Start with one thing:

Also BTW, "injection" is the word YOU are using. — fishfry

The poster asked about a "map into" as a "1:1 relation between".

A word for that is 'injection'.

An injection is a type of function between two sets. — fishfry

That's fine with me. And if you object to saying "injection" rather than "1:1 map" that's fine with me too. I'm not the one asking whether there is a 1:1 mapping (whether you wish to rule out calling that an 'injection') from the set of real numbers into whatever is designated by 'the real world'. Perforce, obviously, I'm not claiming that if there were such a 1:1 relation then its range would be a mathematical set.

First, again, I don't know what the poster means by "the real world" so I don't know what firm and clear notion there is of an injection from the set of real numbers into "the real world".

Also, the argument "There are no relevant experiments regarding surrounding aspects of the reals, therefore there is no such injection" requires the premise, "If there is such an injection, then there are relevant experiments regarding surrounding aspects of the reals". But how would we rule out that there could be an injection but no relevant experiments regarding surrounding aspects of the reals, or that there could be an injection but no known relevant experiments regarding surrounding aspects of the reals?

If the real numbers are instantiated in the real world — fishfry

The question was about an injection. What is the definition of "instantiated in the real world"? Does it just mean that there is the range of an injection from the set of real numbers?

If the real numbers are instantiated in the real world, then questions such as the axiom of choice and the Continuum hypothesis become subject to physical experiment. — fishfry

I don't know that that is the case. Moreover, cutting back to the question of an injection, I don't know that that the lack of someone thinking up an experiment would entail that there is no injection.

Moreover, would entertaining that there is an injection from the set of natural numbers N into the real world entail that there must be some experiment to conduct?

no physics postdoc has ever applied for a grant to study such matters — fishfry

I don't know that that is true.

such questions are so far beyond experimental investigation as to be meaningless. — fishfry

That might be the case. Indeed, even the question alone of the existence of an injection from the set of real numbers into "the real world" doesn't seem to me to have, at least so far, been given a firm and clear meaning.

such questions are so far beyond experimental investigation as to be meaningless. — fishfry

That might be the case; I don't know. But I don't see that to entertain that there might be an injection entails that there must be an experiment to conduct. But again, the question of the existence of an injection from the set of real numbers into "the real world" doesn't seem to me to have, at least so far, been given a firm and clear meaning.

Banach-Tarski — fishfry

I wouldn't think that to entertain that there is an injection from the set of reals into the real world entails that there is a physical version of Banach-Tarski. But again, the notion of such an injection is not definite enough for me to have much of a view anyway (as well as I'm not prepared to discuss details of Banach-Tarski).

I surely don't have a strong opinion on the question of the existence of an injection from the set of real numbers into "the real world", but at least I would want to ponder whether the question is even even meaningful to either affirm or deny.

Supertasks have been used to show there isn't such a mapping for some cases. — Relativist

Zeno's paradox? Other? What arguments are you referring to that there is no injection from the set of real numbers into "the world"?

I don't opine here on that other question. But what do you mean by "maps into"? Do you mean "there is a function into" or do you mean "there is a one-to-one function into"? I surmise you mean the latter.

Therefore there is a gap between all pairs of distinct points of the continuum" — Relativist

It turned out that by 'gap' @MoK meant 'interval'. His thinking is hopelessly confused.

I'm keeping to first order set theory. The point is that in set theory, any two complete order fields are isomorphic. You can think of that as saying that any two complete ordered fields are the same as one another except we changed the names of the elements. In other words, they are structurally the same.

/

d is a Dedekind cut if and only if (1) d is a non-empty proper subset of the set of rational numbers & (2) for all x, if x is a member of d and y<x, then y is a member of d & (3) d has no greatest member

/

one cannot define infinitesimal in the classical continuum — MoK

Be clear. There is a difference between (1) defining the adjective 'is an infinitesimal' and (2) proving the existence of a particular infinitesimal and naming it.

In set theory in which we formulate the system of real numbers, we can define 'is an infinitesimal', and prove that no real number is an infinitesimal, just as I have in this thread. And in set theory we can prove that there is a system with hyperreals (including infinitesimals) and prove that such a system is not isomorphic with the system of real numbers but that that there is a subsystem of hyperreals that is isomorphic with the system of real numbers. However, in plain set theory, even though we can prove the existence of a system of hyperreals, at least the usual methods (compactness theorem or ultrafilters) don't define a particular system of hyperreals.

/

A filter is a certain kind of set. an ultrafilter is a certain kind of filter. ultrafilters are used to prove the existence of a system of hyperreals.

It is folly for you to be trying to figure out what ultrafilters are and how they play into hyperreals when you don't even know page 1 of set theory.

Do you mean that the set of natural numbers is the set of aleph_0? — MoK

Yes. I said it many posts ago in a post you claimed to have read.

aleph_0 is a number. — MoK

Set theory doesn't provide a definition of 'is a number'. Rather, set theory provides individual definitions of such things as 'is a natural number', 'is a real number', 'is an ordinal number', 'is a cardinal number'. In those, 'number' is not taken as a standalone adjective. We could just as well have this terminology in instead: 'is a natnum', is a 'realnum', 'is an ordinalnum', 'is a cardnum'. We don't make proofs by recourse to ascribing any properties to being "a number" in and of itself. Also, instead of 'ordinal number' and 'cardinal number' we may as well just say 'ordinal' and 'cardinal'.

