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Thread: WLC is Wrong: The Argument for Actual Infinities

1. Re: WLC is Wrong: The Argument for Actual Infinities

EDIT: I corrected a confusing point raised by Clive below. Originally Posted by MT
The general reason for my taking down this side track, is that If in the end you wish for an answer to be given in mathematical terms, then I know I will not be able to do so ,because of my personally lacking proficiency with that kind of expression. I will be limiting(and limited) my answers to a more philosophical expression of it. So if that form is inherently not sufficient(even if it were correct), then I would respectfully have to withdraw from the thread. To the extend that you are willing to bare with that limitation.. I will gladly continue on. I have enjoyed the discussion, but I don't want to waste your time if I have no intention of fulfilling your basic request.
I don't care if you don't formally prove statements, but I don't believe that there's an honest way to have this conversation without you being more explicit than you are below. I don't care if you spell out your reasoning here with words or equations, but I do need the reasoning to be explicit and not implicit. In other words, I need you to try to spell out what the operations are that you're writing. Originally Posted by MindTrap028 Infinity - infinty = X

It is true that X = 1, and X = 2
(In fact, X can equal any amount ).
1.) This is vague, and by this I mean that your definitions and reasoning are essentially entirely implicit rather than explicit. You indicate that X = 1 or 2, so I can only assume that you mean that X is a counting number (1,2,3,..., etc, the numbers that can be used to count the number of objects in finite sets). I am assuming then that you mean that "infinity" in the expression "infinity - infinity" is meant to be something that is used to count the number of objects in infinite sets, too. In other words, "infinity" is "aleph-null", the name for the "size" of infinite sets. In which case, I am forced to assume that you mean "-" is supposed to be related to a kind of set difference. I am going to assume these things for now, please correct me if this isn't the case. Firstly, you appear to have an order of operations error. Let's take two infinite sets to illustrate this.

X = {1,2,3,4,5,6, ...}
Y = {2,4,6,8,10,12, ...}

So, let's denote the size of the sets to be given by Size(X) = the size of X. You seem to want to demand something like:

Proposition: "For all a sets U with a subset V, then Size(U - V) = Size(U) - Size(V)."

This isn't a valid theorem; it holds for finite sets, but it doesn't hold for infinite sets. Using the infinite sets defined above, you seem to be arguing something along the the lines of:

Size(X - Y) = Size({1,3,5,7,9,....}) = Size(X) = aleph-null

Now, the crucial failure in your reasoning is that you're assuming that we can do simple arithmetic (addition, subtraction, and so on) with aleph-null, and that contradiction in doing ordinary arithmetic results in us concluding that aleph-null is itself contradictory. But that is incorrect because no one is saying that you can do arithmetic with infinity (add, subtract, etc) in terms of just playing around with the sizes of the set.

So you want us to accept that:

Size(X - Y) = Size(X)

is a contradiction, because you're going to apply (the incorrect) formula on the left-hand side of the equation to get:

Size(X) - Size(Y) = Size(X)

Subtract a Size(X) from both sides to yield:

==> Size(Y) = 0

AND have us assert

Size(Y) = Size({2,4,6,8,....}) = aleph-null =/= 0

And THEN, yes, we would derive a contradiction.

The problem is straightforward:

1.) Size(X - Y) =/= Size(X) - Size(Y) in many instances, including this one. Crucially because:
2.) Aleph-null does not have arithmetic properties; the expression "Size(X) - Size(Y)" is undefined. In other words, you aren't allowed to subtract aleph-null (the size of X) from both sides to get the equation Size(Y) = 0. Aleph-null isn't a counting number, there isn't a way to subtract it, and therefore the proof is erroneous.

2.) Secondly, the valid part of your reasoning seems to stem from the intuition described in the following scenario. The part that may disturb you if you try to do what you're doing and boil everything down to just the size of an infinite set (again, you'll run into similar conceptual problems even for finite sets) is that you think that:

Size(A - B) = 0 <==> A = B.

This is almost true. We need a small clarification, and then it's absolutely true.

