Proving there is a bijection from set of real continuous function on $[0,1]$ denoted by $C[0,1]$ to the set of reals $\mathbb{R}$

Question

I have been trying to establish that there is a bijection from the set of real continuous function on $[0,1]$ denoted by $C[0,1]$ to the set of reals $\mathbb{R}$.

I have been using Cantor-Schröder-Bernstein theorem to prove this. Consider the function: $f:\mathbb{R} \to C[0,1]$ by $f(x)$ equals the constant function $g:[0,1]\to\mathbb{R}$ given by $g(y)=x$ for each $y\in[0,1]$. Clearly, $f$ is an injection. I have been trying to find an injection from $C[0,1]$ to $\mathbb{R}$ but have been unsuccessful in doing so. I suspected that $h:C[0,1]\to\mathbb{R}$ defined by $h(f)=\int_{0}^{1}f$ would work but certainly it doesn't.

Hints would be appreciated.

TL,DR: Find an injection from $C[0,1]$ to $\mathbb{R}$.

Answer 1

A brief sketch of an argument:

Continuous functions are determined by their values on any dense subset of their domain, such as the rationals. The rationals are countable, so pick some enumeration $q_1, q_2, q_3, \ldots$ of $[0, 1] \cap \mathbb{Q}$, and for any function $f$, write out an infinite square table of digits where the $n$th row is the decimal expansion of $f(q_n)$, and then build a unique real number by reading the table in a triangular path.

The issues here are: (1) handling digits before the decimal point in values of $f$, and (2) proving that $.999\ldots = 1$ isn't an issue with uniqueness, but neither of these is difficult.