What are Numbers?
I understand:
Natural: 1, 2, 3...
Whole: 0, 1, 2...
Integer: -1, 0, 1...
Rational: m/n
Irrational: x - m/n
Real: applicable to number line.
What I want to know is how N is defined.
Is there special use of the word 'is'? Natural numbers are N, is incomplete.
A. 1 through 9, are numbers, why?
B. Why does the number system progress, beginning from the left, proceding to the right?
C. Is human number just a tool?
I'm just getting into mathematics...
Sorry for having an intricate view - I'm not trying to distract. My primary question is (A).
Further Edits:
A shadow-argument:
I understand you can count your fingers, 1 - 4, but what says a finger is a 1 and not crossed fingers? The 'whole' of the finger?
In which case it's not a single, there's an organism involved(such as under the skin of the finger), and thus, a finger is not a 1.
I understand 1 is a concept but mathmatically, 1 is a point.
Perhaps, to point at your finger you'll use the number 1 but to define it numerically it's a different number.
Natural: 1, 2, 3...
Whole: 0, 1, 2...
Integer: -1, 0, 1...
Rational: m/n
Irrational: x - m/n
Real: applicable to number line.
What I want to know is how N is defined.
Is there special use of the word 'is'? Natural numbers are N, is incomplete.
A. 1 through 9, are numbers, why?
B. Why does the number system progress, beginning from the left, proceding to the right?
C. Is human number just a tool?
I'm just getting into mathematics...
Sorry for having an intricate view - I'm not trying to distract. My primary question is (A).
Further Edits:
A shadow-argument:
I understand you can count your fingers, 1 - 4, but what says a finger is a 1 and not crossed fingers? The 'whole' of the finger?
In which case it's not a single, there's an organism involved(such as under the skin of the finger), and thus, a finger is not a 1.
I understand 1 is a concept but mathmatically, 1 is a point.
Perhaps, to point at your finger you'll use the number 1 but to define it numerically it's a different number.
Comments (63)
May I ask that you read the 'shadow argument' I edited into the original post?
Mind is, evalutation as well as observation, of shapes in nature amongst other things.
How you formulate a shape is very partially, by drawing a line and at minimal you can formulate dots to help you, that is why we use a rule to draw a straight line, we connect two points. You can imagine two dots and if powerful enough, connect them. Imagining a illusion of a line. This line can take us into a powerful thought network.
The multi-verse may just as well exist in the background of our universe.
I would say all mathematics, non regarding of the basic up, left, right, left, is impartial. I'm not saying it's bad, but it's language holds no bearing over purer symbol.
I need to create symbols, I do, why would I think in symbols that are base 4?
Why not base 8 which would sum up base 4 as 08? Also any symbol or pure symbols ' I am working', per se.
I think our language is bad but lexscribed well, I like more a language like a whisper. Short words, pauses.
I think N is 0.8
If there is a natural number then there's the natural golden number which proves N, which is 0.8, on base 4 you move at 0.8, and on base 8 you move at 64.
Because "rocks" was already taken.
Two ways: via the Peano axioms, in which [math]\mathbb N[/math] is a collection but not a set; and in ZF via the axiom of infinity. In the latter approach 0 is defined as the empty set; 1 is the set containing 0; 2 is the set containing 0 and 1; and in general, n is the set containing 0 through n - 1.
Of course numbers are not literally sets. The set-theoretic definition is just a representation of the abstract idea of the counting numbers.
https://en.wikipedia.org/wiki/Peano_axioms
https://en.wikipedia.org/wiki/Axiom_of_infinity
I believe that when you try to search for the origin of words and the way in which we pour meaning into the world around us - it is most fruitfull to search how those meaning are first discovered, whether it is understanding how a baby generates those concepts or reading the history of how those concepts were created in the early human culture.
A child starts by counting objects in the world - the concept of numbers as something other than an adjective to a group is outside the scope of our first usage of them. This is on par with how we evolve language itself - we use sounds and words that allow us to interact better with the world around us.
In the same way the usage of numbers in the human history evolved as a way to describe actual acts in the world - ways to divide a field of wheat, to make commercial transactions, etc...
It was then used as a way to allow us to make astronomical inquiries - taking part in one of the most early abstract human endeavours: religion. The Pythagorean sect in ancient Greece even evolved mathematics as a religion of its own. It is not weird than that Geometry, so closely related the world of mathematical usage in those days, was the tool by which the Pythagorean advanced their understanding of the world. And an ancient greek would asked then, in the same way you ask now - why do we talk about lines, what are lines?
So my answer to what are numbers - they are signs (verbal/graphic), like words, by which we are able to interact better with the world. The scope in which we use them is sometimes confined to the abstract term - mathematics; But this term, describing a narrower usage of them, isn't all that we do with those signs. We use numbers to describe concepts in every field of human knowledge.
And for a more concrete terminology - they are most widely used to describe a quality of a group of objects, distinguishing that group by the count of object within it. You can basically grow all of the mathematical knowledge from that definition (given a set of logical axioms on such groups).
I tried to look up this concept but the wikipedia pages for peano axioms and natural number do not seem to mention this subtlety. I assume that "a collection but not a set" means that N cannot be an element of another set?
There is also the concept of "proper class":
Quoting Wikipedia on class versus set
So, according to the above, the ordinal numbers are not a set but a proper class.
Quoting Wikipedia on ordinal numbers
So, according to the above, ordinal numbers are not a set in set theory. I couldn't find a reference to the idea of distinguishing between natural numbers and ordinal numbers in Peano arithmetic (PA). It even looks like expressing this distinction requires the full power of the machinery in set theory, such as, for example, by defining Von Neumann ordinals.
Therefore, I am a bit surprised that PA would even be able to introduce this type of subtlety through its axioms.
(Or maybe it actually does, but then implicitly/unexpectedly.)
By the way, I also found this remark on the subject:
Quoting Quora answer on sets versus collections
The answer above even seems to object to using the term "collection" in mathematics, because the term does not naturally emerge from any theory's axioms.
The finger is a representation or symbol representing a mathematical concept.
Like in programming when i name a variable x that represents the number of apples being bought at a store.
Obviously you wouldn't name the variable X using correct coding procedures but a variable represents a number and a number is typically a Asian Indian/Arabic symbol that represents a mathematical concept.
You are really good at this. Are you a team or do you come up with this stuff yourself?
Yes. In PA (Peano arithmetic) each of the numbers 0, 1, 2, 3, ... exists, but not a completed set of them.
Quoting alcontali
Yes in fact in PA, the natural numbers are a proper class. "Too big to be a set," as they say. It's a great example of how you can visualize what that means. It's the axiom of infinity in ZF that bestows set-hood on [math]\mathbb N[/math].
Quoting alcontali
Right. There are no proper classes in ZF even though we use the term informally, but in other set theories the notion can be formalized.
Quoting alcontali
Yes.
Quoting alcontali
Yes. That's the famous Burali-Forti paradox, that the collection of ordinals can't be a set.
Quoting alcontali
Yes. The ordinals are awesomely cool, sadly not as well known as the cardinals.
Quoting alcontali
Yes, you can't define the ordinals in PA because you can't get to the first transfinite ordinal [math]\omega[/math] by successors. You have to take a limit; or what amounts to the same thing, you have to consider the completed set of natural numbers. That in fact is the definition of [math]\omega[/math].
Quoting alcontali
Not sure what you mean. PA can't introduce ordinals or transfinite numbers. You can do a lot of number theory in it but not enough to get the real numbers off the ground. Unless you mean that it introduces subtleties in terms of what it can't do.
Quoting alcontali
Don't know what you mean here.
Quoting alcontali
Yes, collection is what you say when you don't want to imply that something's a set.
Quoting alcontali
Quora giveth and Quora taketh away. I'd like to know who the author is before passing judgment on the sanity of anything math-related on Quora. There are a small handful of math experts there and a lot of people who don't know much. But in general, collection doesn't have a formal meaning. But then again neither does proper class.
But we can define collection. A collection is the extension of a predicate. That is, a collection is all the things in the universe that satisfy some predicate, like the collection of sets. Collections may or may not be sets. I tend to use the word collection synonymously with class. Every predicate defines a class, which may be a set or not. If it's a class but not a set, it's a proper class. But a lot of this usage is informal.
Imagine that sound visualization that creates a zig zag - you probably know what I'm on about.
Imagine a pure symbol that can be anything; much like the sound visualization, picture the symbol fluctuating.
Do I want to restrict my minds capacity to create symbols?
I know that base 4 isn't going to restrict that, but understood in a manner contrary to the way that I understand it, it does.