Anyway, aleph_0 is an ordinal and it is the least infinite ordinal, and it is a cardinal and it is the least infinite cardinal.

How could you treat it as a set? — MoK

Everything in set theory is a set. Natural numbers, real numbers, ordinals, cardinals ... are sets.

By definition aleph_0 = w, and we prove that w = {n | n is a natural number}.

I mentioned that previously.

I am looking for proof that the set of natural numbers that each its member is finite has aleph_0 members. — MoK

fishfry has been giving you the info.

But to understand very well, you need to go in forward direction from page 1 of a book - from the simplest concepts to the more involved concepts that depend on the prior concepts - instead of foolishly trying to work backwards. Anyway:

Thm: If k is a cardinal and j is a cardinal less than k, then j and k are not one-to-one
Proof: By definition of 'is a cardinal', j is not one-to one with k

Thm: If S is one-to-one with a cardinal, then there is a unique cardinal such that S is one-to-one with it
Proof: See above theorem

Dfn. If S is one-to-one with a cardinal, then card(S) = the unique cardinal that S is one-to-one with

Thm. if x is infinite and y is finite, then x and y are not one-to-one
Proof: Put a nickel in the machine

Thm. w is not one-to-one with any member of w
Proof: w is infinite but every member of w is finite

Thm. w = {n | n is a natural number} is a cardinal
Proof: w is an ordinal and w is not one-to-one with any member of w, so w is not one-to-one with any ordinal less than w

Thm. x is one-to-one with x
Proof: the identity function on x is an injection from x onto x

Thm. card({n | n is a natural number}) = aleph_0
Proof: {n | n is a natural number}) = aleph_0. aleph_0 is one-to-one with aleph_0

Your questions arise only because you don't know anything about the subject, and also because (contrary to your claims) you don't carefully read the answers given you. And when you ask question Q1 and get an answer A1, you then have to ask Q2 to understand Q1, on and on. The way to break that regress is to study the subject from page 1 in forward, not reverse, order.

You need a book on the first order predicate calculus, then one on set theory. Nothing else will give you a proper understanding of the subject.

But until perhaps I get some moments to address your latest questions, I'll at least address this one (the answer is trivial, except you don't know anything about the subject to see that it's trivial):

the least infinity namely aleph_0. — MoK

"the least infinity" is not defined by you.

But we prove that there is no infinite cardinal less than aleph_0. Moreover, aleph_0 = the unique infinite cardinal K such that there is no infinite cardinal less than K.

Dfn. if x and y are ordinals, then x is less than y if and only if x is a member of y

Dfn. x is successor-inductive if and only if (0 is in x & for all n, if n is in x then xu{x})

Axm. There exists a successor-inductive set

Thm. There exists a unique set that is a subset of all successor-inductive sets
Proof: Put a nickel in my proof jukebox and I'll supply the proof

Dfn. w = the unique set that is a subset of all successor-inductive sets

Dfn. n is finite if and only if [put a nickel in my definition jukebox and I'll supply the definiens]

Dfn. n is an ordinal if and only if [put a nickel in the machine]

Dfn. n is a natural number if and only if (n is finite & n is an ordinal)

Thm. w = {n | n is a natural number}
Proof: Put a nickel in the machine

Thm. w is successor-inductive
Proof: Put a nickel in the machine

Thm. No ordinal is a member of itself
Proof: Put a nickel in the machine

Dfn. k is a cardinal if and only if (k is an ordinal & k is not one-to-one with a lesser ordinal)

Dfn. x is infinite if and only if x is not finite

Thm. every successor-inductive set is infinite
Proof: Put a nickel in the machine

Thm. Every cardinal is an ordinal
Proof: A cardinal is an ordinal that is not one-to-one with a lesser ordinal. So every cardinal is an ordinal.

Dfn. aleph_0 = w

Thm. w is infinite
Proof: w is a successor-inductive set

Thm. w is an ordinal
Proof: Put a nickel in the machine

Thm. There is no infinite cardinal less than aleph_0
Proof: aleph_0 is infinite. Every member of aleph_0 is finite. So no member of aleph_0 is infinite. So there is no infinite ordinal that is a member of aleph_0. So no infinite cardinal is less than aleph_0.

Thm. If K and L are cardinals, then exactly one of these: (1) K dominates L or (2) L dominates K or (3) K = L
Proof: K and L are ordinals. Put a nickel in the machine for proof that this trichotomy obtains for ordinals (axiom of choice not needed here)

Thm. aleph_0 = the unique infinite cardinal K such that there is no infinite cardinal less than K
Proof: Suppose L not= aleph_0 is an infinite cardinal such that there is no infinite cardinal less than L. So aleph_0 is not less than L. And L is not less than aleph_0. Contradicts previous theorem.

The poster didn't express puzzlement about the fact that there is no finite upper bound to the set of natural numbers.

