The topic is about propositions, but more formally about sentences. Yet it doesn't matter toward the point that 'Kq' does not stand for 'We know of the existence of q' but rather it stands for 'We know that q is true'.

"We don't know that the earth is round"

and

"We believe that the earth is flat"?

The differences are so easy to point out that I don't see the sense in asking about it.

'known' in this context doesn't mean 'we know that the sentence itself exists'. 'known' in this context means 'we know that the sentence is true'.

If one conflates 'it is known that the sentence exists' with 'it is known that the sentence is true', then of course the whole discussion falls apart.

Also, in this context, it is not just sentences that have known to have been expressed but sentences in general (in a formal context, that would be all the sentences of the formal language).

There is no statement in the expositions I've seen of Fitch that 'p' stands for a true sentence.

One may go back and read the exact expositions to see that.

Contrarian, I would think, to the preponderance of philosophers and to everyday common sense.

It is an extraordinarily outlandish view that every truth is known.

Fitch does not claim that all truths are known. That is ridiculous a misunderstanding of him.

What he shows is that

If all truths are knowable then all truths are known. But since not all truths are known, we infer that it is not the case that all truths are knowable.

To believe

For all q, we have q -> Kq

is extraordinarily contrarian.

It should not be overlooked that 'Kq' does not stand for 'q is knowable' but rather for 'q is known'.

Why is

q -> Kq

being stated?

No one believes that as a generalization for all q.

Formally, it's about sentences, regarded in terms of two primitive modal operators, whether true or false, whether known to be true or false or not.

The import is:

It is not the case that for all sentences p, we have p -> LKp.

From that, it does follow that there are true sentences such that it is not possible that they are known to be true.

Moreover, that vitiates certain philosophical views.

What does it mean to say that falsehoods are or are not in the scope of Fitch's paradox? What does being "in the scope" of a paradox exactly mean in this context?

I'm not inclined to go through the earlier posts. But, from what I do see, I don't know exactly what people are claiming about truth in this context.

'p' is a meta-variable ranging over sentences in the object language.

The proof is syntactically correct, as far as I can tell.

Whether any given sentence is true or false in any given semantics for a modal language with two modal primitives, I don't see what that has to do with the correctness of the proof.

And, especially, I don't know what "cover falsehoods' is supposed to mean. Whether 'p' stands for a sentence that is true or is false in any given model, the proof is correct as far as I can tell, thus the conclusion:

It is not the case that for every sentence p we have p -> LKp.

If anyone is claiming there is an incorrect step, then I'd like to know where it is here:

Axiom schemata:

(a) Kq -> q

(b) K(q & r) -> (Kq & Kr)

(c) q -> LKq [this is the axiom presumably to be rejected]

Inference rule:

(d) from |- ~q infer |- ~Lq

Proof:

1. K(p & ~ Kp) assume toward RAA

2. Kp & K~Kp from 1 and (b)

3. ~Kp from 2 and (a)

4. ~K(p & ~Kp) from 1 by RAA

5. ~LK(p & ~Kp) from 4 and (d)

6. p & ~Kp assume toward RAA

7. LK(p & ~Kp) from 6 and (c)

8. ~(p & ~Kp) from 6 by RAA

9. p -> Kp from 8 [note that non-intuitionistic logic is used here]

(1) Just to be clear, Fitch does not hold that

for all p we have p -> Kp.

Rather, the import is that

if for all p we have p -> LKp, then for all p we have Kp,

but since it is not the case that for all p we have Kp, it is not the case that for all p we have p -> LKp, so we reject that for all p we have p -> LKp.

