Bijection between set of real numbers R and the cantor set in R. Is it possible to find a bijection mapping from R to the cantor set?
(We know that the set of all real numbers and the cantor sets are uncountable sets). Give some ideas to construct a bijection mapping between these two sets.!! 
 A: Sure we can! 
First strategy: Cantor-Bernstein. Find a pair of injective functions, one going from the Cantor set $C$ to the reals $\mathbb{R}$, the other going from $\mathbb{R}$ to $C$. The first is easy - just take the inclusion map! For the second, let me sketch instead how to give an injection from $(0, 1)$ to $C$. Given $x\in (0, 1)$, take its binary expansion (the one without an infinite trail of $1$s, if it has two binary expansions) and change all the $1$s to $2$s. Viewing this as a ternary expansion, this yields a real in $C$! (To replace $(0, 1)$ with $\mathbb{R}$ is easy; e.g. precompose with $\arctan$.)
The Cantor-Bernstein theorem then tells us (explicitly!) how to combine these two injections to get a bijection. It's messy, though.
Second strategy: binary expansions. We can also come up with a bijection explicitly, using binary expansions - basically, modifying the construction of the injection $\mathbb{R}\rightarrow C$ given above, appropriately. This is a bit tricky, but not really hard; I'll leave the details as an exercise.
Is there a nice bijection? Unfortunately, not really - there is no continuous bijection between $C$ and $\mathbb{R}$. This follows e.g. from the fact that $\mathbb{R}$ is not compact, but $C$ is (or many other differences between them).
