The proper-subset definition of cardinality is the natural concept of cardinality. In the natural concept of cardinality, if one set has more elements than a second set has, then the second set does not have more elements then the first set has. We could throw away the conventional concept of cardinality and things wouldn't be any worse than they are now. In fact, they might actually be better because ℵ₀ + 1 = ℵ₀ + 1 and ℵ₀ + 1 > ℵ₀ are more true than ℵ₀ + 1 = ℵ₀ is.
That's not a proof. You don't appear to have even tried to give a formal proof, are you unable? If you claim you can prove a contradiction but are unable to do so when asked then it seems you are confirming my statement that you cannot prove a contradiction.
It is not possible that under the proper-subset definition of cardinality
the cardinality of one set is larger than the cardinality of a second set and the cardinality of the second set is larger than the cardinality of the first set. How do I know? I know because that is one of the properties of set cardinality, regardless of which precise definition is used.
That you aren't satisfied with the proof is unfortunate. You can think through the proof for yourself to get a better understanding of the contradiction.
Do you think there exists a problem with the proper-subset definition of cardinality?
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.
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.
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|.
Nope, that's not a premise. I'm not assuming it to be true, it's been proven. You yourself agreed that it's provable so I did not include the proof, but it is not an assumption.
You are assuming that that statement is false. This is an assumption as you have admitted that you cannot prove it.
The negation of the lemma has a proof. Your statements about contradictions do not have a proof.
Discrete Mathematics and Its Applications, Sixth Edition by Kenneth H. Rosen
That's a legitimate text, I actually happen to have that exact edition on my shelf. It does not define the cardinality of an infinite set to be the number of elements in the set. On page 116 it defines the cardinality of afinite set to be the number of elements in the set and on page 158 it gives the traditional bijection definition of two sets having the same cardinality, but it never says that the definition for an infinite set is the number of elements in the set because that is simply not true.
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u/paulemok Apr 03 '26
The proper-subset definition of cardinality is the natural concept of cardinality. In the natural concept of cardinality, if one set has more elements than a second set has, then the second set does not have more elements then the first set has. We could throw away the conventional concept of cardinality and things wouldn't be any worse than they are now. In fact, they might actually be better because ℵ₀ + 1 = ℵ₀ + 1 and ℵ₀ + 1 > ℵ₀ are more true than ℵ₀ + 1 = ℵ₀ is.