(2) SEP gives the argument with quantifiers ranging over sentences, but I'm not sure that is necessary, and so I am not sure that necessarily the argument is second order. I don't see why the argument can't be as proof schema for a language of sentential modal logic with two primitive modal operators 'L' and 'K', letting 'p' and 'q' and 'r' be metavariables ranging over sentences:

Axiom schemata:

(a) Kq -> q

(b) K(q & r) -> (Kq & Kr)

(c) q -> LKq [this is the axiom presumably to be rejected]

Inference rule:

(d) from |- ~q infer |- ~Lq

Proof:

1. K(p & ~ Kp) assume toward RAA

2. Kp & K~Kp from 1 and (b)

3. ~Kp from 2 and (a)

4. ~K(p & ~Kp) from 1 by RAA

5. ~LK(p & ~Kp) from 4 and (d)

6. p & ~Kp assume toward RAA

7. LK(p & ~Kp) from 6 and (c)

8. ~(p & ~Kp) from 6 by RAA

9. p -> Kp from 8 [note that non-intuitionistic logic is used here]

So from the rules (a) - (d) we proved for arbitrary sentence p that p -> Kp. But we don't believe that for all sentences p we have p -> Kp, while (a), (b) and (d) seem eminently reasonable, so we reject (c).

(3)

The SEP article explains some of the arguments against Fitch's argument.

Also, this book looks like a great overview:

Salerno

Suppose some Sn is true.
So Sn+1 is false.
So there is some k > n+1 such that Sk is true.
But Sn is true and k > n, so Sk is false.
So Sn is false.
So for all n, Sn is false.
But then for all n, Sn is true.
So for all n, Sn is false and Sn is true.

If sentence 2 is false, sentence 3 is true. — Banno

How do you infer that?

Sets containing itself are (in)famous. — VincePee

"There is a set that is a member of itself" is not in and of itself contradictory.

The famous contradictory statement is "There is a set of all sets that are not members of themselves".

But that does not even require the notion of 'set' or 'member': For any 2-place relation R, we have the theorem of logic "It is not the case there is an x such that for all y, y bears R to x if and only if y does not bear R to y".

The first video (I didn't watch the second video) is stupid nonsense and disinformation.

In this context, infinite summation is defined only for converging sequences. If the rules of definition are violated, then, of course, contradictions may be derived. There is no mystery or even problem about that.

The person at the blackboard says, "The problem with infinity is all sorts of weird things happen when you're dealing with infinity". First, that doesn't even mean anything. Second, instead of explaining that the fallacy is in using an undefined notion (infinite summation on a sequence that does not converge), the person at the blackboard doesn't even suggest how we may investigate further to see that there is not an actual conundrum.

The video is yet another example of Internet ignorance and disinformation. That person seems to be teaching a classroom. He should be told by the school administrators to clean up his act: If he wants to present mathematical challenges, then he should provide his students with the benefit of techniques and information for solving the challenges rather than obfuscate with "weird things happen".

L is defined by transfinite recursion. L_0 = 0. That is, the empty set is constructible by definition at the base clause.

A conjunction of a statement and its negation is of the form

P & ~P

where P is any statement.

A conjunction of a statement and its negation in the language of set theory is of the form

P & ~P

where P is a formula in the language of set theory.

Set theory is inconsistent if and only if there is a such a conjunction that is a theorem of set theory.

A contradiction is the conjunction of a statement and its negation.

Perhaps in some of the articles cited, it is mentioned that paraconsistent logic is suited for such situations as contradictory data entries. I am not expert in that, but it makes sense to me as useful at least at a glance. Moreover, any philosophical objections to paraconsistent logic are not necessarily extended as reasonable criticisms of formal systems of paraconsistent logic. Such systems have different rules and semantics than do classical systems, so such systems must be regarded in that different context.

There is no need for an axiom regarding intersection. Intersection is just an instance of separation.

None of the axioms of set theory mention 'set'. So objection to any axioms on the basis that they mention 'set' are ill-founded.

/

'set' is not a primitive of set theory. However it, may be defined in the language of set theory:

df: x = 0 <-> Ay ~yex

df: x is a class <-> (Ey yex v x=0))

df: x is set <-> (x is a class & Ey xey))

That definition also carries over to class theory such as Bernays class theory (more commonly known as 'NBG set theory').

