And that quote is not at all tantamount to saying that we take as existing all the sets that are proven to exist according to different set theories. As it stands in and of itself, it could be mean the exact opposite - that there is no single universe that determines the totality of the sets. That what exists depends on each individual theory. He says there is are distinct concepts; yet your notion is that there is a unified concept that is arrived upon by collecting from all the distinct concepts, or from the union of what is proven among an uncountable number of theories. It remains to study Hamkins further to see exactly how his notion works, but at least from that passage, we cannot infer that it works as you claim it does.

But the more basic point is that, no matter your own views (or even Hamkins's, for that matter), it is not the case that "according to set theory, all logically possible (consistent) collections exist".

Set theory does not say that. There is no theorem of set theory that is anything like that. It's not even seen how it could be formulated in set theory.

Again, you conflate your own personal preference about how you wish to (mis)understand set theory with set theory itself. For that matter, I doubt you even know what set theory is.

And please don't dodge. Please say exactly what passages in part 7 you regard as saying that there is a set with cardinality between the naturals that is decided ('settled' in context) in general (not just per particular theories) by the consistency of ZFC+~CH.

And no retraction from you that you falsely put words in my mouth by claiming that I said naive set theory must obey the axiom of regularity. That you falsely put words in my mouth and then elected not to retract when it was pointed out witnesses intellectual dishonesty and poor faith.

And no admission by you that you omitted the key passages in the Wikipedia article that you cited yourself, as those passages directly support my point that naive set theory is commonly understood to be an informal framework that is informally inconsistent (and formally inconsistent if we formally spelled out the comprehension principle). Again, that's witness of your lack of sincerity to understand the subject matter on which you so freely opine and claim.

And no recognition by you that you missed my point, said more than once, about ZFC+~CH.

First, you falsely put words in my mouth:

There is no requirement that a set cannot be a member of itself or that a set must have a minimal member, which you tried to impose on naive set theory. — litewave

Contrary to your mischaracterization of my remarks, I didn't opine as to naive set theory regarding the axiom of regularity, but rather as to set theory:

set theory does preclude certain kinds of sets that otherwise it would be consistent to say they exist. In particular, the axiom of regularity precludes certain kinds of sets that otherwise would be consistent to say they exist. — TonesInDeepFreeze

And that was before you ever mentioned 'naive set theory'.

In a following post, I said:

I'll add: I surmise that naively (informally, intuitively) most set theorists' notion of 'set' includes that sets are not members of themselves, and that, more generally, every set has a minimal member. — TonesInDeepFreeze

And that is true. I don't impose any particular axioms or principles onto naive set theory, since the rubric 'naive set theory' is not definite enough to say what its axioms (if any) are. Whether described as 'naively', 'informally' or 'intuitvely' (let alone formally), the vast number of set theorists regard sets as not being members of themselves.

I said what obtains in set theory and naively, informally, or intuitively with most mathematicians; I didn't say what does or does not obtain in an undefined 'naive set theory'. For that matter, I don't have any need to even mention 'naive set theory' other than that you brought it up.

So, please do not put words in my mouth again.

Wikipedia article on naive set theory just mentions the general concept of a set as a collection of objects, and related general concepts like set membership relation, equality, subset, union, intersection etc. — litewave

First, never take Wikipedia as authoritative on mathematics.

Second, please read the very article you cite. Wikipedia mentions three notions of 'naive set theory':

"Naive set theory may refer to several very distinct notions. It may refer to

Informal presentation of an axiomatic set theory, e.g. as in Naive Set Theory by Paul Halmos.

Early or later versions of Georg Cantor's theory and other informal systems.

Decidedly inconsistent theories (whether axiomatic or not), such as a theory of Gottlob Frege[6] that yielded Russell's paradox, and theories of Giuseppe Peano[7] and Richard Dedekind." [Wikipedia]

So, even according to Wikipedia, it is not a given that we take 'naive set theory' as you stipulate.

Look at Hamkins' paper — litewave

That is a dense article. (And I certainly don't trust that you have sufficient technical background to accurately represent anything it says.) So, please quote the specific passages you contend claim that all sets exist that are "selected" by at least one consistent axiomatic set theory. Tell me the exact formulations you have in mind that Hamkins mentions in his own words. And please cite where Hamkins says there is a set with cardinality between the naturals and the reals and that that is decided by the fact that ZFC+~CH is consistent.

ZFC+CH is not "ruled out", it just defines a part of the multiverse, a part in which there is no set with cardinality between naturals and reals. — litewave

You switched from my point:

since ZFC+CH is now ruled out by your requirement that there exists a set of cardinality strictly between the naturals and the reals, your vague L-theory cannot speak for set theory itself, since set theory itself is not settled as to CH. — TonesInDeepFreeze

The point there, as I mentioned previously, yet you still don't get it, is that you can't speak for "set theory" when set theory in and of itself does not determine CH.

Rather than address refutations head on, and then gracefully admit that there are the errors in your posts I've mentioned, you instead put words in my mouth, cite a complicated and highly technical article (of which I see no reason to think you have even the basics for a background to understand the article), and then also misconstrue my point about CH.

Keeping a running tab (otherwise, the discussion does not progress healthfully as instead the items get buried by your continuing to ignore them or evade them:

You've been corrected on a number of points: What set theory is. What naive set theory is (as you didn't even read the very article about it that you cited). That your own personal glib hand waving is a sufficiently definite notion - mathematically or philosophically. That you conflate your own personal glib hand waving with what set theory and naive set theory actually are. That your own personal glib hand waving doesn't work out the way you think it does. What a definition of a set is as opposed to a definition of a property of sets. And I wonder from what exact specific passage you infer that Hamkins holds that all sets exist that are "selected" by at least one consistent axiomatic set theory. And now, falsely putting words in my mouth about regularity, and evading my point about CH by misconstruing it. And you present no even remotely definite sense of what you mean by 'exist' - whether by syntactical definition, or by membership in a universe for a model, or otherwise. And you show no recognition of the distinction between defining a particular set versus a claim that there exist sets having a certain property - which is a distinction crucial to allowing this subject to be discussed intelligibly.

