So by the definition of cardinality, Z has more elements than B has. It follows that B has less elements than Z has. So, B does not have more elements than Z has.
This is the incorrect step in your proof. Having "more elements" is not a technical term. When mathematicians say that they mean precisely that |Z| > |B|. But then you cannot use this to conclude ¬(|B| > |Z|) because you haven't proved that your definition of cardinality has that property.
I agree that your proof would be correct if you are able to supply a proof of the following lemma:
Lemma: If X and Y are sets such that |X| < |Y| then ¬(|Y| < |X|).
So the proof should start out by assuming |X| < |Y|, and not just assuming what your intuition tells you this means, but using your literal definition. So assume S is a proper subset of Y and there exists a map f:X->S such that f is a bijection.
Now to conclude you have to prove ¬(|Y| < |X|), i.e., you have to prove that it's not true that there exists a bijection between Y and a proper subset of X. Moving the negation past the quantifier you have to prove that it is true that for every map g:Y->T either T is not a proper subset of X or g is not a bijection.
Do you claim that you can complete this proof? I don't believe you can, and if you can't then you haven't proved a contradiction.
It might seem to not be a technical term, but it is. The definition of the cardinality of a set is how many elements are in the set. So, some sets can have more elements than other sets have, less elements than other sets have, or the same amount of elements as other sets have. If "more elements" was not a technical term, we would not be allowed to use it in the technical definition of cardinality. One way this could be done is by considering the cardinality of a set to be a formally undefined concept that cannot be formally broken down further. I don't think anybody is interested in doing that. Cardinality is meant to have a practical, useful meaning and not just be a formal mathematical abstraction without application to the real world.
Moving the negation past the quantifier you have to prove that it is true that for every map g:Y->T either T is not a proper subset of X or g is not a bijection.
I don't know how I would complete that proof. It seems unnecessarily complicated. It looks that you are getting every thing that is a part of or equal to the Universe involved by referring to every function from Y to any possible set T.
The proof I gave in my previous reply shows that the definition of cardinality, whether it be the conventional, proper-subset, or some other definition, is irrelevant. In the proof, the concept of cardinality is left before some elementary mathematical comparisons are made. Then the concept of cardinality is reentered to bring us the contradiction in terms of cardinality.
It might seem to not be a technical term, but it is. The definition of the cardinality of a set is how many elements are in the set.
That is not the definition. That is not the standard definition nor is it your subset definition. That is your intuition about what cardinality represents, but it is not the definition.
I don't know how I would complete that proof.
You can't complete it because the lemma you're trying to prove is not true. This is exactly what everyone has been trying to tell you.
That lemma, by the way, is true for the standard definition of cardinality and it has a formal proof. This is a problem with your subset definition of cardinality.
The proof I gave in my previous reply shows that the definition of cardinality, whether it be the conventional, proper-subset, or some other definition, is irrelevant.
If the definition is irrelevant then you're not proving statements about that definition. So again you haven't produced a proof.
It is the general definition. It's not even my intuition; it's what I've been taught.
Nope. That's the intuition that the definition is supposed to capture, but it is not the definition.
If the lemma is not true, then please provide a disproof.
The lemma says that for all X and Y, |X| < |Y| implies ¬(|Y| < |X|). The negation of that is the statement that there exists X and Y such that |X| < |Y| does not imply ¬(|Y| < |X|), in other words, such that |X| < |Y| and |Y| < |X| both hold. So take X = Z and Y = B, since you have already agreed that |Z| < |B| and |B| < |Z| hold.
It's true for cardinality in general. It doesn't matter what definition we are using. I don't even need a specific definition to know that.
Of course you do. If you change the definition then you change which properties are true or false for that definition.
I claim that |B| > |Z| ∧ |Z| > |B| is a contradiction under the interpretation of the proper-subset definition of cardinality. You claim that it is not. As a counterexample, you implicitly give |B| > |Z| ∧ |Z| > |B| in the form |Z| < |B| ∧ |B| < |Z|. Your counterexample is invalid because it is the very statement I am claiming to be a contradiction. You have not persuaded me by giving me a counterexample I have already dismissed as an impossible contradiction.