I defined the domain D that has this specific property, the number of its members, n, is a member as well. — MoK

Fine. But that's a different domain than the one I used.

Let me ask you this question: Are all members of the natural number set finite? — MoK

Yes, it's a common fact.

how the number of its members could be aleph_0 — MoK

The set of natural numbers is aleph_0. Every set is one-to-one with itself. So the set of natural numbers is one-to-one with alelph_0, so the cardinality of the set of natural numbers is aleph_0.

I am however puzzled how all the members of the natural number set are finite yet it has aleph_0 members. — MoK

There is no contradiction between:

Every member of the set of natural numbers is finite
and
The set of natural numbers has aleph_0 members

There is no contradiction between:

Every member of the set of natural numbers is finite
and
The set of natural numbers is not finite

It seems you have this false premise in your head:

If every member of a set S has property P then S has property P.

That premise is wrong and ridiculous. Get rid of it.

I should not be explaining basic logic to you. Your best bet would be to study a book on the first order predicate calculus and then one on set theory.

This is counter-intuitive to me. — MoK

It's not counterintuitive that there exist functions whose domain is the set of positive natural numbers. Any Calculus 1 textbook has such functions all through the book. Moreover, there are functions whose domain is the set of real numbers. Moreover, there are functions whose domain is a proper subset of the set of real numbers. Moreover, for any set whatsoever, there are functions whose domain is that set (except there is only one function whose domain is the empty set).

What is counterintuitive to me is someone claiming to be familiar with basic calculus but not understanding ordinary mathematical functions.

Consider a function f with the domain D={1,2,...N} where N is a finite positive integer. — MoK

To be clear, 'N' there does not stand for the set of natural numbers, so I'll use 'n'':

Yes, {1 ... n} is a set with n number of members and n is a member of {1 ... n}.

That doesn't vitiate that there are other sets with n number of members but such that n is not a member:

{0 1} has 2 members, but 2 is not a member.

{1 4 7} has 3 members, but 3 is not a member.

the set of natural numbers has aleph_0 members, but alelph_0 is not a member.

the set of real numbers has 2^aleph_0 members, but 2^aleph_0 is not a member.

Could you please explain what happens when N is aleph_0? — MoK

The domain of the function f that I defined is the set of positive natural numbers. aleph_0 is not in that domain.

I mentioned in a previous post that nothing is stopping us from defining a different function that has alelph_0 in the domain. But so what?

the domain of f is the set of positive natural numbers = {n | n is a positive natural number}

f(1) = 1

f(n+1) = f(n)/2

"f(aleph_0)" is meaningless notation since alelph_0 is not in the domain of f

a different function h:

the domain of h = {n | n is a positive natural number}u{aleph_0}

h(1) = 1

h(n+1) = h(n)/2

h(aleph_0) = whatever you want to make it (could be h(aleph_0) = 0, or h(aleph_0) = aleph_0, or h(aleph_0) = pi, or whatever you specify. just say what it is and then we'll know what h is)

The purpose of defining f was to agree with you that the results of dividing by 2 converge to 0. And the fact that there is no finite upper bound to the number of times we can divide by 2 refutes your errant claim that there is a real number such that there is no real number between it and 0, thus disproving your errant claim that you've disproved the existence of the continuum.

Once you define h, you can say what the purpose of doing that is. But any claims about h vis-a-vis real numbers, infinitesimals and the continuum need to be proven and use only terminology already defined or defined by you, not merely imaginistic hand waving.

/

Do you really not understand what a domain of a function is? Or are you trolling me?

The domain of f has aleph_0 members, but aleph_0 is not a member of the domain of f.

The books you've recommended sound very interesting. — Gregory

They are almost entirely mathematics, very little philosophy. Except the introductory chapter in 'Introduction To Mathematical Logic' by Church.

A pretty good book that weaves mathematics and philosophical topics is 'The Philosophy Of Set Theory' by Mary Tiles (but best to have already studied textbook set theory). And some great essays are in 'Logic, Logic, and Logic' by George Boolos and his essay "The Iterative Conception Of Set".

What do you mean by infinity when you talk about the limit in this post? — MoK

I didn't use the word 'infinity' in that post.

I used 'inf' (usually seen as the lemniscate). In the context and sense I used, 'inf' has no meaning by itself; it does not denote a mathematical object. You may think of it is a facon de parler.

As I explained [redacted for context]:

[Suppose] there exists a unique real number x such that for every positive real number y, there exists a positive natural number n such that |f(n) - x| < y.

Then we write:

lim[n = 1 to inf] f(n)

'lim' is a variable binding operator, but it can be reduced to a regular operation symbol:

Df. If f is a function from the set of positive natural numbers into the set of real numbers, and there exists a unique real number x such that for every positive real number y, there exists a positive natural number n such that |f(n) - x| < y, then Lf = the unique real number x such that for every positive real number y, there exists a positive natural number n such that |f(n) - x| < y. — TonesInDeepFreeze

You see there that we unpack lim[n = 1 to inf] f(n) to:

lim[n = 1 to inf] f(n) = the unique x such that for every positive real number y, there exists a natural number n such that |f(n) - x| < y.

