Does the Cantor set have the cardinality of the continuum? I saw somewhere that there are no sets between $\mathbb Q$ and $\mathbb R$ in the sense that there are no set $S\subset \mathbb R$ s.t. $|\mathbb Q|<|S|$ but $|S|<|\mathbb R|$, i.e. all set $S\subset \mathbb R$ s.t. $|\mathbb Q|<|S|$ should have the cardinality of the continuum. Now, what about the Cantor set ? It's a set of measure $0$, but it's uncountable. Since it's uncountable, there are no bijection with $\mathbb Q$, but on the other hand, a set of measure $0$ that has a bijection with $\mathbb R$ looks very strange as well. So, what do you think ? Is the Cantor set having the cardinality of the continuum ? 
 A: Yes.   Cantor set has cardinality of the reals (continuum).
As Cantor Set $\subset \mathbb R$ it's cardinality is at most $|\mathbb R|$ and as it is uncountable it's reasonable that we can't have found a contradiction to "Continuum Hypothesis" and found a cardinality between $|\mathbb Q|$ and $|\mathbb R|$ so it reasonable that Cantor set has the cardinality of the reals.
But to seal the deal we need a bijection between Cantor set and $\mathbb R$.
Following a comment by Arturo Magidin:  
If $x \in [0,1]$ then $x = \sum\limits_{i=0}^{\infty} b_i 3^{-i}$ for some  sequence of $b_i$ where each $b_i=0,1,2$.  If we disallow infinite tailing $0$s  then this sequence is unique.  This is just writing $x$ is decimal in base $3$.  But where all terminating decimals are replaced with tailing $2$s.
Likewise if $y \in [0,1]$ then $y = \sum\limits_{i=0}^{\infty} c_i 2^{-1}$ for some sequence of $c_i = 0,1$.  And if we disallow infinite tailing $0$s 0 this sequence is unique.  This is just the base $2$ decimal.
If $x = \sum b_i 3^{-i}$ is in the Cantor set then none of the $b_i = 1$.  That is because we removed the middle third of all segments and $b_k = 1$ means $\sum\limits_{i=0}^{k-1} b_i 3^{-i} < x < \sum\limits_{i=0}^{k-1} b_i 3^{-i} + 2*3^{-k}$ would mean $x$ is in some middle third.
So let $f(\sum b_i 3^{-i}) = \sum c_i 2^{-i}$ where if $b_i = 0$ then $c_i = 0$ and if $b_i = 2$ then $c_i = 1$.  $f$ is a bijection between the Cantor set and $[0,1]$.

but on the other hand, a set of measure 0 that has a bijection with R looks very strange as well. 

Ah.... not really.  It seems counterintuitive because ... to have measure $0$ no two points can be connected in the set so $\color{red}{\text{for any point there must be a measurable distance before the "next" one}}$ and there can only be countably many such points.  But that clause in $\color{red}{\text{red}}$ is completely erroneous and is based on a naive concept of numbers must "follow each other".  Uncountable numbers don't.  
And the Cantor set exists merely to be a simple counter example.
