How do I prove, that $\!C = \{f:\mathbb{R} \Rightarrow \mathbb{R}\mid f \space \text{ continuous}\}$ is equinumerous with $\mathbb{R}$?

Question

We have $\space \!C = \{f:\mathbb{R} \rightarrow \mathbb{R}\mid f \text{ continuous}\}$. How do I go about proving that it is equinumerous with $\mathbb{R}$?

Answer 1

For any real number $a$ let $f_a(x)=x+a$. Then $a \to f_a$ is an injective map from $\mathbb R$ into $C$.

Let $(r_n)$ be an arrangement of the set of rational numbers in sequence . The map $f\in C \to (f(r_1),f(r_2),...)$ is an injective map from $C$ into the set of all sequences of real numbers which had same cardinality as $\mathbb R$.

Combine these two facts to finish the proof.