When I think mathmatically, I use line-form, icons squares, projected motion, mapping, etc.
I said in a previous thread that thought is partially mathematical in nature. This side of thought is not restricted to base 4. That would be stupid.
Wow. That is very interesting!
Quoting Wikipedia on Burali-Forti
The proof based on Von Neumann ordinals is also straightforward. In that definition, (some interpretation of) [math]\mathbb{N}[/math] itself is also an ordinal. However, with "each ordinal is the well-ordered set of all smaller ordinals", it leads to [math]\mathbb{N} < \mathbb{N}[/math], i.e. the ordinal [math]\mathbb{N}[/math] being smaller than itself. Apparently, Cesare Burali-Forti already managed to prove it without using set-theoretical constructions (but Wikipedia does not explain the older proof strategy).
Quoting fishfry
Yes, I should probably not have used the symbol [math]\mathbb{N}[/math] (above) to designate the ordinal ?. I already sensed this because Wikipedia explicitly stays clear of doing that:
"Let ? be a set that contains all ordinal numbers."
So, in this context, the subtlety is that [math]\mathbb{N}[/math] = { 0, 1, 2, ... } is not complete, while ? = { 0, 1, 2, ..., ? } is complete but then rests on something ultimately contradictory, i.e. a non-well-founded set expression of the type A = { B, A } which is then equivalent with A = { B, { B, A } } = { B, { B, { B, { B, A } } } } and so on, ad nauseam.
It looks like there is a strong (but unexpected) link between and Cesare Burali-Forti's work and what Dmitry Mirimanoff pointed out:
Quoting Wikipedia on non-well-foundedness
So, this result is understandable if we keep the subtlety in mind that [math]\mathbb{N}[/math] is not materialized while ? is materialized. The act of materializing [math]\mathbb{N}[/math] substantially changes its nature. The work of Burali-Forti is quite interesting. In my opinion, it is surprising and even intriguing!
Reading ahead, I surmise that you may making a little more of it than it deserves. Towards the end of your post you seem to draw some metaphysical or philosophical conclusions or associations, and I don't think those are justified. But as it ended up I didn't have time to get to that part of your post tonight. Instead I wrote a little disquisition on [math]\omega[/math] to clarify some technical points.
Quoting alcontali
Yes.
Quoting alcontali
Ah then I boldly go where Wikipedia dares not. Because in fact [math]\mathbb N = \omega = \aleph_0[/math] as sets. These are three different names for the exact same set.
Permit me to explain in order to make this point perfectly clear.
* The axiom of infinity gives us the existence of a set which we commonly call [math]\mathbb N[/math]. Specifically it is the set of the finite von Neumann ordinals, also known as the natural numbers. We have in fact
* [math]0 = \emptyset[/math]
* [math]1 = \{0\}[/math]
* [math]2 = \{0,1\}[/math]
* [math]3 = \{0,1,2\}[/math]
and in general if we denote the successor of [math]n[/math] as [math]n +1[/math], then
[math]n + 1 = n \cup \{n \} [/math].
When you grok that last little bit of notation you will be enlightened! And now finally we define
[math]\mathbb N = \{0, 1, 2, 3, \dots \} [/math]
That last definition relies on the axiom of infinity. In PA there's no rule that lets us form infinite sets. At best we could define set theory for PA, the set theory of finite sets. Everything would be the same as in full set theory except there are only finite sets. In fact PA is equivalent, as a theory, to ZF minus infinity; that is, ZF with the negation of the axiom of infinity.
The axiom of infinity allows us to take the "output of the completed induction," if you think of it that way, and put it all together into a set; which can then be operated on by the rules of set theory.
So now we have defined [math]\mathbb N[/math], the set explicitly given to us by the axiom of infinity.
* Next, we make the observation that since the elements of [math]\mathbb N[/math] are themselves all sets, we can ask whether two given members of [math]\mathbb N[/math] happen to stand in the relation [math]\in[/math] (set membership) to one another.
And now voilà, [math]\mathbb N[/math] is a transitive set well-ordered by [math]\in[/math] so that in fact the pair [math](\mathbb N, \in )[/math] is exactly the first transfinite ordinal [math]\omega[/math].
So [math]\mathbb N[/math] and [math]\omega[/math] are two different names for the exact same set. We use one notation or the other when we want to emphasize the arithmetic properties or order properties of [math]\mathbb N[/math].
* Finally, what about [math]\aleph_0[/math]? I seem to remember I talked about this a while back in some thread or other. The modern definition is that a cardinal is the least ordinal with a given cardinality. That definition has the virtue of making each cardinal a set. In particular, what is the least ordinal cardinally equivalent to [math]\aleph_0[/math]? It's [math]\omega[/math].
So as sets, [math]\aleph_0 = \omega[/math].
Conclusion: The notations [math]\mathbb N[/math], [math]\omega[/math], and [math]\aleph_0[/math] each refer to the exact same set. We use one notation or another when we want to emphasize the arithmetic, order, and cardinality aspects of that set, respectively.
[And I probably should note that NONE of these have anything to do with the [math]\pm \infty[/math] of the extended real number system as used in analysis!]
SO: Not only is it mathematically correct to say these three distinct symbols all represent the same set; it is in fact illuminating to do so. If Wikipedia failed to bring out these connections, I would put that down to Wiki being unilluminating in this instance.
I hope this was helpful. This got long so I'll reply to the rest of your post later.
Your use of complete is nonstandard and I don't know what you mean. Are you trying to go through the proof of Burali-Forti?
Quoting alcontali
No this is the part where you're reading too much into it.
The axiom of regularity, also known as the axiom of foundation, says that all sets are well-founded. There are no infinite downward chains of membership. No infinite regress, if you like.
This is a standard axiom of ZF. It's never mentioned because it's not used for anything. It's simply a given that essentially set theory is the study of well-founded sets.
Now it happens to be the case that the negation of the axiom of regularity is consistent with the rest of the axioms. And therefore it is possible, though quite obscure, to study non well-founded set theory.
So Burali-Forti is a theorem that follows from the axioms of ZF: that the class of ordinals can not be a set. And non well-founded set theory is a thing, but an obscure thing. These two ideas are NOT at some opposite ends of a pendulum or related to one another at all. You are wrong about any important connection or insight here. I'm being dogmatic to emphasize this point.
Quoting alcontali
I don't have any idea what you mean by materialized, nor how [math]\mathbb N[/math] substantially changes its nature. You are not speaking mathematical sense here in my opinion.
I don't want to take away from you the pleasure of discovering Burali-Forti's result, and the nice proof sketch on Wikipedia. But it's not a very important theorem nor does it lead to anything of interest. It's not really surprising once you know that the union of any collection of ordinals is an ordinal. Therefore if the class of ordinals were a set we could take its union to get another ordinal that must be a member of itself. That violates regularity, so there can be no set of ordinals.
But I hope you will see that the study of non well-founded sets is very obscure and far out of the mainstream of set theory. It's not an equal balance go this way or go that way kind of thing. For all intents and purposes, all sets are well-founded. Practically by definition.
I would like to discuss this because it is absolutely not self-evident to me that two sets of different rules, i.e. PA versus ZF minus infinity, would be completely equivalent. The rules do not even look like each other. Just look at the axioms. They are simply different. Still, if their equivalence is provable, then I would consider that to be an amazing result.
In fact, in that case, it should be possible to take the axioms of PA, push them through some kind of algebraic transformation process, and then end up with ZF minus infinity. I cannot imagine what this transformation process could look like.
Quoting fishfry
That is exactly what I mean by "materializing".
The reason why I used this term, is because this is the term used when you fully calculate and store the output of a view formula in relational databases, instead of keeping it around as a merely virtual construct. So, taking the "output of the completed induction" is called "materializing" in that context.
I just accidentally used a term (materialized view) en provenance from another domain.
In fact, it is not a completely different domain, because relational algebra is a downstream domain from ZF set theory. It completely rests on standard set theory. It is only much closer to practical applications:
Quoting Wikipedia on relational algebra
Quoting fishfry
I wasn't aware of the fact that the notation, N = { 0, 1, 2, ... }, is considered complete in set theory (through the axiom of infinity). (It is obviously not considered complete in PA.) So, yes, my use of the term "complete" is not standard in set builder notation in reference to ZF (but not in reference to ZF minus infinity). These things are extremely subtle. It depends on whether the theory in the context of which it is used, has an axiom that can "take the output of the completed induction", i.e. "materialize" it in relational-algebra lingo.
