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I have spent some time (maybe too much) thinking about the CH.

Intuitively, I would say that there is nothing between $\aleph_{0}$ and ${\mathfrak c}$, just as there is no Integer between 0 and 1. - No proof, of course.

So I thought about fractal subsets, based on the idea: "If fractals can have non-integer dimensions, maybe they can help with this too".

Of course, for the Cantor set, there is an easy proof that it's uncountable, so this doesn't help either. Same thing applies for similar sets. Still, I thought that a fractal approach might help with this question.

But then I read up about constructibility and found (paraphrased): Creating fractal subsets of $\mathbb{R}$ would always result in a constructible set, which doesn't prove anything, since we have no proof that all sets are constructible.

Is this correct?

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    I don't have a good answer to your question, but you know that CH is independent of the ZFC axioms, yes? – 2017-02-03
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    What are you trying to prove? CH cannot be proved nor disproved. – 2017-02-03
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    @Crostul: I knew both of that, but until now I didn't totally understand the logic behind it. – 2017-02-03
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    I'm guessing that something went wrong in your paraphrase at the end, since what you wrote doesn't make much sense. Can you provide a more exact quote? – 2017-02-04

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What you're saying is not quite correct.

Fractal, for the most case (that you define them), are just going to be Borel or something close enough. Borel sets are not constructible sets. They are, however, have the perfect set property. Therefore Borel sets cannot be a counterexample to CH in any case.

It is possible, though, that the set of the constructible reals (which itself is a constructible set) is a counterexample to the continuum hypothesis. For example, start with $V=L$, and then add $\omega_2$ Cohen reals.

What you do say, however, is somewhat correct, sets which are produced from a "naive definition over the reals" are unlikely to be capable of serving as a counterexample of the continuum hypothesis.

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    I would be *very* happy to hear some constructive criticism. If the criticism is "don't answer badly written questions", that's also good. – 2017-02-03
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    From Borel sets I have only heard, and I know nothing about Cohen reals yet. Will have to read up about them. As you've seen, I'm an amateur. An amateur who had the idea to try this approach. I didn't really expect to prove or refute the CH this way, but I wanted to see how this works. - When I was talking about fractals, I had the Cantor dust in mind. Which is proven to be uncountable of course, and it's an easy to understand proof too, IMO. – 2017-02-04
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    Other than that, I'm glad to see that a high-level user like you is still looking at my question and taking time to answer it. – 2017-02-04
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    Before trying to understand why CH cannot be proved or disproved, perhaps it is a good idea to actually learn some of the mathematics needed to fully understand something like this. Attacking difficult concepts with a rudimentary understanding of mathematics is just a bad idea altogether. Trust me, I've done that several times, and we see people trying to do that on this site all the time. You end up using the wrong terminology, in order to describe the wrong things, and any reasonable answers that you'd get from someone will probably go above your head. (Case in point.) – 2017-02-04