The falcon de parler 'inf' does not occur in the definiens. There is no resorting to a claim that there is an object named 'infinity'.

Again, we distinguish between the adjective 'infinite' and the noun 'infinity':

Adjective: S is infinite if and only if S is not finite.

Noun: There are various uses of 'infinity' as a noun, including such things as infinity as a point in the extended reals, or infinity as a hyperreal in nonstandard analysis. But my use of 'inf' (the lemniscate) in the context of defining 'limit' makes no use of 'infinity' as a noun.

Have you studied the material of Calculus 1? At least the first week in which the definition of 'limit' is given?

why aleph_1 is uncountable? — MoK

That you ask that question, indicates that you don't know what 'aleph' and 'countable mean' despite that you toss those terms around as if you're making actually meaningful statements with them.

Df. S is countable if and only if (S is one-to-one with a natural number or S is one-to-one with w)

Df. S is uncountable if and only if S is not countable

Df. T strictly dominates S if and only if (there is an injection from S into T & S is not one-to-one with T).

Notation: If j is an ordinal, then j+ is ju{j}.

Definition (by transfinite recursion on the ordinals):
aleph_0 = w
aleph_j+ = the least ordinal that strictly dominates aleph_j
If j is a limit ordinal, then aleph_j = U{alelph_m | m e j}

1 = 0+, so aleph_1 = the least ordinal that strictly dominates w. Since aleph_1 strictly dominates w, aleph_1 is not one-to-one with a natural number nor is S one-to-one with w. So, aleph_1 is uncountable.

/

Please get a book on the first order predicate calculus, then one on set theory.

By infinity, I mean aleph_0. — MoK

Then say 'aleph_0' since, in this context, 'infinity' is ambiguous.

I thought that the value of f(alep_0)=0 — MoK

You think that only if you didn't read my post.

For the third time: The domain of f is the set of positive natural numbers, therefore, aleph_0 is not in the domain of f, therefore "f(aleph_0)" is meaningless.

one could define a sequence g(n+1)=g(n)/10 where g(0)=1 — MoK

What is the domain of g? If the domain of g is the set of natural numbers then:

g(0) = 1
for all natural numbers n, g(n+1) = g(n)/10

So g, just like f, converges to 0.

But g(aleph_0)=0.0...1 — MoK

So the domain of g is not the set of natural numbers. What is the domain of g?

I explained to you before that ".0...1" represents a sequence h on w+1 such that:

For all natural numbers n, h(n) = 0
h(w) = 1

So if aleph_0 is a member of the domain of g, then:

g(0) = 1
for all natural numbers n, g(n+1) = g(n)/10
g(aleph_0) = h

But we don't have a less than relation defined that includes h in the field of the relation, so no definition of "converges".

And are you going to be defining the domain and range of g to include other objects?

and we find g(aleph_0) >0 — MoK

You haven't defined ">" so that it includes g(aleph_0) in the field of the relation.

so f(aleph_0)>0 as well. — MoK

Nonsense. aleph_0 is not in the domain of f. "f(aleph_0)" is meaningless notation.

I am sure you can define things better and provide a better argument. — MoK

Only because my meager knowledge at least includes an understanding of what such concepts as a function, domain, range, value of a function at an argument, sequence, infinite sequence, alephs, etc. mean. Meanwhile, your knowledge about these plainly required things is less than meager.

You need to learn at least the minimal basics of this subject. I mean, you don't even know what "domain of a function" means.

I can only laugh.

An ordinary route:

Prove the existence of a system of naturals, and define a particular system of naturals.

Prove the existence of a system of integers, as there is a subsystem isomorphic with the system of naturals, and define a particular system of integers.

Prove the existence of a system of rationals, as there is a subsystem isomorphic with the system of integers, and define a particular system of integers.

Prove the existence of a system of reals, as there is a subsystem isomorphic with the system of rationals, and define a particular system of reals.

Prove the existence of a system of hyperreals, as there is a subsystem isomorphic with the system of reals, but we cannot define a particular system of hyperreals (at least not with the method of compactness or the method of ultrafilters).

With the real number line, it seems to me it is the line itself that says you cannot. And if you say you can, then it is up to you to show how. — tim wood

As you touch on, the system of naturals can be extended to a system of rationals, and the system of rationals can be extended to the system of reals. And each is a different system. And the system of reals can be extended to a system with infinitesimals. And that is a different system. But when we extend in that way, we do it coherently. We prove the existence of the systems - the sets of numbers and the key operations and relations and the key properties of those). But @MoK's own offering is not coherent - does not proceed by coherent definitions and proofs. He needs to learn at least the basics of the real number system, then he can look up 'non-classical analysis' to see how an actually coherent and rigorous development of infinitesimals is done.

Is f(infinity) a member of the above sequence? — MoK

What do you mean by "infinity" used as a noun?

There is the adjective "is infinite": S is infinite if and only if S is not finite.