In other words, the correct formulas that your intuition seems to be trying to nail down are:

Def: Let "B is the subset of A" mean that "There are no objects inside of B that aren't also in A." (Note: This doesn't exclude B=A, it just checks that there's nothing in B that isn't also in A.)

Theorem: "If B is a subset of A and Size(A-B) = 0, then A = B."

Theorem: "If B is a subset of A, and Size(A-B) =/= 0, then A =/= B."

In set theory, you can prove that this statement is true. And guess what? Infinite sets obey this property.

But what is not true is that:

Non-Theorem: "If B is a subset of A, and Size(A) = Size(B), then B = A."

If A is finite, then B is finite, and then Size(A) is a natural number, in which case operations like "plus" and "minus" are defined. Then, and only then, the proof give in part (1) becomes valid because "Size(A) - Size(B)" is now defined because Size(A) is a natural number and Size(B) is a natural number, and we can define subtraction between the sizes. But if either A or B were infinite, then the formula "Size(A-B) = Size(A) - Size(B)" not only isn't true, it doesn't mean anything because it's not defined.  Reply With Quote

2. Re: WLC is Wrong: The Argument for Actual Infinities

Size(A - B) = 0 <==> A = B.
Technically, |A-B| = 0 <==> A - B = 0 (in the sense of empty set) <==> A \subseteq B, but it looks like you're assuming B \subseteq A, in which case you'd have equality.  Reply With Quote

3. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by CliveStaples Technically, |A-B| = 0 <==> A - B = 0 (in the sense of empty set) <==> A \subseteq B, but it looks like you're assuming B \subseteq A, in which case you'd have equality.
That's why I specified that point. I had other versions of this post that were much longer, including a sentence about this, but as it is pretty long, I decided to just give the right definition and call that correct.

It's interesting, if you frame the set theory operators very carefully, you can get ordinary algebra, which makes sense, but it exclusively works for finite sets. Which also makes sense; the ordinals are not an group under addition. The key place where this fails, of course, is when |A-B| = aleph-null. Actually, maybe a better way to say this is that I think MT is trying to equate ordinal numbers and cardinal numbers. So ordinals form a ring-like structure, but cardinal numbers don't. Thus algebra (rather, MT's heuristics for algebra, but they are probably correct) cannot be done with them. So if you equate them, then you'll derive nonsense with your algebraic intuitions. It might be interesting to try to rephrase his argument to be about sequences and try to show him how ordinal arithmetic works. I'm personally not deeply familiar with it, are you? I could learn about it, I suppose, it is a pretty interesting subject, and I want to learn about resurgence theory a bit (Although from a physics perspective more than a math one, but it'll still be useful).

The third notion of infinity is as a boundary of the naturals/reals, which again you cannot confuse with the other two forms of infinity. This is definitely something that confused me back when I had only a heuristic understanding of infinity for calculus. I spent some time trying to put the boundary of the reals in some kind of algebraic context, i.e. just as the ordinals seem to have, which doesn't really work.

Yeah, viz. Wikipedia:

"A natural number (which, in this context, includes the number 0) can be used for two purposes: to describe the size of a set, or to describe the position of an element in a sequence. When restricted to finite sets these two concepts coincide, there is only one way to put a finite set into a linear sequence, up to isomorphism. When dealing with infinite sets one has to distinguish between the notion of size, which leads to cardinal numbers, and the notion of position, which is generalized by the ordinal numbers described here."

This is why I think he keeps on asking about positions when he's actually referring to sizes. I think he's conflating ordinals (where the position of the object in the sequence is) and cardinals (how large the sequence is up to that object). Because these diverge for infinite sequences, but are equivalent for finite sequences, he's applying, as I said, an algebraic intuition for set equality on the cardinals but using rules that only generalize to the ordinals. And then he's concluding (correctly) that ordinal numbers and cardinal numbers aren't equivalent, but he's interpreting this as the statement that "infinite sequences/sets are contradictory."