No one could predict that by "set theory" you meant your own personal concept (a concept that is not at all what people ordinarily mean by "set theory"). So my original point stands, per an ordinary understanding of "set theory", it is not the case that "according to set theory, all logically possible (consistent) collections exist", as well as we don't have a suggestion as to how that would be rigorous mathematical or even philosophical statement.

But now we have a revision: According to littwave's ersatz concept of set theory (not one ordinarily understood as "set theory"), call it 'L-theory', all logically possible (consistent) collections exist (setting aside how that would be a rigorous or mathematical or even philosophical statement).

I mean set theory in the most general sense. I supposed that this is what is commonly understood as naive set theory, but to clarify, I mean the concept of a set or collection of objects that is elaborated in all consistent axiomatized set theories together. — litewave

And that is not what anyone means by 'naive set theory'. So your notion is not set theory and it's not naive set theory. I'll call it 'L-theory'. Moreover, you skipped my point that you enlist a multiverse view, yet a multiverse view clearly contradicts L-theory, as indeed a multiverse view is the very opposite of L-theory.

In other words: if a set is included at least in one consistent axiomatized set theory, then such a set exists. — litewave

What are all the axiomatized set theories? There is no definitive list, and there is no conceptual limit. For that reason alone your notion is fatally vague. And what does "included" mean? Does it mean that a theory proves a theorem of the form E!xAy(yex <-> P) so that we can name that unique set? That would entail that there are only countably many sets (since there are only countably many names). Otherwise, in what sense does a set "exist" in your framework? Exist as a member of some model of some set theory (as, again, it is not known what theories are allowed to be considered for this purpose)? Or does it mean that there is a proof that there are certain classes of of sets having a certain properties? All and only properties expressible as a formula of set theory? Or sets that are subsets of model theoretic universes? Or what? 'exist' has at least two set theoretic meanings: To be named by a constant symbol in a definition or to be a member of a universe of a model of set theory.

Without providing something a lot more mathematically or philosophically substantive than you have, your notion is arm waving in a "dorm room".

So I include also sets that are members of themselves and sets that don't have a minimal member — litewave

And that is not set theory. And, for consistency then, you have to exclude the ordinary set theories, including ZFC, since they prove that there are no such sets. Your notion is not set theory; it is, at best, L-theory.

Also, how would you know that such a stew that provides that "all sets exist if they exist in at least one theory" is itself consistent?

you stated that there is no such thing as an inconsistent definition, so let me give you an example of an inconsistently defined set: an empty set that has one member. — litewave

That's not a definition of a set. That's a definition of a property, viz. the property of having no members and having a member.

A definition of a set is of the form (where 'c' here is a constant, so it's actually a definition of a constant):

x = c <-> P

where 'P' is a formula in which 'c' does not occur, and at most 'x' occurs free, and we have a previous proof of E!xP.

A definition of a property (actually a predicate symbol) is of the form:

Fx <-> P

where 'F' is a predicate symbol, and 'P' is formula in which 'F' does not occur, and at most 'x' occurs free.

Moreover, any theory that has the axiom of regularity makes inconsistent this formula.

Ex xex

since the axiom of regularity implies ~Ex xex.

And, since you place no limit on how we may extend any of the set theories, we can have many theories that preclude that there exist sets of certain kinds, including sets with cardinality strictly between the cardinality of the naturals and the cardinalities of the reals, or inaccessible cardinals, etc. Indeed, we can have consistent theories that, by dropping some axioms and adopting others, consistently preclude any kind of set we want.

Your thinking about all of this is thoroughly half-baked.

A set theory with CH as an axiom simply selects only certain sets among which a set with a cardinality between naturals and reals is not included. — litewave

No, you skipped what I pointed out:

A set theory (a) proves the existence of certain sets, and certain kinds of sets, having certain properties, and (b) disproves the existence of certain sets, and certain kinds of sets, having certain properties, and (c) for certain kinds of sets, leaves neither proven or disproven that they exist. So, even the most common set theories preclude the existence of certain sets and leave unanswered whether other certain kinds of sets exist. So, again, it is not the case that "according to set theory, all logically possible (consistent) collections exist" (let alone, as mentioned, it is not clear how "all logically possible (consistent) collections exist" could even be exactly stated in the language of set theory or even as a rigorous philosophical claim). — TonesInDeepFreeze

A theory does not "select only certain sets". There are properties such that a consistent set theory leaves unanswered whether or not there are sets having that property. So it is utterly vague to say what sets a given theory "selects" beyond those we know that the theory proves to exist. And even worse to hand wave that there is some universal criterion of existence based on a union of an undetermined set of theories. And again, since ZFC+CH is now ruled out by your requirement that there exists a set of cardinality strictly between the naturals and the reals, your vague L-theory cannot speak for set theory itself, since set theory itself is not settled as to CH.

I mean sets or collections selected by all consistent axiomatized set theories together - that's the multiverse view in set theory. — litewave

I gather that you mean "in at least one" and not "selected by all" [emphasis original]. Using 'all' as you do confuses your own point.

But more importantly: Please cite where Hamkins says anything tantamount to "[all sets exist that are] selected by [at least one] consistent axiomatized set theory."
/

You've been corrected on a number of points: What set theory is. What naive set theory is. That L-theory is a sufficiently definite notion - mathematically or philosophically. That L-theory doesn't work out the way you think it does. What a definition of a set is as opposed to a definition of a property of sets. And I wonder from what exact specific passage you infer that Hamkins holds that all sets exist that are "selected" by at least one consistent axiomatic set theory.

