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I know that the continuum hypothesis has been shown to be neither true nor false.

However, I think I have come across a disproof of the hypothesis. So in relation to this, has there been any concrete disproof of the continuum hypothesis ever given before?

Here's a basic idea of what I thought could disprove the Continuum Hypothesis:

Consider the set B of all possible binary sequences. Then set B is countable (for reference see why is the set of all binary sequences not countable?). Now since this set is countably infinite, we know that

$$|B| = a$$

I have used the symbol $a$ in place of the traditional symbol of aleph-zero.

Now if I try to compute the cardinality of B using permutations, I get

$$|B| = 2^0 + 2^1 + 2^2+... + 2^a$$ (counting all possible strings of all lengths. The last term is $2^a$ because the length of a binary sequence of infinite length is a countable infinity)

So, if I combine the above 2 equations, $$2^0 + 2^1 + 2^2+... + 2^a = a$$ Computing the sum of the geometric progression on the left (there is no constraint in the derivation of this formula which says that it cannot be applied for infinite terms) $$2^0 (2^a - 1)/(2-1) = 2^a - 1$$

However, if the continuum hypothesis were to be true, we have just shown that the cardinality of the set B is an uncountable infinity ($ = 2^a - 1$), which is false. Therefore, the continuum hypothesis cannot be true.

There may be errors in this reasoning, so I welcome any suggestions.

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James Bond
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    "I know that the continuum hypothesis has been shown to be neither true nor false." You can't know this because it isn't true. – aduh Oct 17 '16 at 19:22
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    It has been shown that there are models of $\mathsf{ZFC}$ which satisfy $\mathsf{CH}$ and that there are models of $\mathsf{ZFC}$ which do not satisfy $\mathsf{CH}$. – Hayden Oct 17 '16 at 19:24
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    @JamesBond Read this: https://en.wikipedia.org/wiki/Continuum_hypothesis#Independence_from_ZFC – aduh Oct 17 '16 at 19:24
  • I presume you meant to say "the continuum hypothesis has not been proven to be true nor has it proven to be false". That is not at all the same a "has been shown to be neither true nor false"! Any proposition must be one or the other. – user247327 Oct 17 '16 at 19:27
  • You might want to read: http://math.stackexchange.com/questions/859089/cant-prove-continuum-hypothesis/ and also http://math.stackexchange.com/questions/494099/why-is-ch-true-if-it-cannot-be-proved/ as well as http://math.stackexchange.com/questions/189471/why-is-the-continuum-hypothesis-not-true, but generally there are many disproofs of the continuum hypothesis starting from certain theories (e.g. $\sf ZFC+PFA$ can disprove the continuum hypothesis). If you didn't have any "additional hypotheses", then you are almost likely to be wrong, though. – Asaf Karagila Oct 17 '16 at 19:29
  • In fact, I almost feel like this question should be closed as a duplicate of one of the links I provided. If anyone agrees, feel free to initiate this, and ping me. – Asaf Karagila Oct 17 '16 at 19:44
  • As others said, if ZF is consistent, then there can be no proof of the continuum hypothesis in ZFC. So your proof (assuming it is in ZFC) is either wrong (which most would deem very very likely), or it is a proof that there is a genuine contradiction in ZF. Eitherway, it is best to try to understand what can be false in your proof, or what assumptions aren't theorems in ZFC, so you could post it here, or think for yourself. – nombre Oct 17 '16 at 19:44
  • I think the only way something useful can come out of this question is if you describe your claimed proof that CH fails. – Noah Schweber Oct 17 '16 at 19:45
  • @aduh I made the conclusion "neither true nor false" after reading this: http://math.stackexchange.com/questions/189471/why-is-the-continuum-hypothesis-not-true – James Bond Oct 18 '16 at 09:13
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    So many errors, I have to go and take a nap. – Asaf Karagila Oct 18 '16 at 09:34
  • @AsafKaragila I am sorry if that is the case. Could you please let me know my errors because this has been puzzling me for a very long time – James Bond Oct 18 '16 at 09:36
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    There are several errors. The set $B$ is only countable if it consists of finite binary sequences, so there should be no last term $2^a$. Contrary to what you say, the formula for geometric series does not make sense for infinite sums (for one thing, subtraction of infinite cardinals is not well-defined). Finally, I have no idea what you are trying to say when you say "if the continuum hypothesis were to be true": the continuum hypothesis has nothing to do with the argument you are making. – Eric Wofsey Oct 18 '16 at 09:44
  • I meant to use a proof by contradiction in the last line. If the hypothesis were true, we get $|B| = $uncontable infinity, which is false. Hence, our assumption must be false. You are right about the last term, I agree with that. That is wrong. But I didn't understand your point about the subtraction of infinite cardinals. Aren't all cardinals treated as "regular" numbers, which is why they are used as indices of 2 in the continuum hypothesis? – James Bond Oct 18 '16 at 09:49
  • @EricWofsey Also thank you for your feedback. Learnt a lot of new things – James Bond Oct 18 '16 at 09:52
  • Btw am I right to say that $|B| = a$? It is a countable infinity right? Is there anything wrong in saying this? – James Bond Oct 18 '16 at 09:57
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    $$\Huge\textbf{Cardinals are not real numbers!}$$ – Asaf Karagila Oct 18 '16 at 10:09
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    As @Eric wrote, you only have $B$ countable if you include just the finite strings (as Brian answered in the linked question that you provide as motivation). Moreover, you should know by Cantor's theorem, that $a<2^a$. So it is impossible to have $2^a+\text{other stuff}=a$. Finally, Cardinals are not real numbers, and you cannot apply the same principles to both, especially when dealing with infinitary summations. And since there is no such thing as "regular numbers", I suggest that it might be a good idea to spend a year or two learning about ordinals and cardinals. – Asaf Karagila Oct 18 '16 at 10:18

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The continuum hypothesis has been shown to be independent of ZFC, which means that, assuming ZFC is consistent, it cannot be proved or disproved in ZFC (and that both ZFC + CH and ZFC + ¬CH are consistent). Consequently, any proof or disproof (in ZFC) is incorrect.

arkeet
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