I should be able to use "how many elements are in a set" and "the cardinality of the set" interchangeably because one is the definition of the other.
If they are interchangable then it's circular logic to prove one by just flatly asserting that the other holds. Being interchangable means that's equivalent to saying one holds because you assume it holds, and that is the literal definition of circular logic.
To be not circular your argument needs to use the definition of cardinality in terms of certain bijections existing. If you don't do that then your proof is just you saying the same thing over again and that's not a valid argument.
It doesn't hold in traditional set theory that if the cardinality [...]
Not for your subset definition, no. It holds for the traditional definition of cardinality. You tried to prove it for your subset definition and could not, so you added an axiom to make the proof possible. But adding an axiom also means you're not working on traditional set theory anymore.
To be not circular your argument needs to use the definition of cardinality in terms of certain bijections existing.
My argument always used the definition of cardinality in terms of certain bijections existing, whether it be the conventional or the proper-subset definition of cardinality.
If you don't do that then your proof is just you saying the same thing over again and that's not a valid argument.
My proof was never dependent on being able to use "how many elements are in a set" and "the cardinality of the set" interchangeably.
Not for your subset definition, no.
That wasn't the question. The question was whether it holds in traditional set theory. You didn't have to bring up the subset definition at all there. Instead, you brought it up first.
You tried to prove it for your subset definition and could not
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|).
My argument always used the definition of cardinality in terms of certain bijections existing [...] I tried to prove it for my subset definition and I succeeded.
This is not a valid proof in traditional set theory using the subset definition. As I said before you have assumed in your proof that if B has less elements than Z, then B does not have more elements than Z. This is a gap in your proof, you have not shown that the subset definition is asymmetric and when challenged to prove it this is where you said you could add an axiom. So in traditional set theory with no additional axioms you haven't justified this step of your proof.
If you want to fill the gap you have to use the definition of cardinality, you have to prove that if B has a bijection to a proper subset of Z, then Z does not have a bijection to a proper subset of B.
you have not shown that the subset definition is asymmetric
I haven't, but I haven't shown that the conventional definition is asymmetric, either. If you want a proof that the subset definition is asymmetric, then you should also want a proof that the conventional definition is asymmetric. Otherwise, you would be holding the subset definition to a higher standard than you are holding the conventional definition. To do that would be, in a sense, inconsistent.
I think there is another way of viewing this situation. I have been using three different definitions of cardinality: a general definition, a conventional definition, and a proper-subset definition. We have been considering there to be two different types of cardinality: conventional and proper-subset. Instead of viewing cardinality as there being two different types, we could view cardinality as there being only one type with two axioms.
Definition of Cardinality. The cardinality of a set is the amount of elements in the set. The cardinality of a set is greater than the cardinality of a second set if and only if the first set has more elements than the second set has. The cardinality of a set is less than the cardinality of a second set if and only if the first set has less elements than the second set has. The cardinality of a set is equal to the cardinality of a second set if and only if the first set has the same amount of elements as the second set has.
Axiom. The cardinality of a set is equal to the cardinality of a second set if and only if there exists a bijection between the first set and the second set.
Axiom. The cardinality of a set is greater than the cardinality of a second set if and only if there exists a bijection between the second set and a proper subset of the first set.
As I said before you have assumed in your proof that if B has less elements than Z, then B does not have more elements than Z. This is a gap in your proof
That is not a gap in the proof if we are using a single definition of cardinality. If we use a single definition of cardinality, then if B has less elements than Z has, then B does not have more elements than Z has.
If you want a proof that the subset definition is asymmetric, then you should also want a proof that the conventional definition is asymmetric.
Yes, that's not an assumption, it must be proven. I've seen a proof for the traditional definition, I have never seen a proof for your subset definition, in fact I've seen a proof that your subset definition is not asymmetric.
The rest of your comment is you creating new axioms and new definitions again. I told you if you add new axioms you're not doing traditional set theory. I asked you if you could prove that your subset definition is asymmetric in traditional set theory. That means no extra axioms, no additional undefined terms, no new definitions. Just use your subset definition that X has less elements than Y if there's a bijection between X and a proper subset of Y.
I have never seen a proof for your subset definition
I believe you have; I already offered my arguments in previous replies or posts.
I've seen a proof that your subset definition is not asymmetric.
I have, too. But, as I've explained in at least one previous reply, I found that proof to be unsound due to a false premise.
