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.

With the Continuum Hypothesis, the fact that the Cantor Set has the same cardinality as $\mathbb{R}$ is trivial; without the Continuum Hypothesis, it's no longer trivial, but it's still true. PedroTamaroff has suggested a good approach to the proof in the comments. In fact, though, it's provable in ZFC (that is, without appealing to the Continuum Hypothesis) that all "easily definable" sets of reals (for a particular technical definition of "easily definable") are either countable or of the same cardinality as $\mathbb{R}$. This is actually proved by demonstrating that uncountable "easily definable" sets always contain copies of the Cantor set.