It is also very related to the concept of list comprehension where a similar problem occurs. You can create the list of natural numbers as a virtual construct, but you cannot "materialize" it, because that will cause your system to run out of memory.
Quoting Wikipedia about using virtual constructs that represent the infinite list of natural numbers
Quoting fishfry
if ? is the set of ordinals but ? is also itself an ordinal, then this situation will result in ? being a set that contains itself, and therefore, result in a set that is not well-founded.
Quoting fishfry
Yes, but that is exactly what would happen if ? is the set of ordinals but ? is also itself an ordinal. That is in my impression another reason why ? cannot be termed a set but must be considered a proper class.
Quoting fishfry
Yes, that was indeed the connection that I saw.
The technical condition is that PA and ZF-infinity (read "ZF minus infinity") are bi-interpretable. Here's one link I found:
https://math.stackexchange.com/questions/315399/how-does-zfc-infinitythere-is-no-infinite-set-compare-with-pa
Also see:
https://mathoverflow.net/questions/551/does-finite-mathematics-need-the-axiom-of-infinity
Quoting alcontali
Andrés E. Caicedo's checked answer in the Stackexchange thread outlines the procedure.
Quoting alcontali
I'm not convinced. The completed set of natural numbers is posited, or brought into set-theoretic existence, by the axiom of infinity. If you want to call that materialized, well ok I guess ... but ... Is the powerset of [math]\mathbb N[/math] materialized by the powerset axioms? If I have two sets [math]A[/math] and [math]B[/math], is their union materialized by the axiom of union?
There doesn't see to be much depth to your definition of materialization. It just seems to mean, "brought into existence by a given axiom." In which the entire universe of propositions is materialized by the axiom 0 = 1. Is this what you mean?
Quoting alcontali
Oh I see. I understand the example but it's a bit of a stretch as an analogy for the set of natural numbers being given by the axiom of infinity. But if it works for you I do sort of see what you mean.
Quoting alcontali
It's not really analogous to an axiomatic system IMO but sort of works as a vague metaphor.
Quoting alcontali
Yes, perfectly well aware. But that doesn't make your vague analogy any sharper IMO. But this is not an important matter. Everyone is entitled to their own visualizations, intuitions, and conceptual ideas. Whatever works to understand the material.
Quoting alcontali
It surely isn't. There is no such term of art in set theory. Of course the natural numbers with the usual metric (absolute arithmetic difference) is Cauchy-complete. (Tricky. Why?) It's because every Cauchy sequence of natural numbers is eventually constant. But of course this is not what you meant.
What do you mean? There is no such technical term as saying that [math]\mathbb N[/math] is "completed" by the axiom of infinity. You just said it was materialized. I'd almost buy the latter, because even though it doesn't mean much, at least it's not a totally nonstandard use of the word complete, which has many other meanings in math but none in this context.
Quoting alcontali
Oh, complete as in a completed infinity. Yes well the problem here is that "actual and potential infinity" are terms of art in philosophy, not math.
You seem to be trying to make something out of not much. If what you're saying is that you weren't formerly aware that it's the axiom of infinity that bestows set-hood on [math]\mathbb N[/math], ok now you know. But you can see from statement of the axiom of infinity that this is exactly what it does. It says that there is a set that contains the empty set and, whenever it contains a set [math]X[/math], it also contains the successor [math]X \cup \{X\}[/math].
The negation of the axiom of infinity says there's no such set; and since [math]\mathbb N[/math] is such a set, the negation of the axiom of infinity outlaws its existence.
Quoting alcontali
You're just making stuff up here. I'm trying to figure out why you're meandering down this road. PA is perfectly "complete" in your sense, it contains the conclusions of all its axioms. Or something. Neither PA nor ZF are more or less "complete" than the other. They each contain all and exactly those objects that are permitted by their respective axioms. Yes?
Quoting alcontali
You're reading in subtleties where there are none. From where I sit you are taking a simple fact and tring to give it great significance. PA has all the natural numbers, and ZF via the axiom of infinity has a completed set of them. But if you took ZF minus powerset, then you wouldn't have full powersets. Does the powerset axiom complete ZF with respect to powersets? There's nothing interesting about this, you are seeing complications and subtleties where truly, I say to you as clearly as I can, there are none.
Quoting alcontali
Or make a full powerset. Or a union. So is materializing the same as completing? Why go on about this? Without the axiom of pairing, X and Y could be sets but there's no set {X,Y}. Does the pairing axiom complete, or materialize, pairs of sets? Can you see that you're imagining subtlety where there isn't any?
Quoting alcontali
In Python there's something called a generator. The idea is that I want to iterate through a list but I don't want to pre-create the full list ahead of time. I just supply the algorithm. This I assume is exactly what you mean. I understand the example, I just don't think it's very interesting or meaningful.
Quoting alcontali
Sure whatever.
Quoting alcontali
We've already been through the proof. It's straightforward. First you prove that the union of ordinals is an ordinal, therefore if the class of ordinals is a set so is its union, which violates regularity. Ok done. This is not worth starting a religion over. It doesn't mean anything as significant as what you're trying to read into it.
Quoting alcontali
Yes, and ...?
Quoting alcontali
Yes, that's what regularity prevents. Again, so what? I find myself frustrated. You're trying to convince me that something deep is going on, and there isn't. If your private intuitions help you, all the good. But you haven't written anything of interest; least of all with your materialize and complete terminology. ZF is not complete, you know, at least if it's consistent. You can't say the axiom of infinity completed it using your made-up definition, when it's NOT complete by everyone's standard definition. There's nothing complete about it. It just has more sets than PA because you added another axiom of set existence.
Here's another link that might be of interest, the hereditarily finite sets. This is the class of objects that's described by PA and by ZF-infinity. It's perfectly "complete." It has all the sets and only those sets that it's supposed to as given by the rules of its construction.
To sum up, all I can see is that you're saying that PA is complete with respect to the axioms of PA, and ZF is complete with respect to the axioms of ZF, and ZFC is complete wrt the axioms of ZFC, and so forth. But then ZF is missing some choice functions so it is "incomplete" with respect to the axiom of choice. And ZFC is complete with respect to the axiom of infinity and the axiom of powersets and the axiom of union and the axiom of pairing, but not with respect to the Continuum hypothesis.
There's nothing of interest to this observation; and worse, it's bad terminology because it conflicts with the standard meaning of completeness in an axiomatic system.
This is what I got from your post.
Always had an interest in set theory and foundations.
Quoting jgill
Beg to differ. Functional analysis uses the Hahn-Banach theorem, which is equivalent to a weak form of the axiom of choice. There are many areas of "hard" math in which foundational questions arise. The history of analysis from Newton onward is a two hundred year struggle to get the definitions right in order to have a logically coherent theory. And of course there are many deep foundational questions in probability and measure theory related to the axiom of choice and nonmeasurable sets.
An interesting factoid is this. When Cantor discovered transfinite set theory, he was studying the zeros of the trigonometric functions used by Fourier in his studies of heat flow. If you think about it, such a function could have zeros at each of the integers. Or perhaps in addition, at each integer it has zeros at that integer plus 1/n for each n. And maybe it has chains of zeros hanging off some of the 1/n branches. You need a language, or a notation, for keeping track of the way the zeros of trigonometric functions can occur. That's the ordinal numbers.
Transfinite set theory came directly from physical considerations. If you heat up one end of an iron bar under laboratory conditions and carefully observe the heat flow, you will inevitably discover transfinite set theory. That's how it actually happened.
I found a beautiful-looking article called The Trigonometric Background to Georg Cantor's Theory of Sets behind an academic paywall, Grrrrrrrr! Academic paywalls are evil. @jgill you have a university affiliation by any chance?
ps -- Found a great overview of this material written for a general audience.
How did Cantor Discover Set Theory and Topology?. I think I'm going to give this a read.
Bi-interpretability looks like an interesting subject, but unfortunately the Wikipedia page does not elaborate PA versus ZF-infinity as an example.
Quoting fishfry
Well, you did use the term "complete" in the sense of induction-complete. I clearly used it in the same way, and then you suddenly backtrack to claiming that induction-complete would be "no such term of art in set theory".
Quoting Completeness of the real numbers
Since our conversation had absolutely nothing to do with completeness of the real numbers, I wonder why you mention Cauchy completeness? It just adds to the confusion.
Quoting fishfry
Relational algebra is itself obviously also an axiomatic system. I do not believe that anybody even questions that.
Quoting fishfry
It is used as a term for induction-completing (A term you actually introduced by yourself yourself). It has obviously nothing to do with logic completeness, Cauchy completeness, Dedekind completeness, and so on.
