TITLE: Is there a bijection from $[0,1]$ to the Cantor set $C$?
QUESTION [7 upvotes]: Does there exist a bijective function $\phi$ from the unit interval $[0,1]$ to the Cantor set $C$? If so, how can it be constructed? I could then proceed to build a measure space ($C$, $\mathcal{M}_\phi$, $m_\phi$) where $m_\phi(E)$ = $m(\phi^{-1}(E))$ for $E \subset C$. How would $m_\phi(C) = 1$?

REPLY [5 votes]: We look only at the less interesting part of the question, that of finding an explicit bijection from $[0,1]$ to the Cantor set $C$.
Certainly Cantor-Bernstein-Schroeder is perfectly constructive, and can be used to give an explicit, albeit messy, bijection. 
But the "near-bijection" described by @mixedmath can be tweaked to produce an explicit bijection.
For any real number $x$ in $[0,1]$, let the canonical representation of $x$ be defined as follows.  If $x$ is not a dyadic rational (that is, if $x$ does not have an eventually constant binary representation), the canonical representation of $x$ is the ordinary binary representation.  If $x\ne 1$, and $x$ is a dyadic rational, the canonical representation of $x$ is the one that is eventually $0$'s.  Finally, the canonical representation of $1$ is the representation with all $1$'s.
Now we define the bijection.  If $x$ is not a dyadic rational, map $x$ to the number $y$ whose ternary representation replaces the $1$'s in the binary representation of $x$ by $2$'s, and leaves the $0$'s unchanged.
Suppose now that $x$ is a dyadic rational.  Write down its canonical representation.  
Let $x$ be a dyadic rational whose canonical representation begins with $0$.  Remove the $0$, move the remaining bits one to the left. Then replace the $1$'s by $2$'s. Call the number which has this ternary representation $y$, and map $x$ to $y$.
Let $x$ be a dyadic rational whose canonical representation begins with $1$. Represent it in the "non-terminating form" that is, after a certain point all $1$'s.  Remove the initial $1$, move all bits one to the left. Then replace the $1$'s by $2$'s.  Call the number which has this ternary representation $y$, and map $x$ to $y$.