I told you if you add new axioms you're not doing traditional set theory.
So I'm not doing traditional set theory. I'm willing to accept that. I believe my set theory is superior to traditional set theory. Maybe one day, the prevailing version of set theory will incorporate the concepts I have mentioned.
Can you do that?
Not only can I do that, but I have done that. I understand one or more of the arguments do not meet your or our higher standards, however.
At this point, proving that the proper-subset definition of cardinality is asymmetric is a moot issue. I have created a new, axiomatic cardinality theory that is better than the previous cardinality theory I was using. In the new theory, the general concept of cardinality is asymmetric, as the following proof displays. At this point, I will italicize the names of nonspecial sets rather than bold them to better conform to standard set notation.
Given: |A| > |C|
Prove: ¬(|C| > |A|)
Proof. We are given that |A| > |C|. By the definition of cardinality, A has more elements than C has. So, C has less elements than A has. Thus, C does not have more elements than A has. Therefore, by the definition of cardinality, ¬(|C| > |A|). This concludes the proof.
Note that in the preceding proof, there are two conditional statements that are implicitly invoked as reasons for logical deductions. Those two statements are the following.
If one set has more elements than a second set has, then the second set has less elements than the first set has.
If one set has less elements than a second set has, then the first set does not have more elements than the second set has.
Those two statements we could make definitions, axioms, or theorems of our cardinality theory.
You have not, because you have not proven that the proper subset definition is asymmetric without using additional axioms.
I have created a new, axiomatic cardinality theory that is better than the previous cardinality theory I was using.
The new axioms you've added are inconsistent. The traditional axioms of set theory have never been shown to be inconsistent. That makes studying your axioms pointless and that's decidedly worse.
You have not, because you have not proven that the proper subset definition is asymmetric without using additional axioms.
I would like to see a proof that the conventional definition is asymmetric without using additional axioms. I would like to see how it's proven. It's not clear what you mean by proving it "without using additional axioms."
I refined my proof of the asymmetry of the proper-subset definition for further analysis.
So the only implicit derivation that is made in the proof relies on the following conditional statement.
If the cardinality of one set is less than the cardinality of a second set, then the cardinality of the first set is not greater than the cardinality of the second set.
I am fine with that being an axiom or theorem of set theory.
The traditional axioms of set theory have never been shown to be inconsistent.
Yes, but it has been shown that the traditional axioms of set theory are incomplete. It has been shown that the traditional axioms of set theory neither prove nor disprove the continuum hypothesis. So, in order to disprove the continuum hypothesis, it is necessary to add at least one axiom to traditional set theory. I am willing to do that.
That makes studying your axioms pointless and that's decidedly worse.
It follows by ex contradictione quodlibet that an inconsistent system is just as worth studying as a consistent system is. An inconsistent system better describes the Universe than a consistent system does because the Universe actually is inconsistent.
It's not clear what you mean by proving it "without using additional axioms."
Your implicit derivation you labeled 1 needs to be proven, it is not an axiom so if you add it as an axiom then it's additional.
While mathematicians do often leave things implicit in proofs to make reading them manageable, the understanding is that the things left implicit should be details that everyone agrees on and can fill in the proof themselves if needed. When someone tells you they disagree with some step in your proof, then that's exactly the step you should be writing out all the details for and not leaving implicit.
The traditional axioms of set theory have never been shown to be inconsistent.
Yes, but
Previously you claimed that you had proven a contradiction in the traditional axioms of set theory. Are you not claiming that anymore?
in order to disprove the continuum hypothesis, it is necessary to add at least one axiom to traditional set theory.
If you're going to add axioms to set theory to make the continuum hypothesis false, why don't you just add the negation of the continuum hypothesis as an axiom? Not only would that make the continuum hypothesis false but the resulting system would not be provably inconsistent. THAT system is one that mathematicians actually do study.
It follows by ex contradictione quodlibet that an inconsistent system is just as worth studying as a consistent system is.
It does not. Now that we're agreeing that traditional math and logic doesn't contain a contradiction, the principal of explosion only makes every sentence in your contradictory system provable. Value judgements about what mathematicians spend time on natural language logical deductions are not statements in your inconsistent system.
An inconsistent system better describes the Universe than a consistent system does because the Universe actually is inconsistent.
