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?

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    $\begingroup$ A continuous function is determined by its values on the rationals. $\endgroup$ Oct 26, 2014 at 12:39
  • $\begingroup$ 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) $\endgroup$ Oct 26, 2014 at 12:43
  • $\begingroup$ @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? $\endgroup$
    – Hedgehog
    Oct 26, 2014 at 12:44
  • $\begingroup$ @Hagen von Eitzen yes, your are right, I will edit it now. $\endgroup$
    – Hedgehog
    Oct 26, 2014 at 12:45
  • $\begingroup$ @Hedgehog How many functions from the reals to the rationals? $\endgroup$
    – Hanul Jeon
    Oct 26, 2014 at 12:46

1 Answer 1


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.



  • $\begingroup$ That is a nice hint. $\endgroup$
    – PyRulez
    Jul 11, 2015 at 3:31

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