And there are various infinite sets, such as:

the least infinite ordinal = {n | n is a natural number} = w = aleph_0

the least infinite cardinal = aleph_0 = w = {n | n is a natural number}

the least infinite cardinal greater than aleph_0 = aleph_1

card(set of functions from w into 2) = 2^aleph_0 = card(the power set of aleph_0)

card(set of functions from aleph_1 into 2) = 2^aleph_1 = card(the power set of aleph_1)

There are various uses of "infinity" as a noun, including such things as infinity as a point in the extended reals or infinity as a hyperreal in nonstandard analysis. But you've not specified.

In any case, how can you ask your question when I explicitly defined the domain of f to be the set of positive natural numbers? (Whatever you mean by "infinity" used as a noun, it is not a member of the set of positive natural numbers.)

how could the sequence be an infinite one? — MoK

How can you ask that question when I defined the domain of f to be the set of positive natural numbers? (The set of positive natural numbers is infinite, and if the domain of a relation is infinite then the relation is infinite.)

You really need to learn basic mathematics.

(Also: A sequence is a certain kind of function, and a function is a certain kind of relation, and a relation is set of ordered pairs. So, f(n) is not a member of the sequence f; rather <n f(n)> is a member of the sequence. f(n) is an entry in the sequence (i.e. f(n) is in the range of the sequence).

And still curious whether you understand now that "aleph_1/(2^aleph_1)" is nonsense.
— TonesInDeepFreeze
Yes — MoK

Actually, No:

I simply divide the interval by 2^infinity — MoK

(1) What do you mean by "divide the interval"? At least four options:

(a) Partition the interval

(b) Divide the cardinality of the interval

(c) Divide the distance of the interval

(d) Deploy an operation that divides, then divides the result ... infinitely many times, as a single combined operation that has a final value

Option (d) is meaningless. You are trying to argue against the proof that for every positive real number x (= the distance of the interval [0 x]) there is a positive real number y such that 0 < y < x. Your confused, ill-premised, ignorant and incoherent idea is that by dividing the interval 2^aleph_1 number of times, we get an i such that for any real number y>0, we have 0 < i < y.

But you have not defined such an operation. All you've done now is to claim you understand my explanation of cardinal division while going right past it to repeat your same mistake, but with even greater vagueness by insisting on an undefined "infinity" as a divisor.

You're not only not making progress, but you're going in the opposite direction of progress.

You really really need to learn basic mathematics.

What I have in mind is that I simply divide the interval by 2^infinity in one step. This operation seems to be invalid though to mathematicians. — MoK

Not a question of "validity". It's just that it is not defined by you. Your method is to throw around mathematical terminology without understanding it, thus to combine it in ways that are not coherent.

You really really need to learn some logic, set theory and the mathematics of standard analysis and non-standard analysis.

Greater than any countable number or greater than any finite number?
— TonesInDeepFreeze

It is better to say greater than any finite number given the definition of a countable set in mathematics. — MoK

Then just say "finite". But your particulars involve both countable and uncountable cardinals, all of which are greater than any finite cardinal but some of which are greater than any countable cardinal:

aleph_0 is countable

aleph_1 is uncountable

2^(aleph_1) is uncountable and greater than aleph_1

I have the first edition in print. I don't have an errata sheet for it. But I do have pencil marks for the errata I caught.

There is an errata sheet for the second edition; I think it might be online too.

Anyway, the second edition online is gold, free for the taking.

The course I recommend is, in order:

(1) Logic: Techniques of Formal Reasoning 2nd ed. - Kalish, Montague and Mar
or either of these:
Introduction To Logic - Suppes
Elementary Logic - Mates

(Lately I've been thinking that Suppes is the best choice, especially for its treatment of the subject of definitions.)

(2) Elements of Set Theory - Enderton (the errata sheet might be online)
or
Axiomatic Set Theory - Suppes

(3) A Mathematical Introduction to Logic - Enderton

(4) Introduction to Mathematical Logic (just for the Introduction chapter) - Church (the Introduction chapter is the best overview of the primary considerations I've found)

For a book with both classical and intuitionistic logic:

Logic and Structure - van Dalen

For an overview of many alternative logics:

An Introduction to Non-Classical Logic - Priest

/

And there are so many other great books, especially Smullyan's books 'First-Order Logic' and 'Godel's Incompleteness Theorems' (Smullyan writes so beautifully and his formulations are so clever and elegant).

Two great tomes:

Mathematical Logic - Monk

Fundamentals of Mathematical Logic - Hinman

I mean, they really are tomes. And they are remarkably rigorous with notation and extensive with details as you're likely to find, especially Hinman which goes the whole nine yards in the way it makes explicit which symbols are in the object language and which are in the meta-language.

And more advanced books:

Model Theory - Chang & Keisler (a tome and the OG of model theory textbooks)

Set Theory - Jech (a tome)

Set Theory - Kunen (I like Kunen's "philosophical/heuristic" framework that tends toward "formalism")

These are all beauties.

Many posts ago I addressed the fact that I am naming an expression and so the definiens is in quotes.

You are giving the name "The Pentastring" to "this sentence has five words". You are not giving the name "The Pentastring" to this sentence has five words. — RussellA

For about the 20th time, as you skip each time, and you still don't get use-mention:

Consider:

The Pentastring is this sentence has five words.

