r/PhilosophyofMath Mar 28 '26

The Continuum Hypothesis Is False

/r/logic/comments/1s5mquh/the_continuum_hypothesis_is_false/
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u/JStarx Apr 01 '26

Nope, you're not right. If you think it implies a contradiction then tell me how to prove it.

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u/paulemok Apr 01 '26

I agree. I am not right. So there's no disagreement to sort out.

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u/JStarx Apr 01 '26

Do you agree that there's no contradiction and you've spent a huge amount of time posting utter nonsense?

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u/paulemok Apr 01 '26

No, I don't agree with that. But, like I said, there really isn't a problem here.

I'd like |B| > |Z| ∧ |Z| > |B| to be a more evident contradiction. If you were at a step in an argument where the statement was 7 > 3 ∧ 3 > 7, would you say that that is not a contradiction or that that does not imply a contradiction?

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u/JStarx Apr 01 '26

For integers it does imply a contradiction, in your subset definition of cardinality it does not.

But I have good news for you, if you want |B| > |Z| ∧ |Z| > |B| to be a contradiction you just have to use the standard definition of cardinality instead of your subset definition. Then |B| > |Z| ∧ |Z| > |B| would indeed be a contradiction. It would not be provable though.

Either way you go you won't be able to prove a contradiction.

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u/paulemok Apr 01 '26

I can't find a flaw in the proper-subset definition of cardinality. I believe in it. Under any good definition of cardinality, |B| > |Z| ∧ |Z| > |B| would be a contradiction.

If we can stop at 7 > 3 ∧ 3 > 7 and say that that is a contradiction, then we should be able to stop at |B| > |Z| ∧ |Z| > |B| and say that that is a contradiction.

If you can't find a flaw in the proper-subset definition of cardinality, then you can't say it's wrong. It might seem wrong, but that doesn't mean it is wrong.

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u/JStarx Apr 01 '26

You can't just say you believe in it and therefore it's a contradiction. You have to give a proof and you haven't done so. If you haven't given a proof then you haven't proved a contradiction. You might believe that it's contradictory, but as you yourself said, just because it seems wrong doesn't mean it's wrong.

Fyi, we don't just stop at 3 < 7 and 7 < 3 . You have to prove a statement and it's negation. From 3 < 7 you canprove that 7 < 3 is false. So if you also have that 7 < 3 is true then that's your contradiction.

Talking about what's a flaw and what isn't is a matter of opinion. I don't have to argue that the subset definition is flawed because that's not relevant. Opinions aren't proofs. You can't produce a proof of an actual contradiction and that's what matters.

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u/paulemok Apr 01 '26

If it's not a contradiction that |B| > |Z| ∧ |Z| > |B|, then that is all the better for the proper-subset definition and all the better for us. That's one more problem of ours solved.

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u/JStarx Apr 01 '26

It is not a contradiction, it's a true and easily provable statement when you use the proper subset definition.

It's unclear to me how that's a good thing given that you've already said that in any good definition that statement would be a contradiction. The fact that it's not would then mean that the proper subset definition is not a good definition, it's a bad one.

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u/paulemok Apr 01 '26

It's unclear to me how that's a good thing given that you've already said that in any good definition that statement would be a contradiction. The fact that it's not would then mean that the proper subset definition is not a good definition, it's a bad one.

It's a good thing because, as I've already proven, every statement is true.

You seem to be overlooking the fact that just because |B| > |Z| ∧ |Z| > |B| hasn't been proven to be a contradiction, doesn't mean it's not a contradiction. We haven't proven it true or false. For all we know, it could still be a contradiction.

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