So I think his confusion is that he doesn't understand HOW ordinals and cardinals could ever be allowed to be different, where infinite sets are precisely the case where this occurs. I think you made this point before, but I'm pretty sure now that this is the precise ordering to intuitions that is leading MT astray.  Reply With Quote

4. Re: WLC is Wrong: The Argument for Actual Infinities

Infinity - infinty = X

It is true that X = 1, and X = 2
(In fact, X can equal any amount ).

So, either the above is not true, or you were incorrect to say that showing X= 1 and X = 2 is a disproof.
Well, technically you haven't proved that "X can equal any amount" (which is false; |A-B| can equal any cardinality <= |A|, and not any cardinality > |A|). You just said it. But it's been proved elsewhere in the thread (perhaps not in this strong form), so there's not really a reason to quibble here.

In any case, this example does show that a statement like "infinity - infinity = x" can only be true if the "=" symbol is no longer transitive, i.e. a=b and a=c doesn't imply b=c. If you require that the "=" symbol designate a transitive relation, then "infinity - infinity = x" must be false.

However, this does not show that actual infinities are impossible. To do this, you'd need an argument like this:

(1) If there are actual infinities, then "infinity - infinity = x" must be true, and "=" must be transitive. [Premise]
(2) "infinity - infinity = x" is false, or "=" is non-transitive. [Premise]
(3) Therefore, there aren't actual infinities. [(1),(2), modus tollens]

(1)-(3) is a valid argument, so the only question is whether the premises are true. (2) is true, as you've shown. So what about (1)? What is your argument that (1) is necessarily true?  Reply With Quote

5. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by CliveStaples (2) Why do bijections for infinite sets work differently than for finite sets vis-a-vis proper subsets?

To (2):

If you take an infinite set like N = {0,1,2,3,...} (if you don't like this way of writing down sets, you could also define this as the collection of all constructible non-negative integers) and remove, say, the element 0, you're left with N - {0} = {1,2,3,...}.

There's an obvious 1:1 correspondence between these sets, given by f(n) = n-1 for all n in N-{0}. We know this is a map because of how it's define (it takes as inputs elements of N-{0}, and each of its outputs lie in the set N, and no input is assigned two different outputs). We know this map is a bijection / 1:1 correspondence because we can show that it is both injective (f(n) = f(m) implies n=m) and surjective (for all y in N there exists an x in N-{0} such that f(x) = y).
I think you did a great job explaining the difference (or lack thereof) between the two, thank you Clive. It is the quoted paragraph though that I think highlights the objection. You’ve removed an element {0}, decreasing the cardinality of the set N by 1. Yet, we know that the cardinality has not decreased by one because there is a bijection between N, and N-{0}. It would certainly seem to be a contradiction to be able to hold both positions; “I have reduced the cardinality of the set (transliteration of N - {0})” and “I have not reduced the cardinality of the set (transliteration of f(n) = n-1 for all n in N-{0}.”

Essentially, this is a very round about manner of coming to the same point that Inf-1=Inf. Now this kind of behavior is appealed to as a function of infinite sets, which is another way of restating the conclusion that inf-1=inf. It doesn’t go far to explaining why it is coherent in an infinite set when it wouldn’t be in a finite set or why we should accept such behavior rather than concluding the alternative premise that such a set is incoherent in actuality. Originally Posted by Clive
That, rather than there being some particle that is described by that property, none are. Cf. there is no person described by the property "is currently King of the United States
Which would seem a problem given the nature of the setup in the OP. There is a linear ordered group of particles all of which are can be taken at t=0 to be at a distinct place at a distinct time.

Each particle exists, none are coming into existence, nor is this a process of particle generation. Thus, every particle in that series exists and exists at a definitive point.

There is a boundary on that grouping of particles such that beyond the boundary no member of the set of particles exists.

If we were to introduce a particle moving toward point 1 from point 3, we can ask a valid question, “what particle would it encounter first?” This question is certainly reasonable within the boundaries of physical law, and is coherent within the universe described in the OP.

