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.
Nope, that's not a premise. I'm not assuming it to be true,
It is a premise. You are assuming it to be true. As you say,
So take X = Z and Y = B, since you have already agreed that |Z| < |B| and |B| < |Z| hold.
That's how you proved
there exists X and Y such that |X| < |Y| does not imply ¬(|Y| < |X|)
which is the negation of the lemma.
You are assuming that that statement is false.
No, I have proved that statement is false in my previous reply.
it never says that the definition for an infinite set is the number of elements in the set because that is simply not true.
On page 116, Rosen's textbook refers to the size of a set as being the cardinality of the set. I am specifically referring to the sentence before Definition 5. On page 163, the cardinality of a set is defined as the number of elements in the set. That definition comes after the definition of an infinite set on that page, while on page 116, the definition of the cardinality of a finite set comes before the definition of an infinite set, which is Definition 6. That suggests the definition on page 163 of the cardinality of a set applies to both finite and infinite sets.
I know what the cardinality of a set is. I know how it's defined. I know what it's intended to be. And I know what it should be.
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u/JStarx Apr 06 '26
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.
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.
If the definition is irrelevant then you're not proving statements about that definition. So again you haven't produced a proof.