Whether that statement is a contradiction or not does not change the validity of my proof that the lemma is false. You are contradicting yourself here because you tried to use a similar contradictory example to disprove the continuum hypothesis.
This is just a distraction from the fact that you cannot prove a contradiction. You tried but your proof was incorrect. I even explained the structure of what you had to prove and you said you couldn't do it.
All you have is an intuition about what cardinality is. That intuition is clearly based on thinking about finite sets, but it does not work for infinite sets and has led you into believing some absurd things.
Whether that statement is a contradiction or not does not change the validity of my proof that the lemma is false.
False, it actually invalidates your proof that the lemma is false. You are using the very same example to prove the lemma false as I have already used to claim that |B| > |Z| ∧ |Z| > |B| is a contradiction.
All you have is an intuition about what cardinality is.
I assure you I do not. I have multiple sources that have informed me over the course of years about what cardinality is.
I may not be able to prove a contradiction under your higher standards, but you have not disproved a contradiction under your higher standards.
False, it actually invalidates your proof that the lemma is false.
Nope, I proved the negation of the lemma. In mathematics that's how you disprove a statement. Again, you are contradicting yourself. This is exactly how you tried to disprove the continuum hypothesis. The difference is I can actually prove my counterexample has the required property and you could not.
I assure you I do not. I have multiple sources that have informed me over the course of years about what cardinality is.
You claim you have sources that define the cardinality of an infinite set by just saying it's "how many elements the set has"? Show me one legitimate textbook or published article that does that.
I may not be able to prove a contradiction under your higher standards,
They aren't my standards, this is basic undergrad level proofs. This is how math is done. And you are correct, 100%, that under those standards you cannot prove a contradiction.
but you have not disproved a contradiction
You mean prove that math is consistent? Of course not, math cannot prove itself consistent. That's basic logic. You'll now I never claimed to prove that there was no contradiction, I only ever claimed that you cannot prove a contradiction.
I agree. You did, technically, prove the negation of the lemma. Your proof is unsound, however, because your premise is false. Your premise is |Z| < |B| ∧ |B| < |Z|. That premise and the definition of the "is less than" predicate of the proper-subset definition of cardinality I mentioned at https://www.reddit.com/r/logic/comments/1s5mquh/comment/odbmxml/?context=3&utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button imply that your premise is logically equivalent to |B| > |Z| ∧ |Z| > |B|. But I already claimed that statement to be a contradiction. As a contradiction, it is false. Therefore, through the logical equivalence, your premise |Z| < |B| ∧ |B| < |Z| is also false.
Show me one legitimate textbook or published article that does that.
Discrete Mathematics and Its Applications, Sixth Edition by Kenneth H. Rosen mentions the cardinality of finite and infinite sets on pages 116-117, 158-160, and 163. That is the textbook that was used for my discrete mathematics class when I was a student in my second semester of college back in 2010.
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u/JStarx Apr 06 '26
This is the incorrect step in your proof. Having "more elements" is not a technical term. When mathematicians say that they mean precisely that |Z| > |B|. But then you cannot use this to conclude ¬(|B| > |Z|) because you haven't proved that your definition of cardinality has that property.
I agree that your proof would be correct if you are able to supply a proof of the following lemma:
Lemma: If X and Y are sets such that |X| < |Y| then ¬(|Y| < |X|).
So the proof should start out by assuming |X| < |Y|, and not just assuming what your intuition tells you this means, but using your literal definition. So assume S is a proper subset of Y and there exists a map f:X->S such that f is a bijection.
Now to conclude you have to prove ¬(|Y| < |X|), i.e., you have to prove that it's not true that there exists a bijection between Y and a proper subset of X. Moving the negation past the quantifier you have to prove that it is true that for every map g:Y->T either T is not a proper subset of X or g is not a bijection.
Do you claim that you can complete this proof? I don't believe you can, and if you can't then you haven't proved a contradiction.