The answer would seem to be, still, “there isn’t one.” IE in the system described it is impossible for a particle traveling towards the infinite set described to interact with the set itself. Originally Posted by GP
Ok, but given your metric of how mass is related, the mass of the ‘infinitieth’ particle is zero, which would seem to be an issue with your hypothetical. Now I think an objection you might offer would be that “infinity is not a number.” Perhaps, but it doesn’t matter here, we can ask what the mass is of the particle closest to the position “2,” and given the relationship you’ve described, it would be zero since there are in your hypothetical an infinite number of particles.

Now maybe you would argue that you either weren't aware that if such a particle existed, it would have to exist at a specific point on the continuum and thus you couldn't know that you were comparing two points (I don't think you want to argue that you misunderstood this), or perhaps you'd like to argue that you simply were ignorant of the fact that there doesn't exist a natural number, N, such that ([1 - (1/2)N] == 1 (I don't think you want to admit this either). In fact, I think you might just want to accept the criticism and move on, Squatch, because there's no version of this scenario where you didn't bungle 200-level mathematics.
I re-read this a couple of times to try and see where the error might be, unfortunately, it seems only to have been in, what I can only assume, was your hasty reading of my response.

I asked (in the section you quoted) what the mass of the particle closest to the position “2?” Now maybe you remember the OP differently, but there is no particle at position “2” right? Thus the point 2 isn’t in the continuum you defined. I think you realized this at the time, which is why you defined the continuum as [1,2). And when you said: “i.e. no particle sits at point "2", just arbitrarily close to it, on the interval known as [1,2) by mathematicians”

Rather, I was clearly trying to define a relatively simple question in two different manners to allow you to approach it from whatever angle you felt most comfortable. A question, it seems, destined not to be answered. Originally Posted by GP
They have been easily dismissed, Squatch. Other than the instances where you naysay me and then ask me fill in the details for you, the only valid line of argumentation you've made was rebutted…Here's my quote, referenced to you for the eighth time
Repeatedly offering irrelevant quotes over and over again doesn’t make them any more persuasive. As I noted back on the first page, pointing to possible configurations and modes does little to verify where physics accepts an example of the number of existent, actual things as infinite. To say that there are an infinite number of ways to arrange X and Y does little to show that X and Y actually are arranged in an infinite number of ways. Originally Posted by GP
A clever attempt to engage in shifting the burden of proof, but I'm not biting. I am not saying it is necessarily true that point particles are fully consistent with our understanding of the universe. What I am saying is that WLC's argument only passes muster if it can be applied very generally
This seems to disagree with the nature of your OP.

WLC’s argument is to imply that the concept of an infinite number of sequential events is incoherent. He lays out a series of arguments as to why this would be the case.

You object to one of them by trying to posit a counterexample.

But a basic tenant of argument by counterexample is that you need to show that the counterexample fits the premise described.

For example, if I wanted to use a counter-example for the rule that all squares are rectangles by displaying a picture of a circle, I would need to show that the circle fits the two criteria met:

1) It is a square.
2) It is not a rectangle.

Here, all I am asking for is for you to do the same. WLC’s argument is about this universe, so it is a basic prerequisite of the argument type you are using to show that your counterexample comports with this universe, is infinite, and is an actual infinite. Originally Posted by GP
Two problems with this section. 1) I didn’t argue or ask for you to support that QFT comes from statistical mechanics, so this seems like a bit of a red herring. 2) This is linkwarz. Just pointing to a link and saying “read this” or referencing a term with no explanation of how it supports your arguments is unproductive. Originally Posted by GP
Alternatively, there are attempts to work with causet theory (cutting spacetime up into infinitesimal points) or to try to push the de Broglie-Bohm theory of point-particles into a QFT regime (N.B. Here they keep the point-particle interpretation). As stated, neither of these are promising (very, very far from it), but they are not demonstrably wrong because they haven't been completed as programs and thus haven't made predictions yet (and will probably remain that way).
Interestingly, both examples you are offering are exactly what I pointed to in my initial question about point particles. Neither theory, at least from what I understand or what you’ve shown defines, these particles as literally point particles. Rather, they use points as approximations when the physical extent of the particles is either irrelevant (as with causet theory) or covered elsewhere (BB). In Causet Theory, for example, the elements of spacetime are treated as points because their size is irrelevant. All that matters is their causal and temporal order. If anything there is a reference to these particles having an extent, or at least a minimum size given that the causal order must occupy a light cone. So if we could place these infinitesimally small points infinitely close together, no light cone discussion would be necessary. Originally Posted by GP
then additionally defended causet theory[/URL] (which involves cutting spacetime up into an infinitesimal points, so the degrees of freedom are necessarily point-like) a year ago as plausible and well-founded programs, and you did it for an entire 8 months
Actually, you’ll note that I wasn’t necessarily defending causet theory. I was pointing to causet theory (or rather reacting to Clive’s reference to it) as an example of successive addition. Originally Posted by GP
Squatch, it's already been explained to you both simply and in full technical glory why this isn't the case and that there is no "infinitieth" particle. If you don't understand that there isn't an "infinitieth" particle, then that's fine. But you should be aware that if you continue to assert that there's an "infinitieth" particle, you should know I'm just going to repeat that this argument is a strawman because there does not exist an "infinitieth" particle with zero mass in my scenario.
I don’t recall where it was described in “full technical glory.” Can you please link that response?

