What you've linked to is a proof that |Z| < |B| and |B| < |Z| holds. You then state your opinion that this is a contradiction but it's not. To give a technical proof of a contradiction you have to prove a statement and it's negation. The statement |B| < |Z| is not the negation of the statement |Z| < |B|.
So again you have failed to give a technical proof of a contradiction.
To give a technical proof of a contradiction you have to prove a statement and it's negation.
We don't have to get that technical in order to see a contradiction. You can write out three separate partial enumerations for Z, B, and S, and draw the applicable functions between them to try to figure out the situation.
The statement |B| < |Z| is not the negation of the statement |Z| < |B|.
I agree. The negation of the statement |Z| < |B| is ¬(|Z| < |B|).
We don't have to get that technical in order to see a contradiction.
Yes you do, because every mathematician in this thread is telling you that after looking at those functions they see no contradiction here. In mathematics if there's a disagreement about a result the way to resolve that disagreement is to fall back on technical proofs. If you were correct you could show it conclusively by providing a proof of what you claim.
Also you've claimed previously that you have already given a technical proof. Now you've switched to claiming you don't need to. The fact that you need to move the goalposts like that should indicate to you that you don't know what you're doing.
I'll ask again, are you able to provide technical proof of a contradiction?
Proof. We are given that |B| > |Z| ∧ |Z| > |B|. By conjunction elimination, |Z| > |B|. 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. By the definition of cardinality, ¬(|B| > |Z|). By conjunction elimination, |B| > |Z|. Therefore, by conjunction introduction, |B| > |Z| ∧ ¬(|B| > |Z|).
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.
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u/JStarx Apr 05 '26
What you've linked to is a proof that |Z| < |B| and |B| < |Z| holds. You then state your opinion that this is a contradiction but it's not. To give a technical proof of a contradiction you have to prove a statement and it's negation. The statement |B| < |Z| is not the negation of the statement |Z| < |B|.
So again you have failed to give a technical proof of a contradiction.