/

Every consistent first order theory is extended by a consistent maximal first order theory, in the sense that every sentence in the language is a theorem of the maximal theory or its negation is a theorem of the maximal theory. But it's doubtful whether that would help L-theory: (1) The maximal theory is taken as an infinite union of theories with a sequence, as at each step the consistent alternative is included, but if both are consistent, then either (but not both, of course) may be included. This is nothing like the handwaving of L-theory. (2) From (1), there is not just one maximal theory. Not a definitive union of theories by which it is determined what sets exist or not. (3) The maximal theory is not necessarily recursively axiomatizable, so, to the extent that L-theory might be hoped to provide a formal theory, the theorem of maximal theories does not in and of itself provide a recursively axiomatizable theory.

Your claim was:

according to set theory, all logically possible (consistent) collections exist — litewave

I refuted that claim:

https://thephilosophyforum.com/discussion/comment/766312

Your reply does not refute my refutation, as well as there are other errors in your reply:

First, here are the points in my refutation, most of which you skipped:

(1) It's not even clear how "all logically possible (consistent) collections exist" could be stated as a mathematical statement in set theory.

(2) It's not even clear how "all logically possible (consistent) collections exist" could be stated as a rigorous philosophical principle regarding set theory.

(3) Even if we did have a rigorous statement of such a philosophical principle, it's not a given that it is the consensus of set theorists and philosophers of mathematics that it is true.

(4) There are infinitely many statements formalizable in the language of set theory that state the existence of sets with given properties but such that it is consistent with set theory there exists such a set, but it is not a given that set theorists endorse that any given one of those sets exists.

You did reply to that point, but your reply fails, as I'll explain later in this post.

(5) Set theory does preclude certain kinds of sets that otherwise it would be consistent to say they exist. In particular, the axiom of regularity precludes certain kinds of sets that otherwise would be consistent to say they exist.

Since you did not reply to that, I'll add: I surmise that naively (informally, intuitively) most set theorists' notion of 'set' includes that sets are not members of themselves, and that, more generally, every set has a minimal member. That is especially witnessed as the axiom of regularity is a standard axiom, which is especially relevant since you say that naive set theory is "elaborated upon" by axiomatizations such as ZFC. This is a point blank refutation of your claim that "according to set theory, all logically possible (consistent) collections exist", as indeed both the naive notion of sets and the standard axiomatizations exactly preclude the existence of certain kinds of sets that would not be inconsistent to assert their existence otherwise. That point cannot be skipped and it alone decisively refutes your claim.

Added errors in your reply:

(6) When we say 'set theory' in the last 100 years, we mean one of the axiomatized set theories, not naive set theory. So saying 'according to set theory' would not be understood as 'according to naive set theory'.

(7) 'naive set theory' is ambiguous, as it means different things to different people and in different contexts. In any case, it is not usually understood as "it just says that a set is a collection of objects". And even if that were the meaning of 'naive set theory', it would not follow that according to naive set theory "all logically possible (consistent) collections exist" (whatever exact claim that might be).

The most salient sense of 'naive set theory' is the inclusion of the informal principle that to each property there is the set of all and only the objects that have that property. Or, formally, the axiom schema of comprehension: For every formula P in which 'y' is not free, we have E!yAx(xey <-> P). That schema is famously inconsistent. So taking 'naive set theory' in that sense is of no use to your claim.

And taking 'naive set theory', as you mention, as merely meaning an informal understanding that is nevertheless formalized in a theory such as ZFC also is of no use to your claim, since such theories exactly preclude the existence of certain kinds of sets that would not be inconsistent to assert their existence otherwise.

(8) It is not clear what you intend with "a set is a collection of objects". Is that intended as a definition of 'set'?

As far a quite informal notion, it's perhaps okay though it merely shifts from 'set' to 'collection'. But it is also widely viewed that 'set' is an informal notion that is not defined, especially as you say that naive set theory is explained by axiomatic set theory. Also, formally, in class theory (a variation of set theory), we make take 'is a set' as a primitive (undefined) predicate. But also, even in set theory, we may define 'is a set' as follows:

df. x is a set <-> ((x=0 or Ey yex) & Ez xez)

Or put informally: a set is not a urelement and not a proper class.

Also, the understanding of sets has been refined greatly since early definitions such as Cantor's. Especially we countenance the iterative concept. (For an excellent argument see Boolos's essay "The Iterative Conception of Set" in the great volume 'Logic, Logic, and Logic'.)

Edit: Also, ordinary set theory take sets to be hereditarily sets.

Returning to the main point: Such notions do not entail that all logically possible sets exist. We already saw that the axiom of regularity exactly disputes that all logically possible sets exist, but also it is just a non sequitur to jump from a definition of 'is a set' to asserting the existence of "all logically possible collections". From the definition of 'unicorn' we don't infer that unicorns exist (presumed counterfactual), let alone that all possible unicorns exist. From the definition of 'extraterrestrial creature' we don't infer that extraterrestrial creatures exist (unkown), let alone that all possible extraterrestrial creatures exist. From the definition of 'dog' (known fact but not inferred merely by definition), we don't thereby infer that dogs exist, let alone that all possible dogs exist.

(9) You mention Hamkins's multiverse view. But a multiverse view decidedly contradicts naive set theory (in the sense of the schema of comprehension). As to naive set theory in your sense of informal understanding anticipating formal axiomatization, the multiverse view and your remarks about it actually hurt your claim that "all logically possible collections exist". Indeed, the multiverse notion suits my argument: It depends on what particular theory is considered. For example, if we adopt CH as an axiom, then there does not exist a set whose cardinality is strictly between the cardinality of the set of naturals and the cardinality of the set of reals. But if we adopt the negation of CH as an axiom, then there do exist sets whose cardinality is strictly between the cardinality of the set of naturals and the set of reals. There is not in set theory itself a universal principle that "all logically possible collections exist" (even setting aside, as I've mentioned, that it's not clear how we would rigorously articulate such a principle).

