Set theory and analysis: Cardinality of continuous functions from R to R

Mosis
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Homework Statement


Prove the set of continuous functions from R to R has the same cardinality as R

Homework Equations


We haven't done anything with cardinal numbers (and we won't), so my only tools are the definition of cardinality and the Schroeder-Bernstein theorem and its consequences.

I also don't know any "high brow" mathematical facts about continuous functions.

The Attempt at a Solution


Not much. By Schroeder-Bernstein, we need an injection f:\mathbb{R}\rightarrow C^0 and vice versa. We have the usual embedding map from \mathbb{R} to C^0, but I've no idea how to construct an injection going the other way. Given any continuous function, how do I uniquely identify it with a real number?
 
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I think you need to know at least a few things about cardinality. The main trick is that every continuous function is determined by its values on the rationals, which is a countable set. Does that help?
 
ah yes, then I should be able to associate any continuous function with a real number in [0,1) whose digits correspond to the value of the function evaluated at every q in Q. doing that shouldn't be too hard.

thanks!
 
as an aside: how do I know that every real number is the limit of some sequence of rational numbers? I mean, I "know" that this is pretty much what R is (as the completion of Q), but I'm not sure how to rigorously back that up.
 
Mosis said:
as an aside: how do I know that every real number is the limit of some sequence of rational numbers? I mean, I "know" that this is pretty much what R is (as the completion of Q), but I'm not sure how to rigorously back that up.
Isn't "every real number is a limit of rational numbers" pretty much (among other things) literally what "R is the completion of Q" means?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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