But formally 'is a set' does not occur, rather it is an English phrase to stand for a formal predicate symbol. While 'set' is used to suggest our pre-formal notion of sets, it is not formally required. We could use 'zet' 'Zset' or myriad other words. Same for the phrase 'empty set'. So, while one may choose to argue that the notion of an empty set does not adhere to our usual understanding of what sets are, one should be careful not to charge that set theory is inconsistent on that basis.

The theorem and definition at issue are:

th: E!x Ay ~yex

df: Ax x = 0 <-> Ay ~yex.

No mention of 'set' or 'empty' or 'empty set'.

An objection was made that the axiom of extensionality does not "distinguish" between sufficiency and necessity.

The axiom is:

Axy(Az(zex <-> zey) -> x = y)

that is the sufficiency clause.

The necessity clause comes from identity theory:

Axy(x = y -> Az(zex <-> zey))

So together we have the theorem of sufficiency and necessity:

Axy(Az(zex <-> zey) <-> x = y)

Not sure, but offhand, I suspect it is not the case that ~AC implies there is Tarski infinite but Dedekind finite set. The converse holds though.

About the schema of separation, if we say there is one axiom for each predicate, we need to be careful what 'predicate' means. There is one axiom for each formula. We could say that each formula permits definition of a predicate symbol, so, in that sense there is one axiom for each defined predicate symbol. But we should be clear to say we don't mean 'predicate' in the sense of 'property'.

Absent AC there is a set that is infinite (not bijective with any natural number) yet Dedekind-finite (no proper subset is bijective with the entire set). — fishfry

Absent AC, it is undecided whether there is such a set.

Z is axiomatized by:

extensionality
schema of separation
pairing
union
power
infinity
regularity

It is uncontroversial that AC, ZL, and WO are equivalent in the sense that in Z (perforce ZF) we can prove one from the other. But it is not the case that they are logically equivalent. To be logically equivalent they would have to be provable one from the other using only the pure predicate calculus.

They are equivalent in Z, so, a fortiori, they are equivalent in ZF. But they are not logically equivalent.

Z |- AC <-> ZL & ZL <-> WO & AC <-> WO

But it is not the case that

|- AC <-> ZL & ZL <-> WO & AC <-> WO

AC, ZL and WO are not logically equivalent. But they are equivalent in Z set theory.

Footnote 1 of the SEP article says: "Talk of ‘first’ and ‘last’ members here is just a matter of convention. We could just as well have said that an infinite regress is a series of appropriately related elements with a last member but no first member, where each element relies upon or is generated from the previous in some sense. What direction we see the regress going in does not signify anything important."

Another poster made it appear as if I hold that words (such as 'least') or symbols don't have explicit definitions, and that ambiguity results. My point though is that, other than primitives, symbols do have definitions, with very strict rules as to what constitutes a definitions, and those rules prevent ambiguity. And I gave an explicit definition of 'least'.

I did mention that ordinal less-than is membership:

df: k is ord-less-than j <-> k e j — TonesInDeepFreeze

Df. If x and y are ordinals, then x precedes y (x is less than y) iff x is an element of y. — TonesInDeepFreeze

I mentioned finding a recursive definition of 'prior'. It's simple.

We define the set of symbols prior to a symbol s:

If s is primitive, prior(s) = 0
If s is defined, prior(s) = U{k | Et(t occurs in the definiens for s & k = prior(t))}

Then:

j is prior to s <-> j e prior(s)

I said that my commentary is based only on the clip posted at the top of this thread.

Yesterday I got hold of McGee's paper.

It turns out that his argument does not suppose that the conditionals mentioned are taken in the sense of the material conditional. He says that if the conditionals mentioned are taken in the sense of the material conditional then modus ponens is not impeached by his argument.

Lack of having his paper to know what he actually claimed led to unnecessary disputation about his argument.