Further, I’m more than willing to agree that infinity isn’t a number, which is what you seem to be getting at above. But then how do you explain the statement in your OP: “Suppose this hypothetical universe starts out with an infinite number of particles?” If there are an actual infinite number of particles, how is there not an “infinitieth” particle?

My point is that you seem to be playing a bit loose with your terminology, which allows you to treat infinity as a limit in one sense, but as an existing number in another. Originally Posted by GP
Fine, so there is a point-like particle contained in Bohmian mechanics?
Well no. At least not in the manner you imply in your OP. There is a point like particle and a wave function. Originally Posted by GP
But the wavelength of a particle being the literal the size (uncertainty) of a particle is what de Broglie-Bohm theory is explicitly avoiding in order to remain deterministic. It preserves the uncertainty principle as a statistical form of uncertainty, but it keeps each of the particles deterministic and thus breaks the wavelength-momentum relationship imposed by Fourier analysis.
While I appreciate the slew of wiki links, they don’t seem to support either your statement that “But the wavelength of a particle being the literal the size (uncertainty) of a particle is what de Broglie-Bohm theory is explicitly avoiding in order to remain deterministic” nor the underlying claim that under a BB interpretation particles can placed infinitely close together. Can you offer direct support of either claim?

This interpretation you are using also seems to run into another, related problem. Given that you have localized the particles by confining their locations to arbitrarily small points, you would seem to be restricting the de Broglie wavelength (the possible range of locations the particle could show up in, https://en.wikiversity.org/wiki/De_Broglie_wavelength).

Now, given that de Broglie wavelength and momentum are inversely related, it would seem that as we move to a more and more restricted location for each successive particle, we would need increase the particle’s momentum. That would seem to be a problem with your OP, given that the momentum of each successive particle must be lower in your system.

We could approach this from the opposite angle as well.

You define the mass of each particle in post 10 as: mn = m0 (1/n)2

In the OP you define the velocity of each particle as: vn = (1/n)2

Meaning the momentum (m*v) of each particle is: m0 * 1/(n4)

Given that de Broglie wavelength is: lambda = hbar/momentum we would see that your wavelength is increasing for each successive particle (http://www-physics.ucsd.edu/students...reofMatter.pdf
).

Therefore at some point your wavelength exceeds the range of possible positions defined in your OP for that particle, which is a problem given your statement that “ each particle is also never lying on top of one another; for each n, each particle is at a distinct point.” Originally Posted by GP
Sure, and I've said that I do not believe (outside of a conformal field theory) that you can stack interacting particles into an infinitely
This seems to have been cut off in your response. Could you clarify? Originally Posted by GP
I haven't answered the question. Right now, I'm saying "I don't know." I'm tired of being asked to full in the details of your vacuous speculations, so if you want to prove that there is a problem, then start do some math and show me that there's a problem. If you think the "last" particle contributes the strongest part of the force, them please demonstrate this. GP, this is, after all, your OP and position to defend. It seems odd to become so exasperated over needing to defend a position you offered unbidden.