Quoting fishfry
That is another context in which the term "completeness" is used.
Quoting fishfry
That is about completeness in logic, which is not directly related to the "output of the completed induction". I was not referring to "logically complete".
Quoting fishfry
There is not one definition for completeness. It depends on the context. You may have misunderstood what context I was referring to, but that kind of confusion occurs relatively easy with the term "complete".
In fact, I never used the term induction-complete or induction-completed before. I only used it because you used it first. I tend to use the term "materialized" instead of induction-completed. Furthermore, it is probably better not to further overload the term 'complete' with additional meanings.
Quoting fishfry
No, of course not. That is about logical completeness. I didn't make even one single remark in that context. I don't see why you would understand any of the above in terms of logical completeness.
When a discussion degenerates in "who is smarter than whom", "who knows more than whom", i.e. the typical, ridiculous conversations in the academia, in which they engage because they simply have nothing else to show for, then I tend to back out. That kind of conversations are simply not interesting. In that case, I even prefer -- God forbid -- the slightly less ridiculous conversations in business about whom makes more money than whom, because amounts of money are at least objectively measurable while amounts of knowledge are not.
Not my intention.
Yes, the Zermelo well-ordering theorem comes into play (I took a year of FA fifty years ago), but is not required if the underlying normed linear space is separable (common). I never had any reason to use Hahn-Banach since I didn't do much in soft analysis, but recently I've been intrigued by functional integration and particularly Feynman's path integral (which is not exactly kosher). I dabbled in a generalization of analytic continued fraction theory, so called by an old friend, Wolf Thron RIP. Nothing there hinging upon transfinite set theory (well, to my knowledge!).
Quoting fishfry
No more, unfortunately. Retired twenty years ago.
Yes, it seems to be more subtle than mere logical equivalence.
Quoting alcontali
You are entirely correct. I said, "The axiom of infinity allows us to take the "output of the completed induction," if you think of it that way ..." Guilty as charged. I have apparently been strenuously arguing against my own terminology and blaming it on you. My bad. Thanks for pointing it out.
Quoting alcontali
Ok ... materialization means making a set inductively complete. So if we have a set that contains x, it contains the successor of x and the successor of the successor of x, etc., for all possible applications of finitely many occurrences of the successor operation. A set is materialized if it's closed under taking successors.
In this case I believe you're talking about a limit ordinal. A limit ordinal is not the successor of any ordinal; but is rather the "completion," or closure under the successor operation, of its elements. So [math]0[/math] is the only finite limit ordinal; and [math]\omega, 2 \omega, 3 \omega, \dots, \omega^2, \dots[/math] are all limit ordinals. They're constructed by taking the upward limit of a collection of ordinals.
But why "materialized?" That seems to load the concept with some kind of metaphysical significance. I could live better with "inductively completed," now that I think about it.
Quoting alcontali
Guilty as charged. But I don't love the word materialized either. I think the idea of limit ordinals captures the concept.
Indeed, every consistent theory has a model whose universe is a set. Moreover, every consistent theory that has a model with an infinite universe also has a model with a denumerable universe. Moreover, every consistent theory that has a model with an infinite universe also has a model whose universe is N.
Meanwhile in the standard interpretation of PA there is not even mention of sets, proper classes, or universes for models of PA. So, since the standard interpretation of PA doesn't mention it, and, in PA's ordinary meta-theory, universes are sets, not proper classes, I don't know where there would be an actual mathematical formulation of "N is a proper class as far as PA is concerned."
(2) Usual terms for inductive sets are 'inductive set' or 'closed under induction', The induction may pertain to a relation or operation. In the case of ordinals, we may consider the successor operation ( Sx = xu{x} ), and indeed, as mentioned, limit ordinals are closed under the successor operation.
(3) It is not necessary to use the axiom of regularity or unions to show that there is no set that has all the ordinals as members, as follows:
Lemma (easy to prove without using the axiom of regularity): No ordinal is a member of itself.
Now suppose there is a K such that every ordinal is a member of K. Then let L be the subset of K that has only ordinals as members. Then L itself is an ordinal (easy to prove). But then L is a member of itself. Contradiction.
(4) As mentioned, the notion of showing that two theories (even with completely different symbol sets) "say essentially the same thing" is rigorously captured by showing that there is an interpretation of one theory into the other and vice versa. The mathematical formulation of an interpretation of a theory into another is straightforward and not mysterious.
Correct! Thank you for summarizing all of this bullshit!
If this is in response to something I said, I never claimed such a thing. I use proper class as an informal description of a collection that's not a set. In PA the extension of the unary predicate "n is a natural number" is the collection we call [math]\mathbb N[/math]. However [math]\mathbb N[/math] is not a set in PA. So I say [math]\mathbb N[/math] is a proper class. It's the extension of a predicate that isn't a set.
I can see you objecting that PA doesn't talk about sets. But given the objects of PA, namely the natural numbers 0, 1, 2, 3, ... we can certainly define unions, intersections, pairs, powersets, and so forth; and do finite set theory perfectly well. In other words PA is a model of ZF-infinity. In that case we would find that [math]\mathbb N[/math] is not a set.
My usage was informal but useful in the sense that when we talk about proper classes like the class of all sets, that's hard to imagine; but [math]\mathbb N[/math] is easy to imagine. And if we throw out the axiom of infinity, [math]\mathbb N[/math] is a proper class. Unofficially, of course. But [math]\mathbb N[/math] is the extension of the predicate "n is a natural number" and it's not a set, so it certainly qualifies. Informally.
Quoting GrandMinnow
Yes. Have you a demonstration in PA of the consistency of PA? Of course not, since that would violate Gödel's first incompleteness theorem. The only way to get a model of PA is to wave your hands and say the magic words, "Axiom of infinity!" And now you have a model of PA, along with a consistency proof for PA. But without the axiom of infinity you haven't got a model of PA nor a proof of PA's consistency.
I know you know this so I must be missing something. And you know our friends the ultrafinitists, who doubt the consistency of PA. Sounds crazy but that doesn't make them wrong.
"via the Peano axioms, in which N is a collection but not a set"
"in PA, the natural numbers are a proper class"
Those are not actual mathematical statements. Not even informally. First order PA (throughout, I will mean first order, which is what people usually mean in modern context unless said otherwise, and as your remark about set theory proving model existence also suggests you're not referring to higher order PA) does not speak of N (not as a formula "in" PA, but only by model interpretation in, say, set theory) nor of a distinction between sets and proper classes. And it would be quite remote, along with needing to work out needed details in execution, to justify those two quoted claims by the fact that PA can be viewed as (ZF-Inf)+~Inf.
Quoting fishfry
No such thing takes place in PA there is no unary predicate 'is a natural number'. Usually the meta-theory for making statements about PA is set theory. And in set theory, the notion of an "extension of a predicate" is given by a model. The universe for the standard model is N, which is a set.
Quoting fishfry
PA is a theory, not a model. The correct statement is that PA and (ZF-Inf)+~Inf can be interpreted in each other. But that still doesn't make "N a proper class in PA" a coherent statement.
Quoting fishfry
Then one can say that any axiom of any theory is waving hands and saying magic words. But yes, a model of PA is proven to exist from the set theory axioms that include the axiom of infinity. Again, that does make "N a proper class in PA" a coherent statement.
/
Correct statements include these:
If PA is consistent, then PA does not prove that PA has a model.
Set theory proves that N, which is a set, is the universe for a model of PA.
And that does not involve proper classes.
PA is a certain set of sentences. There is no sentence in PA that one would ordinarily regard as saying "N is a proper class". There is a universal quantifier for PA, and with any model (formulated in set theory) for PA, the universal quantifier stands for the universe of the model. And that universe is always a set. In particular, with the standard model, the universe is N, which is a set.
I think maybe what you were driving at originally might be said this way:
The universe for the standard model of PA is N and has as its members all and only the natural numbers, so N itself is not a member of N. And with the standard model, PA itself does not assert the existence of a set that has all the natural numbers as members.
Couching that in terms of proper classes is off.
We can go further, it is not precluded that other models of PA do include all the members of N and N itself as a member of the universe for the model. Most simply, N+ (i.e. Nu{N}) is a denumerable universe for certain models of PA.
Your use of the terminology is impressive, and I'm not able to determine whether you have advanced knowledge or not; since if you do, my own level of knowledge of these matters would not be sufficient to let me make that determination.
Nevertheless you are wrong on this particular point.
I didn't make up the claim that absent the axiom of infinity, [math]\omega[/math] (or [math]\mathbb N[/math]) is a proper class. I read it somewhere a while back in a reference I can no longer find.