And you say, "every axiomatized set theory selects a limited collection of possible (consistently defined) sets." But that contradicts your own claim that "according to set theory, all logically possible (consistent) collections exist":

A set theory (a) proves the existence of certain sets, and certain kinds of sets, having certain properties, and (b) disproves the existence of certain sets, and certain kinds of sets, having certain properties, and (c) for certain kinds of sets, leaves neither proven or disproven that they exist. So, even the most common set theories preclude the existence of certain sets and leave unanswered whether other certain kinds of sets exist. So, again, it is not the case that "according to set theory, all logically possible (consistent) collections exist" (let alone, as mentioned, it is not clear how "all logically possible (consistent) collections exist" could even be exactly stated in the language of set theory or even as a rigorous philosophical claim).

(10) You say, "For me, as long as such a set is consistently defined, it exists."

First, using the method of formal definition, there is no such thing as an inconsistent definition. (See many a book in mathematical logic for explanation of the method of formal definition, while I think Suppes's 'Introduction To Logic' is the best one on the subject.)

Second, and most telling, that for you something is the case about sets doesn't imply that "according to set theory" it is the case. You overstated. You jumped from your own glib view to a sweeping claim about set theory itself.

/

It is true that set theorists have different perspectives: Some favor a "wider" view of sets; that our theory should allow a more "liberal" acceptance of kinds of sets. And other set theorists favor a "narrower" view of sets. But, again, to understand those perspectives as rigorous requires a lot more work. And, again, since there are such disagreements, it is not the case that "according to set theory, all logically possible (consistent) collections exist".

The axiom of regularity precludes that there exist non-empty sets that don't have a minimal element. Most saliently, the axiom of regularity precludes that there is a set that has itself as a member.

according to set theory, all logically possible (consistent) collections exist — litewave

It is not true that according to set theory all logically possible (consistent) collections exist. First, it's not even clear how that would be stated as a mathematical statement in set theory. Second, it's not even clear how that would be stated as a rigorous philosophical principle regarding set theory. Third, even if we did have in front of us a rigorous statement of such a philosophical principle, it's not a given that it is the consensus of set theorists and philosophers of mathematics that it is true.

Moreover, there are infinitely many statements formalizable in the language of set theory that state the existence of sets with given properties but such that it is consistent with set theory there exists such a set, but it is not a given that set theorists endorse that any given one of those sets exists. For example, it is consistent with set theory that there is a set that has cardinality strictly between the cardinality of the naturals and the cardinality of the reals, but it is not a given that it is the consensus of set theorists and philosophers that such a set exists.

Moreover, even stronger, set theory does preclude certain kinds of sets that otherwise it would be consistent to say they exist. In particular, the axiom of regularity precludes certain kinds of sets that otherwise would be consistent to say they exist.

Also, granted that some writers refer to "consistent collections", but that may cause misunderstanding, since it's not collections of sets that are consistent or not, but rather collections of statements about sets that are consistent or not.

No, it's not like that. You say it is only because you think being a wiseracre suits you.

So we're looking for an infinity bigger than N and smaller than R. — Agent Smith

If you think the continuum hypothesis is false, then you think there is a set with cardinality strictly greater than the cardinality of N and strictly less than the cardinality of R.

For more about the ramifications, you should read more about set theory, to gather the whole context. But perhaps the more pertinent matter is what are the ramifications of added set theoretic principles that would prove CH and of added set theoretic principles that would disprove CH. This is at the heart of contemporary set theory research. I couldn't do the subject justice in posts (and I'm not expert enough anyway).

Yeah, Hilbert enjoined mathematicians to either prove or disprove CH. Godel did half by proving that ZFC doesn't disprove CH and Cohen did the other half by proving that ZFC doesn't prove CH. So a kind of "stalemate": Hilbert's challenge can't be answered on the terms Hilbert had in mind. This is a fascinating, profound and remarkable intellectual turn of events.

No. The way you wrote it is wrong. The continuum hypothesis is that the cardinality of the set of reals is aleph_1. This point keeps getting lost. Don't just take it for granted that the cardinality of the set of reals is aleph_1, when that is the very point that is in question with the continuum hypothesis.

But this part that is suggested (though mangled) by you is correct: The continuum hypothesis is equivalent to the assertion that there is no set that has cardinality strictly greater than the cardinality of the set of naturals and strictly less than the cardinality of the set of reals.

One more time. Here are three equivalent ways of saying the continuum hypothesis [here 'N' stands for the set of naturals]:

(1) aleph_1 = 2^aleph_0

(2) card(R) = aleph_1

(3) There is no set that has cardinality strictly greater than card(N) and strictly less than card(R)

Those are three ways of saying the continuum hypothesis.

If you say card(R) = aleph_1, then you are asserting the continuum hypothesis, which is an assertion that can't be proven in ZFC nor disproven in ZFC.

Or, since this point seems not to be getting through, I'll say it this way:

Don't just assume that card(R) = aleph_1. "Is card(R) = aleph1 ?" is the QUESTION. Don't just assume its answer is 'yes'.

thm: n is a natural number <-> (n is finite & n is an ordinal)

dfn: card(x) = the least ordinal k such that x is 1-1 with k

dfn: c is a cardinal <-> there exists an x such that c = card(x)

thm: n is a natural number <-> (n is finite & n is a cardinal)

thm: x is finite <-> card(x) is a natural number

thm: x is infinite <-> card (x) is infinite

/

So, yes, if x is finite, then its cardinality is a natural number.

But if x if infinite, then its cardinality is not a natural number but rather is an infinite cardinal.

In either case, every set is 1-1 with its cardinality.

/

dfn: aleph_0 = the set of natural numbers

dfn: aleph_1 = the least infinite cardinal that is strictly greater than aleph_0

dfn: R = the set of real numbers

dfn: x is denumerable <-> (x is 1-1 with the set of natural numbers & x is infinite)

thm: card(R) = 2^aleph_0 (in other words, the cardinality of the set of real numbers is the cardinality of the set of denumerable binary sequences)

The great question of set theory: Is the cardinality of the set of real numbers aleph_1? Put another way: Is it the case that there is no set that has a cardinality strictly greater than the set of natural numbers but strictly less than the cardinality of the set of real numbers? Put another way: Is aleph_1 = 2^aleph_0?