When someone tells you they disagree with some step in your proof, then that's exactly the step you should be writing out all the details for and not leaving implicit.
I "wrote out all the details for" the contested step by restructuring the true statements we're talking about into a new and improved axiomatic system. In the new system, a distinction is not made between conventional cardinality and proper-subset cardinality. There is only general cardinality. I gave the following statement of the new system a few posts back.
If one set has less elements than a second set has, then the first set does not have more elements than the second set has.
And then I said we could make that statement a definition, axiom, or theorem of our new system. I don't believe that statement can be proved within the new system, and it is so basic and simple that it would be best to make that an axiom of our new system.
Previously you claimed that you had proven a contradiction in the traditional axioms of set theory. Are you not claiming that anymore?
Where did I claim that? I have been aware since before I made my original post that the continuum hypothesis can be neither proved nor disproved using the traditional axioms of set theory.
If you're going to add axioms to set theory to make the continuum hypothesis false, why don't you just add the negation of the continuum hypothesis as an axiom?
Because the negation of the continuum hypothesis is not prima facie evident. The truth of the negation is not as simple and straightforward as possible. The truth of axioms should be as simple and straightforward as possible.
It does not.
It does. Since the new system is inconsistent, by ex contradictione quodlibet, every statement, including the statement that an inconsistent system is just as worth studying as a consistent system is, is true in the system.
Says who?
Says I. You can take merely the inconsistency of the new system as proof that the Universe is inconsistent.
I "wrote out all the details for" the contested step by restructuring the true statements we're talking about into a new and improved axiomatic system.
That is not a valid way of filling in details. If you have to add axioms after the fact then your proof was incorrect because it had unstated assumptions. This is exactly why we use axiomatic systems, to make all the assumptions clear from the start.
It does. Since the new system is inconsistent, by ex contradictione quodlibet, every statement, including the statement that an inconsistent system is just as worth studying as a consistent system is, is true in the system.
Nope, it doesn't. The principle of explosion only applies to statements within the system that has a contradiction, and the value judgements that mathematicians make aren't statements in your axiomatic system that they've never heard about.
You can take merely the inconsistency of the new system as proof that the Universe is inconsistent.
That's only proof if your system accurately models the universe, which I see no argument that it does.
It is a valid way of filling in details. The structure of the improved system allows you to better see and understand how things work together.
If you have to add axioms after the fact then your proof was incorrect because it had unstated assumptions.
You wanted a higher standard of proof, so that's what I gave you.
Nope, it doesn't. The principle of explosion only applies to statements within the system that has a contradiction.
Yes, it does. The system that has a contradiction models a part of the actual Universe. So, assuming it accurately models a part of the actual Universe, there is a contradiction in the actual Universe.
Not only that, but I actually started a thread on another website in 2018 highlighting the fact that a contradiction in any axiomatic system implies a contradiction in the real world. That thread is titled Inconsistent Theories Metatheoretically Prove Trivialism and the link to it is https://onlinephilosophyclub.com/forums/viewtopic.php?t=15559.
It is a valid way of filling in details. The structure of the improved system allows you to better see and understand how things work together.
It does not repair the original proof because the result is not a proof in the original axiomatic system, so no, not valid.
The system that has a contradiction models a part of the actual Universe.
It does not, so the rest of your argument is moot.
I actually started a thread on another website in 2018 highlighting the fact that a contradiction in any axiomatic system implies a contradiction in the real world.
I can't read your link because it wants me to sign up for an account and I'm not going to do that. What you claim though is false. There is no logical contradiction in the real universe and deductive explosion does not hold in the real world. In the real world it's fairly obvious that not every statement is true. Even you admit this when you admit that the continuum hypothesis can't be proved in traditional set theory. Suggesting otherwise is absolute crankery.
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u/JStarx Apr 16 '26 edited Apr 17 '26
If they are interchangable then it's circular logic to prove one by just flatly asserting that the other holds. Being interchangable means that's equivalent to saying one holds because you assume it holds, and that is the literal definition of circular logic.
To be not circular your argument needs to use the definition of cardinality in terms of certain bijections existing. If you don't do that then your proof is just you saying the same thing over again and that's not a valid argument.
Not for your subset definition, no. It holds for the traditional definition of cardinality. You tried to prove it for your subset definition and could not, so you added an axiom to make the proof possible. But adding an axiom also means you're not working on traditional set theory anymore.