# 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.

• From a set theoretic point of view, the Cantor set is $\{0,1\}^{\mathbb N}$ (and from a topological point of view too, using the product topology). Can you show this set has cardinality continuum? Consider the binary representation of numbers. – Pedro Tamaroff Oct 27 '16 at 4:58
• @PedroTamaroff Hi Pedro. I have seen that proof before. I guess I'm not trying to find "a" proof, but just trying to understand what exactly makes the one I mentioned work. – user124910 Oct 27 '16 at 5:00
• @user124910 I don't think you're think of the same proof as Pedro. You're thinking of a proof that the Cantor set is uncountable; he's thinking of a way of setting up a bijection between $\mathbb{R}$ and the Cantor set. – Reese Oct 27 '16 at 5:03

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.

• Ok yes, I think I understand. I guess what I was trying to ask is, if you're only given ZFC, then is it true that the argument I suggested is no longer sufficient? Pedro provided an argument that does not require one to accept or reject the continuum hypothesis. – user124910 Oct 27 '16 at 5:12
• The argument you suggested is missing a massive part. You're relying on the existing fact that perfect sets of reals are uncountable; this is true, but requires its own proof. A simple diagonalization argument is enough for that, but doesn't show anything more. If you start from the theorem that perfect sets of reals have cardinality equal to that of $\mathbb{R}$ (which is provable in ZFC using Pedro's argument, not diagonalization) then your argument shows in a straightforward way that the Cantor set has that cardinality. – Reese Oct 27 '16 at 5:16

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.