The assertion "the cardinality of the set of real numbers is aleph_1" is called 'the continuum hypothesis' ('CH'). Cantor thought CH was true, but he couldn't prove it. Hilbert proposed that finding a proof, one way or the other, was a priority of mathematics. Later, Godel proved that ZFC does not prove the negation of CH, then later Cohen proved that ZFC does not prove CH.

So the great question of set theory does not have an answer per merely ZFC. So mathematicians have been proposing and studying axioms that they consider to be intuitively true and could be added to ZFC to settle CH. Some mathematicians propose axioms that prove CH, and other mathematicians propose axioms that prove the negation of CH. There is no consensus.

So it is silly just to say "the cardinality of the set of real numbers is aleph_1", without citing a context for belief in the assertion, when it is a profound open question.

/

PS: The generalized continuum hypothesis (GCH) is that for any x, aleph_(x+1) = 2^aleph_x.

Of course, GCH implies CH, and the negation of CH implies the negation of GCH. Godel actually proved that ZFC does not prove the negation of GCH. And Cohen, by proving that ZFC does not prove CH perforce proved that ZFC does not prove GCH.

ℵ1 (the reals?) — Agent Smith

I addressed that about half a dozen times in posts above.

The statement "aleph_1 is the cardinality of the set of real numbers" is the continuum hypothesis. It is not derivable in ZFC and its negation is not derivable in ZFC.

Yes, x is a natural number iff x is a finite cardinal.

And aleph_1 is not a finite cardinal.

And the poster is asking about finding a certain natural number, so aleph_1 is not an answer.

You want a greatest practical natural number. aleph_1 is not a natural number; it's an infinite cardinal.

aleph_1 is the least cardinal greater than aleph_0.

That is the case by definition.

"aleph-1. The the infinite cardinal of the real numbers"

That is the continuum hypothesis. The continuum hypothesis is "aleph_1 is the cardinality of the set of real numbers" (or equivalently, "aleph_1 = 2^aleph_0") and it is not a theorem of ZFC nor is its negation a theorem of ZFC.

When you construct "a denumerable binary sequence not in the range of f", aren't you deriving that contradiction? — ssu

When you use reductio ad absurdum, you construct a denumerable binary sequence not in the range of f, which contradicts the assumption that f is a bijection between the set of natural numbers and the set of denumerable binary sequences. But Cantor didn't do it that way.

the set of natural numbers and rational numbers have the same cardinality, there is a direct proof (the set of rational numbers can be well ordered — ssu

The proof doesn't rest on showing a well ordering. Showing a well ordering of the set of rational numbers is not adequate for showing that the set of rational numbers is countable. (Even uncountable ordinals are well ordered.)

Here's the argument, which is not reductio ad absurdum:

Let f be a function from the set of natural numbers to the set of denumerable binary sequences.

Construct a denumerable binary sequence not in the range of f.

Conclude there is no function from the set of natural numbers onto the set of denumerable binary sequences.

/

He also could have used reductio ad absurdum, but he didn't:

Assume there is function from the set of natural numbers onto the set of denumerable binary sequences.

Derive a contradiction.

Conclude there is no function from the set of natural numbers onto the set of denumerable binary sequences.

Cantor proved that you cannot make a bijection between the natural numbers to the reals, hence the [cardinality of the] reals [is not] aleph_0 like for example [the cardinality of the] rational numbers. — ssu

Correct.

When you first assume that there is a bijection between the natural numbers and reals — ssu

Cantor didn't make that assumption.

the "axiom of infinity": For any natural number n, there exists at least one set with n members. — alan1000

That is not the axiom of infinity.

The statement that for any natural number n there is a set with n members is a trivial theorem, not even an axiom.

The axiom of infinity is that there exists a set such that 0 is a member and for any x, if x is a member then xu{x} is a member .That axiom, with other axioms, implies that there exists an infinite set, and implies that there exists a set (which is infinite) whose members are all and only the natural numbers.

my wording might not be rigorous — ssu

Not merely a lack of rigor. Rather, your statement "the continuum hypothesis is that from aleph_0 the next is aleph_1" is plainly false.

the sequence of cardinal numbers: aleph_0. aleph_1, aleph_2, aleph_3 and so on. The question is if this hierarchial system holds — ssu

There is no question of whether it "holds". It doesn't even make sense to say that a sequence holds or not. What hold or not are statements.

if there is a cardinality or not between the naturals or the reals — ssu

Yes, the continuum hypothesis is the claim that there is no cardinality between the set of naturals and the cardinality of the set of reals. That's another way of saying what I've been telling you.

The continuum hypothesis is that the reals is the next aleph, that there isn't anything else. — ssu

The continuum hypothesis is that the cardinality of the set of reals is aleph_1. That is equivalent to saying that there is no uncountable subset of the set of reals that is not 1-1 with the set of reals. Of course, no matter the continuum hypothesis, there are cardinals greater than aleph_1.

the diagonal argument — ssu

The diagonal argument given by Cantor was not a reductio ad absurdum.

The cardinality of the set of real numbers is aleph_1 — TonesInDeepFreeze

By the way, I did not say that the cardinality of the set of real numbers is aleph_1. I said that "The cardinality of the set of real numbers is aleph_1" is the continuum hypothesis.

Correct, alelph_1 is not countable.

It seems that you wish not to recognize that:

aleph-1. The the infinite cardinal of the real numbers — ssu

is the continuum hypothesis

and that

the continuum hypothesis is that from aleph_0 the next is aleph_1 — ssu

is incorrect, since "from aleph_0 the next aleph is alelp_1" is not the continuum hypothesis.

/

what can be put into one-to-one correspondence with the set of natural numbers and the reductio ad absurdum proof that this cannot be done with the set of reals. — ssu

Cantor's proof was not by reductio ad absurdum.