I did find this discussion supporting my claim. It's in a Stackexchange thread called The purpose of the Axiom of Infinity.
I will quote from the checked reply by Andrés E. Caicedo. Professor Caicedo is a well-known professional set theorist. I linked his home page so that you can determine for yourself his stature within the set theory community. (Click on the link Notes and Papers). He says:
"In particular, the axiom of infinity goes well beyond the Peano axioms (and not simply in terms of consistency strength or expressive power). The Peano axioms are provable in ZF without the axiom of infinity. In this theory, you cannot prove that [math]\omega[/math] is a set, but you can prove that as a (perhaps proper) class, it satisfies both first and second order PA."
Perhaps you can take a look at his full reply, which is of interest beyond our conversation and worth reading in general, and put it into context for me. If it means something other than what it plainly seems to say, then I'll concede the point. Else you'll need to.
ps -- In the same Stackexchange thread see Asaf Karagila's comment:
"In a model of ZFC with the negation of the axiom of infinity instead, the natural numbers are just the ordinals of the universe."
In other words, in ZFC minus infinity, the natural numbers are the proper class of ordinals. You may object to my terminology (and Professor Caicedo's), but if the extension of a predicate is not a set, what is it? It's a proper class, right?
(1) There is a difference between ZF-Inf and (ZF-Inf)+~Inf. I'll call the later 'HF' (the theory of hereditarily finite sets).
The language of HF is the language of ZF (i.e. the language of set theory).
PA and HF can be interpreted in each other.
The usual universe for HF that we have in mind is the set of hereditarily finite sets. And of course N is also a universe for HF.
(2) Most textbooks take 'is a set' as informally primitive, but we can be precise in the language of set theory:
x is an urelement <-> (~ x=0 & ~Ey yex)
x is a class <-> ~ x is an urelement
x is a set <-> (x is a class & Ey xey)
x is a proper class <-> (x is a class & ~ x is a set)
In set theory, we can prove:
Ax x is a set (though, as mentioned, most textbooks don't bother with something so basic).
(3) The language of class theory (such as Bernays style class theory, which I'll call 'BC') has a primitive predicate 'is a set' (or a many-sorted language is used, which is essentially the same as using a primitive 1-place predicate), so in BC 'is a set' is not defined but instead certain axioms are relativized to sets.
In BC we prove:
Ex x is a proper class
(4) I explained why "N is a proper class in PA" [or whatever paraphrase] is, on its face, not coherent. But I allowed that one is welcome to adduce some particular mathematical statement instead. And I explained why it would not be a correct statement in set theory (and I would add, not even in BC). So maybe we turn to HF.
Since HF is in the language of set theory, in HF we can define any predicate of set theory, and we can define any operation of set theory for which we can prove existence and uniqueness in HF.
HF proves ~ExAy(y is a natural number -> y e x). So there is no definition of 'N' (in the sense of the set of natural numbers) in HF.
So, while HF can have predicates 'is a natural number', 'is a set', and 'is a proper class', still HF can't have the definition N = the set of natural numbers.
As far as I can tell, the best we could do in NF is this theorem:
If Ax(Ay(y is a natural number -> yex) -> x is a proper class). But that holds vacuously, since there we have ~ExAy(y is a natural number -> yex).
So, as far as I can tell, we are still thwarted from making sense of "N is a proper class in PA" or even "N is a proper class in HF".
And in set theory (and even in BC, if I'm not mistaken) the universe for a model is a set, not a proper class.
(5) Caicedo says, "in ZF without the axiom of infinity [...] you cannot prove that w is a set, but you can prove that as a (perhaps proper) class, it satisfies both first and second order PA."
I don't know why he says 'perhaps' there. And without more explanation, I don't understand what he's saying.
I do understand that, in ZF-Inf, there is not a proof that there is a set of which all natural numbers are a member (that's another way of affirming the independence of the axiom of infinity).
But when he says "you can prove", does he mean prove in ZF-Inf? Proof of satisfaction with models takes place in set theory, not in ZF-Inf nor in HF. And in set theory, universes of satisfaction are sets, not proper classes.
What is understandable to say is:
ZF-Inf does not prove there is an x such that all natural numbers are a member of x.
HF proves there is no x such that all natural numbers are a member of x.
PA and HF are mutually interpretable.
The set of natural numbers N is a universe for a model of ZF-Inf or of HF.
But saying "in Pa (or in HF), N is a proper class" makes no sense.
(6) Quoting fishfry
No, absent Inf, it is not a theorem that N is a proper class. Indeed, absent Inf, there is not even possible a definition N = the set of natural numbers. Rather, absent Inf, there is not a proof that there exists an x such that all natural numbers are in x, and there is not a proof that there is no x such that all the natural numbers are in x. In other words, "there is an x such that all natural numbers are in x" is independent of ZF-Inf. However, (ZF-Inf)+~Inf does prove "there is no x such at all natural numbers are in x", but still, it does not say anything about such a thing (which does not exist anyway in NF) being a proper class or not.
(7) Quoting fishfry
In HF, we can define the predicate 'is an ordinal' and for any finite ordinal, we can define a constant for it. But, as you mention, we can't define a set that has all the finite ordinals as members.
But even in set theory, there are specific ordinals that don't have a definition (there are more than countably many ordinals, but only countably many definitions we could form).
That is okay as an understood informality when speaking of set theory. But it does not transfer to PA or HF in the same way. The reason is that PA and HF already have a meta-theory in which everything is a set. So saying "the proper class that is the extension of the predicate" (i.e. the predicate carved out by the formula) is a way of saying that there is no set of all things that have the predicate but for convenience, and recognizing this as merely informal, we can speak of the formula itself as if it specifies a "class". However in PA or HF, any formula specifies not a proper class but a set. Even the universal predicate specifies the set that is the universe of whatever given model for the theory.
Example:
With set theory, "the proper class of ordinals". This is an informal way of saying "those x such that 'x is an ordinal' holds". It is a breezy way of speaking, when more formally we would refer only to the formula 'x is an ordinal'. In other words, saying "If x is in the proper class of ordinals" is an informal locution for "If 'x is an ordinal' holds". Or for example, "relativize F(x) to the proper class of finite sets" is a locution for "Ax(x is finite -> F(x)).
With PA, "the proper class of natural numbers". No, doesn't work, because the usual meta-theory for PA is set theory where there is the SET of all natural numbers, which encompasses those x such x is a natural number.
(A)Numbers are just words and abstract concepts used to label and understand the world around us. Words are created whenever someone decides to label something around them (which hasn't been labeled as such before) and enough people agree to the convention that it sticks either because it is useful or because people just like it.
(B)By convention we think of the number system proceeding from a point of origin going to the right, but that is only when we are counting. When dealing with things such as geometry something to the right of somethings isn't necessarily more positive than it.
(C)I believe one can say that words, languages, numbers/abstract concepts, and mathematics are more or less tools that we use to define the world around us.
Quoting Qwex
Because you can only call your finger or crossed fingers '1' and not make it '1' since one is only an abstract concept and your fingers are instances of various abstract concepts. Labels and abstract concepts are created whenever you decide to call something by some name,describe some aspect or attribute of it but these labels and mental projections are not the things in and of themselves. Only the actual physical instances of things are things in and of themselves.
And on that subject numbers are never a physical instance of anything because they can only describe an attribute of a physical thing and all physical things have more than a single attribute to them.
Then I think it would be efficent to use base 8, or whatever base math is better. Definitely not base 4. Base 4 is tiny on the ratio. Like pointing a subject using selectable squares only. 08 in base 8 IS base 4. Powerful stuff.
I think the best base is containing math, the rules of containment.(or something along these lines).
Thanks. You wrote a really interesting post and I have a lot of questions and comments. Before going into them I have to say that my stance remains unaltered: which is:
Regardless of whether it's technically accurate, it can sometimes be a useful visualization of proper classes.
So it doesn't matter if you're right on the technical part. If I myself find it a helpful visualization, that is my right. We are all entitled to our visualizations.
So at best you could possibly argue that even though I have the right to my own private visualization, I should not mention it aloud in polite company. I'll take that under advisement.
As an analogy, suppose an engineer takes calculus class and spend the rest of his professional life believing that dx is an infinitesimal. We could of course correct him mathematically, but it's a harmless belief for an engineer or physicist and a highly effective conceptual aid. Even many mathematicians who technically know better privately think of dx as an infinitesimal when analyzing a calculus problem. I hope you take my analogy as on point.