That does not parse because, in this case, the words after "is" form a sentence itself and must be mentioned not used.

[noun] is this sentence has five words

makes no sense.

Instead,

[noun] is "this sentence has five words"

makes sense.

These are correct:

"The Pentastring" refers to "This sentence has five words"

The Pentastring is "This sentence has five words"

These are obviously gibberish:

"The Pentatsring" refers to this sentence has five words.

The Pentastring is this sentence has five words.

You've been making that mistake, and using it in your phony arguments, for a long long time in this thread.

Bigger than any countable number. — MoK

Greater than any countable number or greater than any finite number?

According to TonesInDeepFreeze there are mathematical systems with infinitesimal. — MoK

Don't take my word for it. Look up 'non-standard analysis', 'hyperreal', 'infinitesimal', 'internal set theory'.

Better, read Enderton's beautifully written 'A Mathematical Introduction To Logic' in which he has a wonderfully clear and concise section on non-standard analysis. (It's been a while since I studied it, so I might not be able to immediately answer all questions about the details. Anyway, to get up to the section on non-standard analysis, one needs to first comprehend the material leading up to, starting from the first page, which is the best favor anyone could do for oneself if one were sincerely interested in topics such as this one.)

You didn't provide this argument before. Did you? You just defined infinitesimal! — MoK

That is blatantly false and with an exclamation mark that is a cherry on top of falsehood. I wrote:

One more time: No non-zero real number is an infinitesimal. The proof that no non-zero real number is an infinitesimal is immediate from the fact that for every real number x there is a positive real number y such that y < |x|. We don't need to keep going over this over and over. — TonesInDeepFreeze

This latest time I just made that proof even more explicit. That proof essentially had been given you at least a few times by other posters, but you still didn't understand (chose not to understand?). So I gave it again for you. And then you still skipped it and asked me to give a proof even though I already had. And so I gave it to you even more explicitly. And that explicitness is itself more than you would need, as the matter is so extremely simple to begin with:

No non-zero real is an infinitesimal, since, unlike an infinitesimal, every non-zero real is such that there is real between it and 0.

/

And still curious whether you understand now that "aleph_1/(2^aleph_1)" is nonsense.

How could you have an infinite sequence of divided results without infinite division!? — MoK

An operation symbol takes only finitely many arguments.

For example, the operation of division (x/y) takes only two arguments (x and y), and is defined accordingly:

Df. If y not= 0, then x/y = the unique z such z*y = x.

Every operation symbol, whether, primitive or defined, takes only finitely many arguments. The reason is that every formula is finite in length, so every term is finite in length, so no operation symbol can take infinitely many arguments.

So we have division, which is a binary operation. But we also prove that for every positive real number r, there exists a function f whose domain is the set of positive natural numbers and such that, for every positive natural number n:

f(1) = r
f(n+1) = f(n)/2

f is a function from the set of positive natural numbers into the set of real numbers.

Notice that it is trivial to prove that for no n is it the case that f(n) = 0.

Then we prove:

There exists a unique real number x such that for every positive real number y, there exists a positive natural number n such that |f(n) - x| < y.

Then we write:

lim[n = 1 to inf] f(n)

'lim' is a variable binding operator, but it can be reduced to a regular operation symbol:

Df. If g is a function from the set of positive natural numbers into the set of real numbers, and there exists a unique real number x such that for every positive real number y, there exists a positive natural number n such that |g(n) - x| < y, then Lg = the unique real number x such that for every positive real number y, there exists a positive natural number n such that |g(n) - x| < y

You see that 'L' is an operation symbol that takes only finitely many arguments - in this case, one argument.

The argument itself is an infinite set (an infinite sequence in this case), which is okay, because the operation symbol takes only finitely many arguments - in this case, one argument.

And we prove, regarding the function f we previously defined:

Lf = 0

In everyday parlance, "the limit of f is 0" and for no n is it the case that f(n) = 0.

There's no operation of infinite division (no operation takes infinitely many arguments), but there are infinite sequences that converge to 0, defined by, for example, halving each previous entry in the sequence. But if we call taking the limit of such a sequence f "infinite division" then that's okay as long as we are clear that that is what we mean and not some other undefined notion and that we recognize that for no n is it the case that f(n) = 0.

Moreover, the point is sustained that the fact that f converges to 0 does not refute that no non-zero real number is an infinitesimal.

My fundamental problem is that it is logically impossible to go from knowledge about the content of an expression, such as "The Pentastring is this sentence has five words", to knowledge about something that may or may not exist in the world, such as The Pentastring. — RussellA

"The Pentastring is this sentence has five words"

No one has anything to say about any "content" of that other than that it's gibberish.