As to the other poster, the current question of the continuum hypothesis does not stem from the definition of 'countable'. Rather, the current question stems from the fact that the continuum hypothesis is independent of ZFC.

Yes, the continuum hypothesis is about the first two infinite cardinals. Meanwhile, what I said stands:

aleph-1. The the infinite cardinal of the real numbers — ssu

That is the continuum hypothesis.

In other words,

The cardinality of the set of real numbers is aleph_1

is the continuum hypothesis.

I think the continuum hypothesis is that from aleph_0 the next is aleph_1. — ssu

That is not the continuum hypothesis.

In other words

aleph_1 is the next aleph after aleph_0

is not the continuum hypothesis.

Actually, I think the continuum hypothesis is that from aleph_0 the next is aleph_1. — ssu

It is correct that the next aleph after aleph_0 is aleph 1. That follows trivially from the definition of the alephs. Since the alephs are indexed by the ordinals, and 1 is the next ordinal after 0, It is trivially the case that the next aleph after aleph_0 is aleph_1, regardless the continuum hypothesis. Meanwhile, actually, it is incorrect that "the next aleph after aleph_0 is aleph 1" is the continuum hypothesis.

The continuum hypothesis is that the cardinality of the set of real numbers is aleph_1. Or, equivalently, that 2^aleph_0 = alpeh_1.

What did I say? There's still something for us to understand with infinity. — ssu

You said there's more for us to understand regarding infinitude. Indeed, whether the cardinality of the set of real numbers is aleph_1 is something more that is not clearly understood. Again, settling that question is to settle the continuum hypothesis.

aleph-1. The the infinite cardinal of the real numbers — ssu

card(reals) = aleph_1 is the continuum hypothesis. It is not provable in ZFC. It is thought to be true by some mathematicians and false by other mathematicians - an unsettled question.

It's not about going into the meanings of "appearance", "filtered way", "unfiltered way", or "our action". Rather, it's about the logical form, no matter the meanings of those are.

Is the relevant Russell set incoherent? — Agent Smith

Your question is incoherent.

No sentence is everywhere true — Sunner

What does that mean?

Do you mean that there is no sentence that is true in all models?

But there are sentences that are true in all models.

every sentence is both true and false — Sunner

No, there are sentences that are true in all models, sentences that are false in all models, and sentences that are true in some models but false in other models. And, for any given model, there is not sentence that is both true and false in that model.

he understood the opposite of the correct answer! — fishfry

You haven't shown that my answer is incorrect. Nor has anyone said what other answer is "the" correct answer.

The problem that deserved an answer (in my words):

Set theory says, "For all sets, it is not the case that it has every set as a member" (i.e. there is no set of all sets). But that refers to all sets. So isn't what is referred to by "all sets" a set that has every set as a member, which contradicts that there is no set of all sets?

My answer is basically: The meaning of "for all" is per models, such that for any given model there is a universe that is a set, but it is not the non-existent set of all sets.

That is perfectly correct and it shows that the seeming contradiction is not actually a contradiction.

To refute my answer would require, refuting at least one of these claims:

(1) "for all" is understood per models. Correct. Just look it up in any textbook in mathematical logic.

(2) The universe for a model is a set. Correct. Just look it up in any textbook in mathematical logic.

(3) The universe for a model is not the non-existent set of all sets. Correct. There is no such set, so no set, including a universe for a model, is the non-existent set of all sets.

I've been feeling the same way about you. — fishfry

I haven't, as you have, gone on and on challenging someone who is making correct mathematical statements. I have no trip, so nothing for you to wonder about.

And your criticism is belied by the fact that the poster himself explicitly said that my answer was clear and helpful, and his followup questions do show that he basically understood my answer.

First you challenged me on substantive points in my posts, and you were incorrect. Most particularly, a universe for a model is a set.

Then you griped that my answer to the poster was too technical and not helpful, which contradicts what the poster himself said.

So, I really don't know what your trip is.

We frequently quantify over proper classes — fishfry

That's your informal understanding. I can't comment with real definiteness, because your informality doesn't provide a clear, definite meaning of 'quantify over'. Meanwhile, I have given the rigorous sense in which I use 'quantify over'.

your responses to the recent OP Sunner were too detailed and technical to be of use at the level the question was being asked. — fishfry

So we've moved from your substantive criticisms, which were incorrect, to a criticism of the pedagogy.

My answer to the poster's question is somewhat (though not terribly) technical to avoid confusions from oversimplification or vague hand waving. Meanwhile, you quoted me about that, why I was as technical as I was, but replied merely to say "too technical". Argument by mere assertion is what that is. Moreover, if a poster lacks familiarity with set theory and mathematical logic, then we can bet that there is no ideal answer for him; no answer that wouldn't be either too vague to be responsible to the mathematics or too technical that he wouldn't quickly understand it. (Though, this particular poster did basically grasp the explanation.) I choose to err on the side of being correct, and then to leave it to the poster to follow up himself by learning more about the subject.

Not sure what contradiction "mathematics indeed does not tolerate." The referent of this paragraph is unclear. — fishfry

It refers to the original question, as I said exactly that's what it refers to - in the very first sentence of the paragraph.

https://math.stackexchange.com/questions/2724236/how-do-i-quantify-over-the-proper-class-of-all-the-cardinal-numbers — fishfry

(1) I don't know exactly which post(s) and passage(s) he is saying "no" to.

(2) I understood everything in that thread prior to his post. And, modulo details of phrasing, I agree with the replies that, in set theory, we simply use the predicate 'is a cardinal'. That is what I said myself in my posts above.

(3) "bounded quantifiers" there is just another way of saying what I said: The antecedent of the conditional stipulates that a set has a certain definable property. And I said (though I don't know whether that poster agrees) that the quantifier itself ranges over the universe, not just those sets having the property specified in the antecedent of the conditional.

So, if I understand that poster, he's mostly in agreement with me, not you.

You had the notion that the quantifier ranges over groups. But the quantifier ranges over the universe.

As I understand him, his remarks mostly coincide with mine:

"quantifiers quantify over everything."