Also I did find at least one professional set theorist who was willing to make an admittedly offhand remark in agreement with my thesis. And I agree that it's not even clear what he meant. But he referenced the idea. If my intuition matches professor Caicedo's, I consider my point to have been made to my own satisfaction.
All that said, you wrote a lot of really interesting things so I'll try to comment.
Quoting GrandMinnow
Oh yes right away you have actually identified a couple of points of confusion or ignorance in my mind.
In my reading the last few days I do keep seeing the notation ZF-infinity +(not infinity) and this is puzzling me greatly.
For example in discussions of the axiom of choice we see ZF or ZFC. I've never seen anyone say, "ZF + not-AC". The not-AC is emplied when you say ZF. And in a given model of set theory, there either is or isn't a choice function on every collection of nontempty sets. It has to be one way or the other, we don't have to say it twice.
But when it comes to infinity, lately I keep reading ZF-infinity + not-infinity and I have no idea why they're doing that! I'd think that likewise, in any model of ZF-infinity, there either is or isn't an infinite inductive set containing the empty set and its successors.
So why do we need to say "+ not-infinity?"
Quoting GrandMinnow
Right, the hereditarily finite sets. Why are they considered interesting? I just thought they were the usual finite von Neumann ordinals, ie the natural numbers. Is there something else special about them?
Quoting GrandMinnow
Ok so interpretation is a technical term that I think I don't know. I know what it means to interpret an axiomatic theory, ie assigning meaning to the symbols or at least assigning elements of some model. But I'm not sure about thie bi-interpretability business.
Quoting GrandMinnow
Ah. How are HF and N different?
Quoting GrandMinnow
Ok, very interesting. Haven't seen those defined formally before. I learned the "informally primitive" way.
Quoting GrandMinnow
Ok I never actually looked at BC but I've heard of it. Also I believe Morse-Kelley has classes.
Quoting GrandMinnow
I have seen no such explanation; or if I have, I haven't understood it. A proper class is the extension of a predicate that's not a set. That's how I understand the term. I haven't thought about this interms of BC or Morse-Kelley so maybe you're right, I don't know.
Quoting GrandMinnow
I admit I haven't followed the argument but that's probably my fault.
Quoting GrandMinnow
Ok.
Quoting GrandMinnow
Oh I think I see where you're going. You're saying I can't form the predicate that I think characterizes N. Am I on the right track at all?
Quoting GrandMinnow
In other words, if I'm understanding, then I can't form the collection of all numbers because I haven't got the language to do that. So my argument fails.
NOTE: At least the TECHNICAL argument fails; but my intuition still likes it!!
Quoting GrandMinnow
I don't understand the details but perhaps this is a good argument. In which case I'm technically wrong that N is a proper class but it's still a useful intuition; and if not for everyone, at least it is for me. You can't enjoin me from thinking my thoughts, misguided though they may be.
Quoting GrandMinnow
Yes, models are supposed to be sets. That's my understanding.
Quoting GrandMinnow
Yes. Perhaps he is speaking INFORMALLY because he has a similar intuition as mine. Perhaps he even has a context in mind where the statement could be made rigorous. Perhaps not. But we DO know that at least one professional set theoriest thinks it's something you can say in this context. That means a lot to me.
Quoting GrandMinnow
Right. Because if their were, it would witness the axiom of infinity! But I take this differently, not in terms of proofs. I imagine that in any model of ZF-inf there is not a set of all natural numbers. Because if there were it would prove the axiom of infinity, and therebey contradict ZF-inf. Am I wrong about this?
Quoting GrandMinnow
I'm afraid I have no idea what his point may have been; only that he used the magic words proper class.
Quoting GrandMinnow
Now I do not understand this. ZF-inf says: There is no x such that all natural numbers are a member of x. You said we can't prove there isn't; I'm saing that there isn't. This seems subtly different.
Quoting GrandMinnow
Ok the distinction between ZF-inf and HF is totally lost on me. If I understood it perhaps I'd be enlightened.
Quoting GrandMinnow
Ok. But not quite the same in some way I can't grasp.
Quoting GrandMinnow
Yes I believe that.
Quoting GrandMinnow
a) If you say so; and
b) But can you forbid me from thinking it? What if I promise not to tell anyone else to think it? But taken informally, it's a good visualization of what is meant by proper classes.
Quoting GrandMinnow
Of course it's not a theorem, in ZF there is no such definition or thing as a proper class. I thought I made it clear that I understand this point. So it should be obvious that I'm speaking vaguely and metaphorically and not literally.
But in fact we're at this same "theorem" impass. It's true that there's no theorem. But morally, N is a proper class!! This is the nub of our disagreement.
Quoting GrandMinnow
Yeah yeah. You're technically right and morally wrong. When the Peano axioms say, "O is a number," what exactly is "is a number" if not a legal predicate?
Quoting GrandMinnow
I'm losing my train of concentration here, don't think I got much out of this para. My fault.
Quoting GrandMinnow
Ok we found something we agree on!
Quoting GrandMinnow
Ok. I concede all your points even though there are some I don't understand, and among those I understand, some I disagree with.
But what of it? It's obvious that since there are no proper classes in ZF, when I speak of proper classes in ZF I mean metaphorically. You can go into any set theory book and they'll tell you that the set of all sets is a proper class, even as they say it's not technically so because we're working in a set theory that doesn't have proper classes!
Ok my head is officially confused. There is much I don't understand about these matters. But I retain my private intuition that N is morally a proper class with respect to PA; and that if I were to share my intuition in print, I would be helping more people than I'd be hurting.
Ok for that!! More words than this subject is worth. Not a hill I need to die on. I'll consider myself suitably chastized for my belief that N is a proper class with respect to PA. As Galileo whispered as they forced him to recant his belief that the earth moves around the sun: And yet it moves.
That is not correct.
ZF-Inf is ZF but without the axiom of infinity. (The '-' here means 'without'; it doesn't mean 'the negation of'.)
(ZF-Inf)+~Inf is ZF but with the axiom of infinity replaced by the negation of the axiom of infinity.
Quoting fishfry
No, the finite ordinals are a proper subset of the set hereditarily finite sets. For example, {0 2} is an hereditarily finite set but it's not an ordinal.
Quoting fishfry
They may be of interest for many reasons, but for starters, they are the usual universe for a model of "finite set theory" = (ZF-Inf)+~Inf = HF.
Quoting fishfry
This is a different sense of 'interpretation' (but closely related). Simplifying here: We interpret a theory T into a theory T' by defining the symbols of T in the language of T' so that every every theorem of T is a theorem of T' plus the added definitions. And we say the theories are equivalent when there is such an interpretation from T into T' and vice versa. (This deserves a sharper statement, but it's too many technical details for a post.)
Quoting fishfry
HF is a theory. N is a set.
Quoting fishfry
Defining a predicate symbol is not a problem. But there is no definition of a constant symbol (such as 'N') such that N = {y | y is a natural number}, since HF does not prove that there exists an object that has as members all the natural numbers.
In any language, in any theory, we can define whatever predicate symbols we want. It's only function symbols (including constant symbols, where a constant symbol is a 0-place function symbol) that require the supporting existence and uniqueness theorem
Quoting fishfry
In HF, you have the language, but you don't have the existence theorem ExAy(y is a natural number -> y e x).
Quoting fishfry
No, the sentence ExAy(y is a natural number -> y e x) is a theorem, but that doesn't preclude what the members of the universe for the model may be.
For every infinite cardinality, there is a model of ZF-Inf with a universe of that cardinality. And that universe can have as members any sets whatsoever. Same for (ZF-Inf)+~Inf. Same for PA.
For example, we can have a model of PA whose universe is {w, w+1, w+2} ['w' for "omega" = the set of natural numbers"] and each of those members of the universe is infinite.
But wait, (ZF-Inf)+~Inf has a theorem ~Ex Ix [where 'I' is a defined 1-place predicate symble we are read in English as "is infinite"], so how can the universe of a model have a member that is infinite? Well, because for such a model, the predicate symbol 'e' is interpreted not as the ordinary membership relation but rather as some other "bizarre" relation and so also my 'I' be interpreted differently from "is infinite". When we talk about models in general, we can't presume that any given model of a theory "captures" the way we ordinarily "read off" the theorems of the theory. If we want to narrow the discussion to only models that adhere to the way we "read off' the theorems, then we should confine to talking about standard models.
Quoting fishfry
In ZF, we may define:
x is a proper class <-> Ey y e x & ~Ez x e z
And we may prove:
~Ex x is a proper class.
Quoting fishfry
First order PA doesn't have a primitive 'is a natural number'.
Peano's historical own formulation should not be conflated with first order PA.