As for the challenge, I’ve already demonstrated this in earlier posts. A particle moving from point 3 into the interval you describe would seem to never encounter a first particle. You invoked “field effects” in an attempt to resolve that problem, I’m simply asking you to defend your rebuttal. Originally Posted by GP
However, in the real world there's a compensating electrostatic force that repels particles, which tend to balance each other (hence why you're not flying up into the air due to electric forces or flying down to the earth due to gravity). So you'd probably want to include a repulsive force, as well, so that might be avoided. Still, it's not obvious to me that you can't use a delicate balancing to charges and masses to compensate each other to yield finite energy, momentum, forces, and so on, to prevent the system from pathologies.
Perhaps, it is your OP, and you were the one invoking field forces into the system you described. “I don’t see why not” is not much of a defense. Do you have a defense of your claim that field forces within the scenario you describe are coherent? Originally Posted by GP
I wouldn't (I'm just claiming that the energy and momentum is finite). But you're the one claiming that this system is necessarily impossible/there's something necessarily pathological about this scenario.
Well, this seems a bit like shifting the burden. It is your scenario. Field effects are your defense for an initial objection. It would seem pretty obvious that you would, at least, need to put a modicum of effort into actually forming that defense rather than just appealing to it.

Remember, these effects were brought up to reject the objection that a particle travelling from point 3 into your interval apparently never encounters a first particle. Your point was that particles don’t actually strike each other. Fine, if a little off the mark, how does the fact that they interact via field effects resolve the issue described then?  Reply With Quote

6. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by Squatch347 I think you did a great job explaining the difference (or lack thereof) between the two, thank you Clive. It is the quoted paragraph though that I think highlights the objection. Youve removed an element {0}, decreasing the cardinality of the set N by 1. Yet, we know that the cardinality has not decreased by one because there is a bijection between N, and N-{0}. It would certainly seem to be a contradiction to be able to hold both positions; I have reduced the cardinality of the set (transliteration of N - {0}) and I have not reduced the cardinality of the set (transliteration of f(n) = n-1 for all n in N-{0}.
It seems you're using two different notions of "cardinality reduction". In the first sense, you mean only that you've removed an element. In the second sense, you mean that there's a bijection between the set after the element is removed and the set before the element is removed. So there's an equivocation.

So it seems perfectly sensible to say "Under one definition of 'cardinality reduction', there has been a cardinality reduction. Under a different definition of 'cardinality reduction', there has not been a cardinality reduction." There is only a contradiction here if the first definition implies the second definition. Is that what you're alleging?

Essentially, this is a very round about manner of coming to the same point that Inf-1=Inf. Now this kind of behavior is appealed to as a function of infinite sets, which is another way of restating the conclusion that inf-1=inf. It doesnt go far to explaining why it is coherent in an infinite set when it wouldnt be in a finite set or why we should accept such behavior rather than concluding the alternative premise that such a set is incoherent in actuality.
I mean, I don't know what you're stuck on. Here's a wikipedia page with much of the relevant definitions. Which of the following do you dispute:

(1) There exists a bijection between N and N-{0}

(2) There exists a bijection between N and N U {1.5}

(3) For cardinals x,y,z: x - y = z iff z + y = x

(4) For cardinals x,y: For two disjoint sets A,B such that |A| = x and |B| = y (if A and B are not disjoint but |A| = x and |B| = y, then we can replace A with A x {1} and B with B x {0} and have disjoint sets; the existence of A and B can be shown by constructing the ordinals, which can be done without choice by using Hartogs numbers), x+y := |A U B|

(5) For cardinals x,y: x < y iff for every set A,B such that |A| = x and |B| = y, there exists an injection f:A -> B but there exists no injection g:B -> A

(6) If A is finite, then there is no injection from A U {1} to A

(7) If A is infinite, then there is an injection from A U {1} to A

(8) If A is finite, then |A U {1}| > |A|

(9) If A is infinite, then |A U {1}| = |A|

(10) If A is finite, then |A| + 1 > |A|

(11) If A is infinite, then |A| + 1 = |A|

(10) and (11) follow from the given definitions. It would be a contradiction to say that (10) is both true and false (with all the symbols and words defined the same way) or that (11) is both true and false. But this is not what is being claimed; rather, the claim is that (10) is true and (11) is true.