If 'everything' means 'everything in the domain of discourse' then I agree.

"All sets."

I don't know what 'all sets' means. But if it means 'all sets in the universe', then I agree.

"Bounded quantifiers are shorthands to make it clearer that we are only interested in a particular set."

That seems to be what I'm saying about the antecedent of the conditional.

"But you can bound them to a class just as easily."

I don't know what he means by 'bound them to a class'. But, as I keep saying, we can relativize (use a conditional) to a proper class through a defined unary predicate. Also, another poster there suggested using a defining formula, which is tantamount to what I've said.

my usage is informal. — fishfry

And my rigorous, mathematical and standard use and explanations are not refuted (or whatever your disagreement is supposed to be) by your own informal usage.

The original question was informal. The original question was in invitation to explain a seeming contradiction. That merits a response that is rigorous and definitive, in order to appreciate that mathematics indeed does not tolerate such a contradiction, not just by informal hand waving, so that when we look at the matter with exactness, we do show that the seeming contradiction actually is avoided.

But there is no set of all groups! — fishfry

Why are you exclaiming that to me? Of course I agree.

The class of groups is a proper class. So you seem to be conceding my point. — fishfry

There is no class of all groups in set theory. Set theory has only classes that are sets. In class theory, there is the class of all groups.

To say I "concede" that is like saying I concede that 0+0=0.

In any event, much of the rest of your post is pretty technical and I'm not sure how it bears on the question. — fishfry

My post is nothing that isn't in, or discernible from, introductory mathematical logic.

And it bears on the questions in exactly the way I explained.

/

I can't comment on that quote without a link to it.

Except, as you've presented it alone, "quantifiers quantify over everything", I say:

Please rigorously, mathematically define "quantifier over" and "everything".

I mentioned a rigorous, mathematical usage. My posts are correct in that context. And that context is definitely relevant to addressing the original question in a rigorous, mathematical way.

Anyway, if we have a model of class theory (or even of set theory for that matter), then, yes, the sentences "Ax x is a class" or "Ax(x is a set or x is a proper class)" are true in the model. But, ironic though it may be, a model of class theory has a set as its domain of discourse. Class theory is a first order theory; and a model of a first order theory has a set as its universe.

As far as whether a domain of discourse must necessarily be a set, I don't see how that can be possible. When we make general statements about sets or groups or cardinals, we are quantifying over proper classes. — fishfry

I explained that exactly in my post.

Moreover, in an earlier post, I proved that allowing a universe for a model to be a proper class implies a contradiction. You can go back to look at that.

Whether you call that a domain of discourse or not seems like a question of semantics. — fishfry

I guess you mean "question of semantics" in the sense of how we use words.

I don't presume to say what people may mean by 'domain of discourse'. I just gave an answer to the original question by using 'domain of discourse' in the sense of a universe for a model. The poster himself didn't use 'domain of discourse'. Rather, I first used it, and in the sense of rigorous mathematics as developed in mathematics. If someone else means something else by 'domain' of discourse' or 'quantify over' then of course I can't ensure that my use agrees, while meanwhile I would say, "Then what are your rigorous, mathematical definitions of 'domain of discourse' and 'quantifies over'?"

For example, in set theory, we have a defined predicate 'is an natural number' (I'll use 'B'). So theorems such as:

Ax(Bx -> (x is even or x is odd))

And in the intended interpretation, 'B' maps to the set of natural numbers that is a subset of the universe for the model. But for another model, it might be a different universe, and 'B' might map to a different set from the set of natural numbers.

Or consider first order PA.

In the intended interpretation, the universe is the set of natural numbers, so the quantifier ranges over the natural numbers. But there are non-standard models (not just for the language but even of PA), so the quantifier ranges over a set very different from the set of natural numbers.

It is only by per a model that the domain of discourse is definite, and so it is only per a model that it is definite what the quantifier ranges over.

We're not talking about models here, though. We're talking about domains of discourse. — fishfry

model theory is not relevant to this conversation as I understand it. — fishfry

The question was:

wouldn’t the statement or conclusion «there is no set of all sets» be all-inclusive in one way or another if it really is true? — Sunner

"all inclusive in one way or another" is not definite. My choice was to give a mathematically definite framework for the the question. I surely do not presume to guarantee that I know what the poster in particular had in mind, but I responded correctly vis-a-vis the best way I know to make the question mathematically definite.

in general, there is an operation of unrestricted complement. — fishfry

Not in set theory (as I see you agree). And (just to be clear) in class theory, only with sets, not proper classes.

The complement of the set {1,2,3}, within set theory, is the collection of all sets that are not {1,2,3}. That complement is a well-defined collection, but it's not a set. — fishfry

It is a proper class (as you mention also) in class theory. In set theory, it doesn't exist. Clearly, the context of my posts was set theory.

I looked up Domain of Discourse on Wikipedia, and they did indeed say a domain of discourse is a set. I assumed they were mistaken, and were simply using "set" in its everyday, casual meaning, without regard for the issues of set-hood versus proper classes.

So I agree that if I looked it up, I'd find at least one source, namely Wiki, that claims a domain is a set. I just think they're wrong, and gave many examples to show why. — fishfry

Wikipedia is unreliable for mathematics (and other subjects). But it happens to be correct on this matter. And you don't need to wonder whether it's just a fluke of Wikipedia. Look in any textbook in mathematical logic or model theory. Or any PDF book or class notes on the Internet. The universe (aka 'the domain of discourse') for a model is a set.

any model of set theory has as a set, not a proper class, as its domain of discourse. For any model, the universal quantifier ranges over the members of the domain of discourse, which is a set.
— TonesInDeepFreeze

You are saying that when I make a statement such as, "Every set has a powerset," I am really saying:

1. I assume ZF is consistent.

2. By Gödel's completeness theorem, if ZF is consistent it has a model, which is a set.

3. The powerset axiom is implicitly quantifying over that set. — fishfry

No, I am definitely not saying that.