Quoting fishfry
There is not a set of all sets, not even in class theory. There is the class of all sets, and it is a proper class. And I explained why referring to proper classes in discussion about set theory can be understood as an informal rendering for an actual formal notion in the background, but that is lacking here in saying N is a proper class in discussion about PA.
And N is a set, which is not needing exceptions in view of the fact that in PA there can be no definition N = {x | x is a natural number}. If one wishes to say "N is a proper class with respect to PA" but not formulate the exact mathematical meaning of "with respect to" or even to a clearly articulate an intuitive/heuristic notion that is still consistent with the ordinary mathematical result that N is a set, and hopefully has value as a metaphor rather than confusing the subject with impressionistic use of terms, then, of course, I cannot opine whether or not in one's own mind it somehow makes sense nevertheless. But I do say, and have explained, that it makes no sense to me.
I just wrote a post about the bi-interpretability of PA and HF. The worst problem I had in investigating this idea was to find examples of how to translate standard set builder notation into arithmetic-set notation.
For example, { 1, 4, 7 } in set builder notation is equivalent with ?(x):= 1 - sgn((x-1) (x-4) (x-7))² in arithmetic-set notation.
Now if you want to find even one publication where they give such example, good luck, l because I could not find even one!
https://projecteuclid.org/download/pdfview_1/euclid.ndjfl/1193667707
[If the link doesn't work, do Internet search for 'richard kaye hereditarily finite' and it will be the first result.]
Note that the article warns against overlooking certain technicalities, as doing so leads to a merely "folktale" understanding. Indeed, the subject of interpretation of theories into another, and especially the exact sense of "equivalence", especially with PA and HF, is a lot of technical details that are doomed to be mangled in the confines of short posts.
Ok. In general terms, I noticed you didn't engage with my point that "N is a proper class in PA" is a personal visualization that I find helpful; even though it is not literally true and, according to your thinking, is so flagrantly false that I shouldn't even think it.
If you didn't engage with this point I assume you accept it and are just explaining some of the technical points I brought up, which I appreciate.
Quoting GrandMinnow
Right, perfectly well understood. But why don't they do the same thing with ZF-infinity +(not-infinity)? Nobody every does that. Rather, ZF-infinity means ZF plus the negation of the axiom of infinity by default. Why is that?
Quoting GrandMinnow
Oh ok it's all the finite sets. I think I knew that then forgot. All good.
Quoting GrandMinnow
Right.
Quoting GrandMinnow
I didn't follow all that but it's ok, I'll check out the def one of these days. Especially since bi-interpretability came up the other day. PA is bi-interpretable with ZF-infinity + (not-infinity) but apparently I'm not getting the subtleties of that statement.
Quoting GrandMinnow
Aiiiyyyy now I'm confused. You just explained to me that HF are all the finite sets in ZF. So if I'm understanding you, HF is the inductive definition of the collection of all finite sets in ZF; and that collection of finite sets is a model of HF. But it's wrong to also call that HF?
Quoting GrandMinnow
So your argument comes down to saying that since N is not a definable symbol in PA, I can't say "N is a proper class" because I have no idea what N is. Is that right?
Quoting GrandMinnow
Now if by PA we mean what is described by Wikipedia, then I am frankly right and you are wrong. So it must be that by PA you mean something else. Is that the source of our misunderstanding? I will stipulate that they shouldn't have used the word set, but rather collection. Or (ahem) proper class!
Quoting GrandMinnow
Ah this is what I meant a moment ago. You don't think PA is what is described by Wiki, and I and probably millions of other people are confused about this. Is that what you are saying?
Quoting GrandMinnow
I'm terribly sorry, of course I know better. See how you have me confused!! I meant that the collection of all sets is a proper class. Just misspoke myself.
Quoting GrandMinnow
It's still a good conceptual metaphor IF not literally true. But again, according to Wiki your point is not so clear. So you must be objecting to the common Wiki understanding of PA.
Quoting GrandMinnow
There is no SET such as that because first, PA doesn't have sets, and second, even if it did that would not be a valid specification of a set since it violates the axiom schema of specification.
But "is a natural number" must be a predicate, what else can it be? It's a thing that classifies all the objects in the universe into "yes it's one of these" and "no it's not." So the number 3 is a number and a tunafish sandwich is not a number.
Therefore I can form the COLLECTION, or "predicate satisfier," or as it's officially called the extension of the predicate, N = {x | x is a natural number}. N is a class and it's not a set. So it's a proper class.
Quoting GrandMinnow
But it makes perfect sense to me! And you haven't explained what "is a number" could be, if not a predicate. And for any predicate whatever you can always form its extension consisting of all and only those things that satisfy it. Which, if it's a set, is a set; and if it isn't, is a proper class.
tl;dr: By PA do you mean something other than what's called PA on Wikipedia? That's the only way your post makes sense.
Quoting GrandMinnow
You say that like it's a bad thing! LOL.
But there is also a relative sense. Perhaps we can say that a collection is a proper class relative to some theory T if: (i) there is a predicate P such that x belongs to the class iff P(x), (ii) T proves that there is an x such that P(x) but (iii) T proves that there is no y such that x belongs to y iff P(x). For example, in ZFC, the class of all sets is a proper class, since x=x is a predicate satisfied by every set, but ZFC proves that there is no y such that x belongs to y iff x=x. Whether a collection is a proper class, then, would be relative to the strength of the theory. Perhaps a weaker theory (say, Kripke-Platek) would consider a proper class something that some other theory (say, ZFC) would consider a set.
Unfortunately, it seems that even in this relative sense, the natural numbers are not a proper class relative to PA. Now, there seems to be some confusion here about whether to take first-order or second-order PA. Notice that, when people take about PA, they generally mean first-order PA. In any case, that's the theory that is bi-interpretable with ZF-Inf (to save me from typing ZF-Inf+~Inf all the time, let's just adopt the convention that ZF-Inf means the latter. I'll rarely make mention of this theory anyway, so no confusion will ensue). Moreover, second-order PA can't be at issue, because second-order PA proves that there is an infinite set, namely the set of all natural numbers, so the natural numbers cannot be a proper class relative to this theory. What about first-order PA?
Here, the problem is that there is no predicate N such that N(x) iff x is a natural number expressible in PA. For suppose there is. Add a constant c to the language and consider the set of all formulas An:= n
The computer has to solve logic, this logic, N, is opposite to P which is the objective. If P is placed before N, we commit a fallacy and our stance is as logic is already countable; we have frozen time, we expect that P = NP when NP is 2NP in regards N requiring it's own formation.
We assume N has no answer, to say P = NP. What is N in NP if not 2NP?
More properly phrased P = N(P)? Yes it does, P is the solving of a problem, via 'calculator' N, but whether the problem can be solved 'quickly' is innacurate, it's more fluency, it could take days, it depends on N. P = NP is a fallacy.
Read further down in the Wikipedia article, and you will see the axioms for first order PA. There is no predicate 'is a number'.
/
Regarding your notion about improper sets relative to PA as personal visualization, I didn't ignore it - you even quoted me remarking on it. I said I don't opine as to what does or does not make sense in your mind. But I said your notion makes no sense to me. And I would add that I think it does muddle discussion. But I didn't say you shouldn't think it.
Quoting fishfry
Some people use it that way. And I have seen it lead to misunderstandings as casual readers fail to note that, in this context, we need the negation of the axiom of infinity and not just leaving out the axiom of infinity.
Quoting fishfry
No, not all the finite sets. Only the hereditarily finite sets.
Quoting fishfry
I defined 'HF' to stand for the theory (ZF~Inf)+~Inf.
But now I realize that writers often use 'HF' to stand for a class. So my choice to use 'HF' as the abbreviation was not good. From now on, I won't use it to stand for the theory (ZF~Inf)+~Inf. Instead I'll use:
TF = (ZF~Inf)+~Inf
Quoting fishfry
To be more precise, whatever symbol 's' we pick, TF does not support a definition:
s = {x | x is a natural number}
because the theory does not prove that there is a such an object.
Quoting fishfry
In class theory, it is well understood that a proper class is a class that is not a member of any class. All I'm doing is pointing out that we can also say that in set theory and conclude in set theory that there are no proper classes. It might be annoying, because it's not a very useful series of formulations. But it its technically correct, and I find that it sharpens the picture. Especially it goes against a common misconception that we can define a predicate symbol only to stand for a relation (sets are 1-place relations) that has members. No, we can always define an empty predicate. For example:
dfn: Jx <-> (x is odd and x is even)
is allowable, even if rather pointless.
