Direct bijection between $C[0,1]$ and $[0,1]$ By applying the Schroeder-Bernstein theorem one can state that there exists a 
bijection between $C[0,1]$ and $[0,1]$. But is it possible to construct a  bijection between $C[0,1]$ and $[0,1]$? Thanks in advance for any help.
 A: All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. 
There is an obvious injection from $[0,1]$
to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[0,1]$. 
Here is an example: 
Any $f\in C[0,1]$ is determined by its restriction to the rationals. Fix an explicit enumeration $(a_n)_{n\in\mathbb N}$ of $\mathbb Q\cap[0,1]$. 
The simple continued fraction of a real $x\in(0,1)$ has the form $1/(c^x_0+1/(c^x_1+\dots))$ where the $c_i$ are positive integers. It is unique unless $x$ is rational, in which case it has exactly two representations, that differ on their last term (this is essentially a matter of convention, and some presentations pick one from the beginning); pick the shortest one if that is the case. The sequence $(c^x_0,c^x_1,\dots)$ is infinite unless $x$ is rational, in which case we will extend it to an infinite sequence by setting $c^x_i=0$ for all $i$ past the last index where the sequence was defined. 
Each real $r$ corresponds to a unique sequence $(b^r_n)_{n\in\mathbb N}$ of naturals as follows: 


*

*$b^r_0$ is $1$ if $r=0$, it is $2$ if $r>0$, and it is $3$ if $r<0$. 

*Let $b^r_1=\lfloor |r|\rfloor+1$.

*Let $r'=(|r|-\lfloor|r|\rfloor)/2$ if $|r|-\lfloor |r|\rfloor$ is in $(0,1)$, and let $0'$ and $1'$ be your favorite (distinct) irrationals in $(1/2,1)$. Now let $b^r_n=c^{r'}_{n-2}+1$ for all $n\ge2$.


The assignment $r\mapsto(b^r_n)_{n\in\mathbb N}$ is an injection.
Fix an explicit bijection $\tau:\mathbb N\to \mathbb N\times\mathbb N$, denote $\tau(n)$ by $(\tau(n)_0,\tau(n)_1)$.
Assign to $f\in C[0,1]$ the following infinite sequence of positive integers $(d^f_n)_{n\in\mathbb N}$:


*

*$d^f_n=b^{f(a_{\tau(n)_0})}_{\tau(n)_1}$.


Since everything is explicit so far, note that from $d^f_n$ we can reconstruct all the values $f(a_n)$, $n\in\mathbb N$, and therefore $f$. 
Finally, there is a unique irrational $y_f\in(0,1)$ whose simple continued fraction is $(d^f_n)_{n\in\mathbb N}$. The map $f\mapsto y_f$ is an explicit injection.
The explicit bijection provided by your favorite explicit proof of the Bernstein-Cantor-Schroeder theorem and the two explicit injections specified above do the trick.