Phrases such as "quantifier ranges over" are not definite. To pin them down to a definite mathematical formulation, we turn to the mathematical logic. The method of models gives us definite formulations. A quantifier ranges over a universe. But what is that universe? Well, a universe is the carrier set for a model. So, per any given model, the quantifier ranges over the universe of that model.

In what I said, there is no need for an assumption that set theory has a model or is consistent. There is a crucial between (1) a model for the language of a theory and (2) a model of the theory:

(1) M is a model for a language L iff [fill in the definition here, in outline: M is a non-empty set together with a mapping of the non-logical symbols to (elements of the universe, n-ary functions on the universe, and n-rary relations on the universe)]

(2) M is a model of a theory T iff (M is a model for the language of T & every theorem of T is true in M)

Of course, if a theory is inconsistent, then there are no models of that theory. But whether or not a theory is consistent, there are models for the language of the theory.

To mention that a quantifier ranges over a universe per a model requires only (1) and no consideration whether any given model is or is not a model of the theory.

set theory can not prove its own consistency. — fishfry

Yes, if set theory is consistent, then set theory does not prove that set theory is consistent. Nothing I've said contradicts that.

The claim that every set has a powerset is true whether or not set theory has a model. All that is required is the axiom of powersets. — fishfry

No, a sentence is true or not depending on what model we're looking at. Truth is defined per models. The power set axiom is true in some models but false in other models (as you agree). It is true in any model of set theory (since set theory includes the power set axiom) but it is false in other models (ones that are not models of set theory).

Indeed, "Every set has a powerset" is NOT a semantic claim; it's a syntactic one. It follows from the axiom of powersets. There are models lacking the axiom of powersets where the claim is false. — fishfry

A sentence is an uninterpreted syntactical object. Aside from what the word 'claim' means, a sentence is interpreted per models.

And the phrase "models lacking the axiom of powersets" doesn't make sense. Models don't have axioms. Rather, axiomatizations of theories have axioms. What you might mean is "models in which the power set axiom is false".

And "every set has a powerset" is just an English way of saying the power set axiom.

Perhaps we're arguing about syntactic versus semantic domains. — fishfry

I have never read of a "syntactic domain". I don't know what you mean by it.

On the other hand, we can always define unary predicate symbols. For example, in set theory, we can have the predicate symbol 'G' defined by

Gx <-> [fill in the requirements for x being a group]

Then we have universally quantified conditionals:

Ax(Gx -> P)

Note the quantifier ranges over the universe, but it happens that the formula it applies to is a conditional in which x being a group is the antecedent.

So that is a relativization of P to groups.(That's a simplification. Relativizations are recursive so that P and its subformulas are themselves relativized.]

For example, we define a unary predicate symbol 'L' and 'V':

Lx <-> [fill in the requirements for x being constructible]

Vx <-> x=x

So the relativizations such as:

Ax(Lx -> P)
read as "P holds for constructible sets".

and

V = L
for
Ax Lx

And per a given model, 'L' will map to a subset of the universe, and 'V' will map to a subset of the universe (and if the model is a model of "Vx <-> x=x" then V maps to the subset of the universe that is the universe itself).

that impressively buzzword-compliant paragraph — fishfry

I didn't use the terminology to impress anyone with buzzwords. And they are not buzzwords. They are terminology of mathematics.

there is no mention of the domain of discourse. So again, none of this is relevant. — fishfry

It's about models. A model is a domain of discourse along with a function on the nonlogical symbols. So where I mentioned a model, there is a domain of discourse associated with that model. And the paragraph was not meant to address the original question, but rather to give an idea of how notions of proper classes as models are (as I hope I recall correctly) reducible to the syntactical approach of relativizations.

You're perfectly correct that ZFC is consistent if ZF is, but what has that got to do with the conversation? — fishfry

It was added merely as an illustration of an important theorem that comes from relativizations.

There also is the notion of proper classes as models, or more specifically, inner models. However, I think (I am rusty on this) that when we state this formally, it actually reduces to the syntactical method of relativization, so that when we say L is an inner model of set theory, we mean something different from the plain notion of a model. If I recall correctly, roughly speaking, relative to a theory T, saying 'sentence P is true in "class model" M' reduces to: In the language for T, we define a unary predicate symbol 'M', and P relativized to M is provable in T. So, for example, when considering the consistency of the axiom of choice relative to ZF, we find that the axiom of choice is true in the constructible universe L ("L is a model of AC"), which, in one way of doing this, reduces to: In the language of set theory, define a unary predicate symbol 'L', then we show that AC relativized to L is a theorem of ZF. So if we have a model D of ZF, then the submodel that is D intersected with L (the intersection of a set with a proper class is a set) is
a model of ZFC. That entails the consistency of ZFC relative to the consistency of ZF. But I am rusty here, so I may be corrected.

There are set theories in which classes are formalized — fishfry

Right. But even with those theories, the domain of discourse for a model for the language of the theory is a set.

Even a class theory such as NBG has only models that have a domain of discourse that is a set.

Moreover, if we tried to allow a proper class to be a domain of discourse, we'd get a contradiction:

For example, suppose we are doing model theory in a class theory in which there are proper classes. Okay, so far. Now suppose U is a proper class and, for simplicity, we have a language with just one nonlogical symbol. And let R be the relation on U that, per the model, is assigned to the nonlogical symbol.

Then we have the structure <U R>. But then, unpacking the ordered tuple by the definition of tuples (such as Kuratowski), we get that U is a member of a class, which contradicts that U is a proper class.

If you look at textbooks in mathematical logic, model theory, and set theory, you will see that without exception the definition of a model stipulates that its domain of discourse is a set.

Note: I put strikethrough there to accommodate the following post:

Well, we are quantifying over the collection of all sets. — fishfry

Not formally. Formally, any model of set theory has as a set, not a proper class, as its domain of discourse. For any model, the universal quantifier ranges over the members of the domain of discourse, which is a set.