This has been a waste of my time and the time of people reading my posts. If I knew from the onset that he's not talking about the material condition, then I wouldn't have unnecessarily bothered.

PPS

Two iconic books that handle forcing in detail are:

'Set Theory' - Jech

'Set Theory: An introduction To Independence Proofs' - Kunen

Jech's book is a great tome. But I might prefer Kunen, because, though I'm not a formalist, I find that a formalist sensibility contributes to good exposition.

However, both those books are at a graduate level. For an introduction to set theory, I always recommend:

Elements Of Set Theory - Enderton, which is a great read

supplemented with

Axiomatic Set Throry - Suppes.

PS. We don't need definitions of 'cardinality' an 'ordinal' to state CH and GCH. We can state it equivalently:

GCH.

If S is infinite, then is no set X such that all these hold:

S injects into X
There is no bijection between X and S
X injects into 2^S
There is no bijection between S and 2^S

And CH is just a special case of GCH, where S = w.

I'm not a set theorist, but I have some thoughts.

I haven't seen articles before that give a layman's explanation of forcing and of axioms for proving CH. So I appreciate that.

I have my own question. I kinda got the idea behind the explanation of the filter, but I wonder if this is correct:

That filter proves the existence of a certain real number. Then we use other filters to prove the existence of other real numbers. By doing that some uncountable number of ways, we prove that there are sets of real numbers that have cardinality between the cardinality of the set of natural numbers and the cardinality of the power set of the set of natural numbers. Is that correct?

(IMPORTANT. In this post, I take "ZF is consistent" as a background assumption. For example "ZFC is consistent" in this post means "If ZF is consistent then ZFC is consistent".)

'AC' stands for the axiom of choice.
'CH' stands for the continuum hypotheis.
'GCH' stands for the generalized continuum hypothesis.
'ZFC'stands for ZF+AC.

I know nothing about forcing other than this:

(1) Cohen used forcing to prove:

ZF+~AC.

ZFC+~CH is consistent. So, a fortiori, ZFC+~GCH is consistent.

(2) Forcing involves ultrafilters and/or Boolean algebra.


Re (1), Godel had previously proved:

ZFC is consistent. [*]

ZFC+GCH is consistent. So, a fortiori, ZFC+CH is consistent.

Godel did it with the notion of the constructible universe.

Combining Godel and Cohen, we get:

AC is independent from ZF.

CH is independent from ZF.

GCH is independent from ZF.

[*] But what about Godel's second incompleteness theorem that entails "Set theory does not prove its own consistency"? Well, actually the second incompletess doesn't entail that. The second incompleteness theorem does entail, "If set theory is consistent then set theory does not prove set theory is consistent". And, since I put "ZF is consistent" as a blanket assumption for this post, "ZFC is consistent" in this post, stands for "If ZF is consistent then ZFC is consistent". That qualification applies, mutatis mutandis, to both the Cohen theorems and both the Godel theorems above.


FOR REFERENCE:

df. A set of sentences S is consistent iff S does not prove a contradiction.

df. A sentence P is independent from a set of sentences S iff (S does not prove P and S does not prove ~P).

th. A sentence P is independent from a set of sentences S iff (S+P is consistent and S+~P is consistent.

th. If a set of sentences S is consistent, then there is a model in which every sentence in S is true (this is Godel's completeness theorem).

So:

To prove "ZFC is consistent", it suffices to prove there is a model of ZFC. But "ZF is consistent" entails "ZF has a model". So it suffices to prove that "ZF has a model" implies "ZFC has a model".

To prove "ZFC+GCH is consistent", it suffices to prove there is a model of ZFC+GCH. But "ZF is consistent" entails "ZF has a model", which, Godel proved, entails "ZFC has a model". So it suffices to prove that "ZFC has a model" implies "ZFC+GCH has a model".

To prove "ZF+~AC is consistent", it suffices to prove there is a model of ZF+~AC. But "ZF is consistent" entails "ZF has a model". So it suffices to prove that "ZF has a model" implies "ZF+~AC has a model".

