AKG]

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.