The intuitive reason why maps from A U {1} to A fail to be injections when A is finite is because you "run through" all the elements in A, and there's no element of A 'left over' to assign to 1.

The intuitive reason why some maps from A U {1} to A are injections when A is infinite is because you can "set aside" an element of A to assign to 1 and still have 'enough' left over to assign to all the elements of A.

If you'd like to see some mathematical proofs:

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Let A be a set with |A| = n for some n in N = {0,1,2,...}. If n=0, then A is the empty set, A U {1} = {1}, and there are no maps f: {1} -> emptyset. In particular, there are no injections from {1} to A.

If n > 0, then there exists a bijection f: A -> {1,2, ... ,n}. Then the map g: A U {1} -> {1,2, ... , n , n+1} where g(x) = f(x) for x in A, and g(1) = n+1 is a bijection. If there is an injection h: A U {1} -> A, then f o h o g-1 is an injection from {1, 2, ... , n, n+1} to {1,2,..., n}.

(lemma) For all n in N+, there is no injection from {1,2,...,n,n+1} to {1,2,...,n}.

(Base case) n=1
There is only one map f:{1,2} -> {1}, and f(1) = f(2) = 1, so f is not injective.

(Inductive case)
Suppose the lemma holds for some k >= 1.
Then every map from {1,2,...,k, k+1} to {1,2,...,k} is not injective.

Consider g: {1,2,...,k, k+1, k+2} -> {1,2,...,k+1}.

Suppose g is injective. Then either k+1 is in g({1,2,...,k,k+1}) or it isn't. If it isn't, then we have an injection from {1,2,...,k,k+1} to {1,2,...,k}, a contradiction.

So k+1 is in g({1,2,...,k,k+1}). Since g is injective, there is a unique x in {1,2,...,k,k+1} such that g(x) = k+1. Additionally, g(k+2) != g(y) for any y in {1,2,...,k,k+1}. Thus g(k+2) != k+1. So g(k+2) is in {1,2,...,k}. We can then construct h:{1,2,...,k,k+1} -> {1,2,...,k} where h(a) = g(a) for a != x, and h(x) = g(k+2). Then h is an injection, which is a contradiction.

Thus g is not injective.

Thus there is no injection from A U {1} to A.

(9)

If A is infinite, then there is a bijection f: A x A -> A by Zermelo's theorem (which requires the axiom of choice). Since |A| > 1, we can find x,y in A with x != y. There is an obvious injection from A U {1} to A x {x} U A x {y}, namely g(a) = (a,x) for a in A, and g(1) = (y,y). Since A x {x} U A x {y} is a subset of A x A, g can be extended to be an injection from A U {1} to A x A. Then f o g is an injection from A U {1} to A. Since clearly there is an injection h:A -> A U {1}, by CSB theorem there exists a bijection between A and A U {1}, so |A U {1}| = |A|.

Perhaps the "incoherence" is that, under the given definitions, you could have cardinalities x,y,z such that x + y = x + z but not y = z. This is because cardinal addition is not always left-cancellative. But there's no contradiction here; there are plenty of definable operations on a set that fail to be left-cancellative, or non-commutative, or non-associative, or whatever. Left-cancellation is a sort of precondition for having a left inverse (or for an element being embeddable in a larger structure in which it has a left inverse). But there's no contradiction here, i.e. there aren't elements x,y,z such that x+y = x+z and y = z and y != z.

Which would seem a problem given the nature of the setup in the OP. There is a linear ordered group of particles all of which are can be taken at t=0 to be at a distinct place at a distinct time.

Each particle exists, none are coming into existence, nor is this a process of particle generation. Thus, every particle in that series exists and exists at a definitive point.

There is a boundary on that grouping of particles such that beyond the boundary no member of the set of particles exists.

