# Proof that the cardinality of continuous functions on $\mathbb{R}$ is equal to the cardinality of $\mathbb{R}$.

Proof that the cardinality of continuous functions on $\mathbb{R}$ is equal to the cardinality of $\mathbb{R}$.

I think is should be proved with the help of Cantor-Bernstein theorem. It is easy to show that cardinality of of functions continuous on $\mathbb{R}$ is at least continuum - this set contains constant functions and there is natural bijection between them and $\mathbb{R}$. So we have one injection for the Cantor-Bernstein theorem. Can you please help with another? Or maybe there is another way, without Cantor-Bernstein theorem?

• A continuous function is determined by its values on the rationals. Oct 26, 2014 at 12:39
• So more precisley the quation is to prove that the cardinality of continuous functions on $\mathbb R$ is equal to the cardinality of $\mathbb R$. (The title is not completely clear about that) Oct 26, 2014 at 12:43
• @David Mitra, thanks, I see how to apply this argument to get second injection, but I have now idea on how to prove it :( Can you please suggest anything like references, keywords or maybe sketch of the proof? Oct 26, 2014 at 12:44
• @Hagen von Eitzen yes, your are right, I will edit it now. Oct 26, 2014 at 12:45
• @Hedgehog How many functions from the reals to the rationals? Oct 26, 2014 at 12:46

As David Mitra said, a continuous function is completely determined by its restriction to $\Bbb{Q}$. Hence, the following map is injective

$$\Gamma : C(\Bbb{R}) \to \Bbb{R}^\Bbb{Q}, f \mapsto f|_\Bbb{Q}.$$

This implies

$$|C(\Bbb{R})|\leq |\Bbb{R}^\Bbb{Q}|.$$

I will let you take it from here.

Hint:

$|\Bbb{R}|=2^{|\Bbb{N}|}$

• That is a nice hint. Jul 11, 2015 at 3:31