Can I assume the continum hypothesis in a proof

Question

I am showing that the cantor ternary set has the same cardinality as $\mathbb{R}$

I want to use the fact that it is uncountably infinite and a subset of $\mathbb{R}$. ($|N| < |C| \leq \mathbb{R}$)

If I assume the continum hypothesis, the proof is done, but can I really do this? Is it a valid proof?

Notes:

The continum hypothesis: There is no set whose cardinality is strictly between that of the integers and the real numbers.

Definition of Cantor Ternary Set: https://en.wikipedia.org/wiki/Cantor_set

Answer 1

You may prove it under the continuum hypothesis, but in your case CH in unnecessary. You can construct a explicit bijection $f$ between the Cantor set and $\mathcal{P}(\Bbb{N})$, namely $$n\in f(x)\iff \text{$n$th digit of tenary expansion of $x$ after dot is 2}$$ (The tenary expansion of $x$ may not be unique, if $x$ is rational. In that case, choose the expansion which contains no 1. Such representation is unique.)

If possible, you should avoid using the continuum hypothesis. Our usual mathematics does not assume the continuum hypothesis and there are lot of (philosophical) disputes whether the continuum hypothesis is 'correct' axiom or not. If you prove something under continuum hypothesis, people who does not accept the continuum hypothesis (or even accept the axiom contradicted with the CH) would not accept your result.

Answer 2

Since the continuum hypothesis is not provable or disprovable you can complete the set theory with CH or it's negation and get a new axiom system. Within that (first) system it's valid to use CH in a proof as it's a system where CH is an axiom.

Most mathematicians are only concerned with the standard ZFC set theory (including the axiom of choice). Therefore a statement with proof requiring CH would not concern most, even if you can transform the statement without needing to add the axiom (for example "If CH holds then the Cantor set has the same cardinality os $\mathbb R$).

So they would not discard it as being invalid (given that you assume CH as an axiom or prerequisite in your statement), they would discard it because it doesn't interrest them.

Your example is a bit bad since you can prove the stament without having to use CH and therefore the proof would probably be discarded anyhow as relying on unrequired assumptions. Most would probably strive to drop as many requirements, making the prerequisite in a theorem as generic as possible

Since the CH is independent of the standard axiom set any concrete set you may have created without using the negation of CH will have the property that it will either be countable or have cardinality of at least $|\mathbb R|$ (or maybe that it's cardinality can't be decided - I would like to see such a set if it can be described). This is because if you could construct a set without using the negation of CH that has cardinality between $\mathbb Z$ and $\mathbb R$ you would have disproved the CH.

So hopefully for any uncountable set you encounter it should be possible to prove that it has cardinality of at least $|\mathbb R|$ without having to use the CH.