Uncountable subsets of $\mathbb{R}$ is bijective to $\mathbb{R}$ The context of the question is the following: We can show that the cantor set is uncountable by showing it is non-empty, perfect, and complete. What I'm not clear on is, to show that it has the same cardinality of $\mathbb{R}$, do you have to invoke some extra axiom into set theory? Please forgive me if the question is very niave, but this is something I've just used intuitively and never really given it much though: the continuum hypothesis says that there is no set whose cardinality is between the cardinality of the integers, and the cardinality of the reals. Wikipedia says that the continuum hypothesis is independent of ZFC, so does that mean that the result (there is a bijection between cantor set and reals) if we adopt the continuum hypothesis, and the proof is not sufficient if we reject the continuum hypothesis? Thanks.

No extra axioms other than those in ZFC are required to prove that there is a bijection from the Cantor set to the set $\mathbf{R}$. 
Consider the following facts:
It can be proven (in ZFC) that a Cantor set has the same cardinality as the set $\{0,1\}^{\mathbf{N}}$. 
This set of all countable sequences of $\{0,1\}$ can be show to have the same cardinality as the set $\mathbf{R}$.
Therefore it can be proven that there is a bijection from the Cantor set to the set of reals within the ZFC axiom system. The Jech book contains the details of the proofs.