To prove "ZF+~CH is consistent", it suffices to prove there is a model of ZF+~CH. But "ZF is consistent" entails "ZF has a model". So it suffices to prove that "ZF has a model" implies "ZF+~CH has a model".

Those are examples of relative consistency. "If theory T is consistent, then theory Y is consistent".

Some years after Godel's results just mentioned, Sierpenski proved that ZF+GCH proves ZFC. So:

Proving "ZF+GCH is consistent", a fortiori, proves "ZFC is consistent".

Proving "ZF+~AC is consistent", a fortiori, proves "ZF+~AC+~GCH", which, a fortiori, proves "ZF+~GCH is consistent". So, "ZF+~AC is consistent" proves "ZF+~GCH is consistent".


BACKGROUND:

What is the axiom of choice?

What is the continuum hypotheis?

What is the generalized continuum hypothesis?


df. 0 = the empty set.

df. PS = the set of subsets of S.

df. Y\Z = the set whose members are all and only those members of Y that are not members of Z.

C (the axiom of choice) is the statement:

"For every S, there is a function on the PS\{0} such that for every x in PS\{0}, f(x) e x". We call such an f "a choice function for S".

To visualize the above:

Imagine a nation made up of provinces (and possibly there are infinitely many provinces). From each province we can choose a representative who is a resident of that province.

If S is finite, without the axiom of choice, by a trivial induction on the cardinality of S, we prove there is a choice function for S, so we don't need the axiom of choice to prove there is a choice function for S.

"For every S, there is a function on the PS\{0} such that for every x in PS\{0}, f(x) e x".

The axiom of choice is equivalent with a number of other theorems, especially "Every S has a well ordering" and "Every S is equinumerous with an ordinal".


df. S^x = the set of funtions from x into S.

df. x and y are equinumerous iff there is a bijection between x and y.

df. card(S) = the least ordinal k such that S and k are equinumerous.

df. for ordinals, x < y iff x e y.

df. w = the set of natural numbers

df. R = the set of real numbers

th. card(R) = card(Pw) = card(2^w)

th. There is no surjection from S onto PS. (Cantor's theorem)

th. card(w) < card(R)

CH is the statement:

"There is no S such that card(w) < card(S) < card(R)".

GCH is the statement:

"If X is infinite, then there is no S such that card(X) < card(S) < card(PX).

Cantor failed to prove CH. Hilbert wanted somebody to prove it. Godel proved that we can't disprove GCH, a foritori that we can't disprove CH. Cohen proved that we can't disprove ZFC+~CH, so, a fortiori, we can't dispove ZFC+~GCH.

So some set theorists, who feel that ~CH fits their concept of 'set' have been trying to discover a set theoretic statement that is even more convincingly true to their concept of 'set' and that proves ~CH. Other set theorists, who feel that CH fits their concept of 'set' have been trying to discover a set theoretic statement that is even more convincingly true to their concept of 'set' and that proves CH.

FOR REFERENCE:

Z is basic infinitistic set theory (first order logic with identity, extensionality, schema of separation, pairing, union, power, infinity, regularity). ZF is Z with the axiom schema of replacement added.

The axiom schema of replacement is the statement [I'm simplifying]:

"If R is a function class, then, for every S, there is the T whose members are all and only those y such that there is an x in S such that <x y> in R."

R is a proper class. It is a proper class of ordered pairs. The axiom of replacement is: If R is functional (i.e. if <x y> in R and <x z> in R, then y=z), then for any set S, there exists the set T that is the image of S by R. I.e., if you have a set S, and a functional relation, then the "range" of that relation restricted from the domain S is a set.

That mentions proper a proper class, though Z proves ~Ex x is a proper class. So the actual axiom schema of replacement is a set of axioms, with each axiom mentioning a formula instead of a proper class. The formula "carves out" the proper class.