These are all in world. They are not merely in the mind of one person:

""The Pentastring" refers to "This sentence has five words"" (stipulated by definition)

"The Pentastring is "This sentence has five words"" (true)

"This sentence has five words" (true, as far as I can tell)

"The Pentastring has five words" (true)

""This sentence has five words" has five words" (true)

"New York City has five boroughs" (true)

""New York City has five boroughs" has five words" (false)

"This guy is in love with you" (true when sung by Herb Alpert about Lani Hall, or by anyone about someone they love)

""This guy is in love with you" has five words" (false)

"The Pentastring is this sentence has five words" (gibberish)

It is logically impossible to go from knowing that "unicorns are grey in colour" to knowing whether unicorns do or not exist in the world. — RussellA

(1) If we interpret as "All unicorns are grey" then usually, we do not infer "There exists a unicorn". (2) but the analogy is irrelevant. I didn't propose a predicate such as 'is a unicorn' and then infer that there is something of which that predicate is true.

Rather, merely, there does exist the expression "This sentence has five words". And I named that expression. I named it "The Pentastring". The expression "I shall return" exists. I can name it "Mac'sPromise" or whatever I want to name it.

There is no logical connection between "This sentence has five words" was named "The Pentastring" and The Pentastring is "this sentence has five words". — RussellA

When we give the name "Y" to X, we then say such things as "Y is X". When you give the name "Buppy" to your dog, you then write "Bubby is my dog". When I give the name "The Pentastring" to "This sentence has five words" I then write "The Pentastring is "This sentence has five words"".

You SKIP the examples:

A puppy was born on August 30, 2024 at 8:00 AM in the house at 100 Main Street in Smalltown, Kansas. That puppy was named "Noorbicks". Noorbicks is the puppy born on August 30, 2024 at 8:00 AM in the house at 100 Main Street in Smalltown, Kansas.

The prop comedian Scott Thompson was named "Carrot Top". Carrot Top is the prop comedian Scott Thompson.

"Thou shall not steal" was named "The Eighth Commandment". The Eighth Commandment is "Thou shall not steal".

"This sentence has five words" was named "The Pentastring". The Pentastring is "This sentence has five words".

As you said:
"London" is a city. (false - "London" is a word, not a city)
— TonesInDeepFreeze — RussellA

Yes.

"This sentence has five words" is an expression. "The Pentastring" is an expression that is a name of the expression "This sentence has five words". The Pentastring is "This sentence has five words".

"All men are created equal" is an expression. "Jeffy'sBigCredo" is an expression that I now use to name the expression "All men are created equal". Jeffy'sBigCredo is "All men are created equal". I could name it "Flookimims" if I wanted to. Flookimins is "All men are created equal". The silliness of that is just to impress the point that names are stipulative and that when we give the name "Y" to X, we then say "Y is X". And contrary to an argument that you revive now, but that had been refuted many many posts ago, such naming does not cause us to assert that certain things have certain predicates other than, of course, they then have the predicate of being named a certain way.

I can anticipate an objection: But if you name your goldfish "Winston Churchill" then you would say "Winston Churchill is my goldfish", which is not true since Winston Churchill was a person, not a goldfish. But that merely reflects that natural languages have ambiguity. Sally Jones is a lab researcher. Another Sally Jones is not a lab researcher. We don't protest that "Sally Jones is a lab researcher" and "Sally Jones is not a lab researcher" is a contradiction, as we recognize that in natural languages names may have different referents. Even in formal languages. The symbol "+" might name standard addition of natural numbers or it might name modular addition for a given modulus. So we recognize that denotation is per an interpretation of a language and that truth is relative to such interpretations.

I would get therapy. — AmadeusD

Your put-downs are lame-o-rama. If I were you, I'd get someone to help me write better insults.

Tones listed some of your errors for your benefit. — Banno

There was only one (combined) error. I mentioned several facets of the matter just to provide an ample understanding of it.

No recognition that you were just blowing smoke when you referenced a source that you didn't even read to see that it says the OPPOSITE of your claim?:

I googled and I found two references about the division of cardinal numbers. You can find the references here.
— MoK

That source itself points out that when the numerator is less than the denominator, there is no definition of numerator/denominator. The very source you point to disputes your claim that (aleph_1)/(2^aleph_1) is properly defined. And you should have proven that for yourself when you first thought of it [here 'X' stands for the Cartesian product]:

K*L = card(K X L). And we have the theorem that if L <= K and K is infinite and L is non-zero, then K*L = K.

The definition of x/y:

x/y = the unique z such that z*y = x. If there is no such unique z, then x/y is not properly defined.

Suppose here that L <= K and K is infinite and L is non-zero:

If L = K, then there are Z such that Z*L = K, but there is no unique such Z. So L/K is not properly defined.

If L < K, then the unique Z such that Z*K = L is L. So L/K = L.

If K/L were properly defined, then K/L would be the unique Z such that Z*L = K. But there is no such Z at all, let alone a unique one. So K/L is not properly defined.

You did an Internet search but didn't even bother to read what you found, instead recommending that other people read it even though, unknown to you since you didn't even read it, it says the exact opposite of your claim that you made up out of thin air.

"(aleph_1)/(2^aleph_1)" is pure bunk. — TonesInDeepFreeze

I looked at all your posts and didn't find the proof that no non-zero real number is an infinitesimal. — MoK

Clearly, you're not reading.

I gave you a proof in the very post to which you have now replied:

The proof that no non-zero real number is an infinitesimal is immediate from the fact that for every [non-negative] real number x there is a positive real number y such that y < |x|. — TonesInDeepFreeze

You need it spelled out for you again?