Quoting fishfry
Was a typo of omission; I meant {w, w+1, w+2 ...}
Quoting fishfry
Yes, we can have a predicate 'is a natural number' in TF. And upon an interpretation of the language, it has an extension (a subset of the universe for the model) and that extension is a set, not a proper class.
Hard to discuss a counterfactual here.
So let's turn to TF.
It's not a matter of being consistent with the axiom schema of specification.
Instead, in the absence of the axiom of infinity, we do not have a supporting existence theorem for a definition:
N = {x | x is a natural number}
I don't begrudge anyone from that notion, but, for me, it's too vague. What is "too big"? And which universe?
Quoting Nagase
I might suggest saying (only a subtle difference with yours):
For a theory T with 'e' in the language, we say a formula F(x) in the language for T "invokes" a proper class relative to T
iff
(i) T proves Ex F(x). (ii) T proves ~EyAx(F(x) -> xey).
So there is no proper class mentioned there. Instead there is a two place relation between a theory T and a formula F. Okay, fair enough, as maybe that's what fishfry has in mind. And it does work where T is set theory itself, indeed as it is common even in informally stating such things as the axiom schema of replacement where we refer to a 'function class'.
But I warn against thinking conflating this with the one place predicate 'is a proper class', except quite informally.
But worse, this manifestly clashes with ordinary terminology.
Let T = NBG (I call it 'BC' for Bernays class theory). For example:
BC proves Ex x is an ordinal. But BC does not prove ~EyAx(x is an ordinal -> xey), indeed BC proves the negation of that. But we do say, with BC, that the class of ordinals is a proper class.
Quoting Nagase
Yep.
No. I was giving practical advice to not overlook that when we read natural language renderings of formulas, then we can't expect that how we naturally take such locutions in English is preserved with every interpretation (model) for the formal language.
Yes ok. Agreed.
That made me chuckle.
Quoting GrandMinnow
Yes ok I get that you assert that.
But when Wikipedia says that the first axiom of PA is, "0 is a natural number," are you asserting that "is a number" is something other than a predicate?
My chain of reasoning is:
P1: I take Wiki's account of PA as reasonably accurate (whereas I suspect that perhaps you don't);
P2: Wiki says that the first axiom of PA is "0 is a natural number";
C: If "is a number" is not a predicate, what the heck is it? In which case N = { x : x is a natural number} is the extension, or "predicate satisfier" as I think of it, of the predicate "is a natural number." It's not a set. What is a predicate satisfier that is not a set? It's a proper class.
Can you please tell me where you disagree?
It goes back to Frege. A predicate defines a class. Of course he thought that a predicate defined a set, which was falsified by Russell. But a predicate still defines a class.
This is how I understand what's going on.
I wasn't even aware of this before opening the brochure I got in the mail today from the Princeton Press.
:worry:
For more on the limitation of size idea, I strongly recommend reading the summary in Incurvati's book. Still, here are two ways of making the idea more precise: (i) A collection A is too big iff there is an injection from the class of all ordinals into A (Cantor); (ii) A collection A is too big iff it is in bijection with the collection determined by the formula x=x (von Neumann).
On NBG, I thought it proved that there is no set of all ordinals? If so, then it does not prove the negation of your formula, since typically we would employ (à la Bernays) two sorts of variables, one for classes, one for sets, and lower case letters would correspond to sets... or else we could explicitly have two predicates, say M(x) for x is a set and C(x) for x is a class, and then adapt the formula.
We should not overlook that ‘class’ does not mean just ‘proper class’. Some classes are sets and other classes are proper classes. Everything in NBG is a class.
Yes, NBG does not prove that there is a set of all the ordinals. To be clear, with NBG we don’t have variables for just proper classes. With the multi-sorted version we have general variables for all objects - all classes, i.e sets and proper classes. And we have another special variable for just sets. Where ‘x’ and ‘y’ are general variables (class variables) we have the theorem
EyAx(x is an ordinal -> x e y)
And where ‘s’ is a set variable, we have the theorem
~EsAx(x is an ordinal -> x e s)
As you mentioned, having two kinds of variables is equivalent to having a primitive predicate M for ‘is a set’ and relativizing. (But a primitive predicate C for ‘is a class’ turns out to be the universal predicate: Cx <-> x=x.) So the second theorem is
~Ey(My & Ax(x is an ordinal -> x e y))
But the first theorem itself shows that your formulation doesn’t work unless you revise it in some way, perhaps with such relativization, and we would have to see exactly what that formulation would be.
I would add a quantity precisely defined to every other quantity (with that we exclude the problem of quantity defined being a "heap" of something, for example).
That would be a great definition, but it ain't.
We don't allow the infinitesimal or infinity to be numbers. :roll:
Numbers are not the only names for quantities. They are names for quantities nonetheless. So... that criticism misses the mark.
If you consider NBG as a two-sorted theory, and take the lower case variables to range over sets, then there's no need to revise my statement, since it employed only lower case variables...
I'll use M, because it stands out better.
So:
For a theory T with 'e' in the language, a formula F(x) in the language for T "invokes" a proper class relative to T
iff
(i) T proves Ex F(x). (ii) T proves ~Ey(Mx & Ax(F(x) -> xey)).
But that depends on the theory having M as primitive or defined (or being in a 2-sorted logic) And M not to be just an arbitrary predicate, I guess the theory has to prove Ax(Mx <-> Ey xey).
So:
For a theory T with 'e' in the language, and having a 1-place predicate symbol 'M' as mentioned below, a formula F(x) in the language for T "invokes" a proper class relative to T
iff
(i) T proves Ex F(x). (ii) T proves ~Ey(Mx & Ax(F(x) -> xey)). (iii) T proves Ax(Mx <-> Ey xey).
This is not a problem for any language with 'e', as such an 'M' can be defined if not primitive.
So 'x is an ordinal' invokes a proper class relative to NBG.
But PA is not eligible for applying this definition, because PA does not have 'e'. So what about (ZF-Inf)+~Inf as a surrogate for PA?
We want to see whether 'x is a natural number' invokes a proper class relative to (ZF-Inf)+~Inf.
I guess it works. (?)
If you mean the part where it lists the axioms only referring to the successor function [math]S[/math] and the symbol [math]\in[/math], why can't I define
[math]\mathbb N = \{x : x = 0 \lor \exists y (x = S(y))\}[/math]?
Quoting GrandMinnow
Yes I think that appeared in your post after I wrote that. But I really don't get why you think the idea is so terribly wrong.
Quoting GrandMinnow
I'm still confused on this point, since nobody ever describes ZF as ZF-C + (not-C). I sort of see your point that "ZF" is being noncommittal on the point, but generally from context you can tell when they explicitly mean ZF-C + (not-C). So perhaps people should be more formal about that.
Quoting GrandMinnow
As above, what of
[math]\mathbb N = \{x : x = 0 \lor \exists y (x = S(y))\}[/math]?
Surely I can form that class directly from the symbols 0 and S referenced in the PA axioms.
Quoting GrandMinnow
We're agreed on all this but I don't see how it invalidates my point. But I'll concede that if I understood you better you might well be right.
Quoting GrandMinnow
Yes sorry, obvious in retrospect but now I don't remember what point was being made ...
Quoting GrandMinnow
You're saying that my definition of [math]\mathbb N[/math] is a set? I confess I don't follow your reasoning but I admit I don't necessarily disagree with it. I just don't follow it. Maybe I haven't the background to follow your argument. I don't see now this could be a set in TF.
Quoting GrandMinnow
Right, but I gave the correct formula based on the formal axioms you pointed me to in the Wiki piece. We can't form a set with that definition but it's certainly the extension of a predicate. And if it's that, and it's not a set ... that's what I call a proper class. Maybe the people who told me that were speaking more loosely than I realized, that would be your point.
No the part that has the headline:
First-order theory of arithmetic
That is first order PA. (In this context, By ‘PA’ we mean first order PA.)
And PA does not have set abstraction notation.
That is not ZF.
I’ve never seen that theory discussed (though maybe it comes up somewhere.)
I take ‘-‘ to mean ‘without’ and ‘+’ to mean ‘also with’.
So ZF-Inf is ZF but without the axiom of infinity.
And (ZF-Inf)+~Inf is ZF without the axiom infinity and also with the negation of the axiom of infinity.
Two very different theories.
TF proves there is no such set. But meanwhile set theory proves there is that set. The set is the universe for a model of TF. The set itself is not a member of that universe.
You're agreeing with me. You simply haven't made your case IMO. If it's the extension of a predicate and it's not a set in TF then it's a proper class. It may well be (and of course is) a set in some more powerful theory such as ZF.