Definition: x is an infinitesimal if and only if, for every positive real number y, |x| < y.

Theorem: No non-negative real number is an infinitesimal.
Proof: Suppose x is a non-negative real number. Since x is a non-negative real number, |x|/2 is a positive real number and |x|/2| < |x|. So it is not the case that |x| < |x|/2|. So it is not the case that for every positive real number y, |x| < y. So x is not an infinitesimal. So no non-negative real number is an infinitesimal.

The distance between two points is zero if the number of divisions is strictly infinite so there cannot be a point between two points in this case. — MoK

You skip that I addressed that. There is no operation of infinite division in the real number system. There are infinite sequences of divided results, and the one you have in mind converges to 0. It is very crank (confused, ignorant and fallacious) to conflate the limit of a sequence with an out-of-thin-air claim of an operation of infinite division. And the fact that the limit of the sequence is 0 does not refute that between any two distinct real numbers there is real number strictly between them. Indeed, the convergence to 0 itself depends on the fact that between any two distinct real numbers there is a real number strictly between them.

Followup with the liar bot:

I wrote:

"That's wrong. soundness does not follow from completeness" - TonesInDeepFreeze

ChatGPT replied:

"You're correct. I misspoke. The soundness and completeness theorems for first-order logic are indeed distinct results, and soundness does not follow from completeness.

Here’s a more accurate description:

Soundness Theorem: This theorem states that if a formula can be derived using a formal system (i.e., it is provable), then it is also valid (true in all models). The proof of soundness generally involves showing that the rules of inference in the formal system preserve truth.

Completeness Theorem: Godel's completeness theorem, proven in 1930, states that if a formula is valid (true in all models), then it is provable within the system.

Godel proved the completeness theorem, but the soundness theorem was established in the context of formal logic as well, often attributed to earlier foundational work in formal systems, including work by other logicians and formalists who developed and refined these ideas. Soundness was addressed as part of the broader development of formal systems and logical proofs." - ChatGPT

/

And that is still a quite incorrect statement of the theorems. And slop in the summation paragraph too.

"Godel’s completeness theorem, which he proved in 1930, shows that if a formula is semantically valid, then it is provable within the system, while the soundness theorem follows as a corollary: if a formula is provable, it is valid." ChatGPT

That's great. ChatGPT caught, in flagrante, lying about the pair of central theorems of mathematical logic.

The soundness theorem does not follow as a corollary of completeness.

Trivially, an inconsistent system is complete and not sound!

It is SO discouraging to look at the Internet.

Not authoritative, of course — Banno

Time Magazine's Understatement of the Year

(1) It is possible that Godel first proved the soundness of the predicate calculus. I mentioned in my post that I don't know who first proved it.

(2) Source other than an AI bot?

(3) If G |-P then G |= P is soundness not completeness. Even if Godel was the originator of the soundness proof, still soundness and completeness are different, being converses of each other.

/

I have no comment on discussion of "existence is not a predicate" in this thread. But, whether or not this suits the context of that discussion: In the most basic sense of predicate logic, of course existence is not a predicate, but existence is formulated as a predicate in certain modal logics.

Of course, I never took you to be stating that it is your own view that Russell's paradox undermines the basis of all mathematics.

My point is that if 'undermines' is taken in more than a quite weak sense, I don't see many, or even a significant number of mathematicians or philosophers (if even any by the time of the systems that came out shortly after the turn of the century) regarding Russell's paradox as undermining the basis of all mathematics. Or are there logicists who said that if logicism fails then there is no, or very little, basis for mathematics?

By the way, Russell's paradox and the liar paradox are, of course, closely related, but they are also very much not identical. (Though, the barber paradox is just an anecdotal rendering of Russell's paradox.)

I have no problem with "this sentence has five words" being named "The Pentastring". In other words, "The Pentastring is this sentence has five words". — RussellA

No, only in your own ridiculous words.

Consider:

The Pentastring is this sentence has five words

That's gibberish.

Consider:

"The Pentastring is this sentence has five words"

That's quote marks around gibberish.

Consider:

The Pentastring is "This sentence has five words".

That's a sensical statement.

Consider

"The Pentastring is "This sentence has five words""

That's quote marks around a sensical statement.

No, that's the converse of the completeness theorem.

You're right. My silly mistake. I made a note of the edits in my post. Thanks.

From the Wikipedia article on Formalism (Philosophy)
Formalists within a discipline are completely concerned with "the rules of the game," as there is no other external truth that can be achieved beyond those given rules. — RussellA

An example of Wikipedia promulgating sloppy misinformation.

What is called 'game formalism' or 'extreme formalism' regards mathematics as merely execution of rules for strings of symbols. But formalism in the philosophy of mathematics has variations that are not game formalism. Indeed, it seems that game formalism is not widely accepted while other forms of formalism have more acceptance.

This goes along with the fact that, contrary to the Internet (and even printed) meme, Hilbert did not say that mathematics is just a game played with symbols, as indeed Hilbert did very much view finitistic mathematics as contentual and infinitistic mathematics as applicable even though ideal.