If we were to introduce a particle moving toward point 1 from point 3, we can ask a valid question, what particle would it encounter first? This question is certainly reasonable within the boundaries of physical law, and is coherent within the universe described in the OP.

The answer would seem to be, still, there isnt one. IE in the system described it is impossible for a particle traveling towards the infinite set described to interact with the set itself.
Sure, the question is reasonable--i.e. it can be formulated and isn't nonsense. It isn't clear that the response "There isn't one" is at all problematic, except to your particular intuitive understanding.  Reply With Quote

7. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by Squatch347 I re-read this a couple of times to try and see where the error might be, unfortunately, it seems only to have been in, what I can only assume, was your hasty reading of my response.

I asked (in the section you quoted) what the mass of the particle closest to the position 2? Now maybe you remember the OP differently, but there is no particle at position 2 right? Thus the point 2 isnt in the continuum you defined. I think you realized this at the time, which is why you defined the continuum as [1,2). And when you said: i.e. no particle sits at point "2", just arbitrarily close to it, on the interval known as [1,2) by mathematicians

Rather, I was clearly trying to define a relatively simple question in two different manners to allow you to approach it from whatever angle you felt most comfortable. A question, it seems, destined not to be answered.

[...]

Further, Im more than willing to agree that infinity isnt a number, which is what you seem to be getting at above. But then how do you explain the statement in your OP: Suppose this hypothetical universe starts out with an infinite number of particles? If there are an actual infinite number of particles, how is there not an infinitieth particle?

My point is that you seem to be playing a bit loose with your terminology, which allows you to treat infinity as a limit in one sense, but as an existing number in another. Originally Posted by Squatch
Repeatedly offering irrelevant quotes over and over again doesnt make them any more persuasive. As I noted back on the first page, pointing to possible configurations and modes does little to verify where physics accepts an example of the number of existent, actual things as infinite. To say that there are an infinite number of ways to arrange X and Y does little to show that X and Y actually are arranged in an infinite number of ways.

You are not engaging with the arguments that I've given you because you don't understand the even the very basics of the source material, despite your protestation to the contrary. It is absurd that you think modes are "not extent, actual things" or that "an actual infinite number of particles" means there's "an infinitieth particle." At least, it's farcical that you can fail to understand issues like these but still think that you can still create a coherent argument regarding these subjects. If you were more mathematically adept or if you were more adept with the physics, then I could work with that. But I can't see any point in re-repeating what I've already said, so I'll leave it here.  Reply With Quote

8. Re: WLC is Wrong: The Argument for Actual Infinities

An interjection... Originally Posted by GP(..SPEAKING TO SQUATCH)
Squatch, it's already been explained to you both simply and in full technical glory why this isn't the case and that there is no "infinitieth" particle. If you don't understand that there isn't an "infinitieth" particle, then that's fine. But you should be aware that if you continue to assert that there's an "infinitieth" particle, you should know I'm just going to repeat that this argument is a strawman because there does not exist an "infinitieth" particle with zero mass in my scenario.
Well, there most certainly is an "infinitieth" particle... namely the fist one when counting backwards.

Counting down, as with an infinite past, or counting backwards of the OP's set of particles.
(from infinity).... 10, 9 , 8, 7, 6, 5, 4, 3, 2, 1..

There are an infinity number in the set. (cardinality...is that right?), and 1 is the "infinith" member of that set.
The observable fact is that we can only observe the infinith set by counting down.

So it would seem to me, that any set that has defined limits on either end.. must have an "infinith" member.

Squatches observation is that on a measuring stick (from 1-2) that is given infinity points, but is limited to not reaching 2, when we know that 2 does meet, then there must also be a "infinith" particle on that end.. else the theory does not reflect a real world.
In other words, if the theory precludes meeting, when we observe a meeting, then the theory violates whatever law that observation of actual meeting reflects.

-----
In terms of the hotel.
This is to say, that if there are two hotels with a front door on opposite ends, and each with an infinit number of rooms. The theory of infinit sets, would suppose that there is no room of one hotel adjacent to a room from the other hotel, even though they are part of the same line (IE observably meet). The theory contradicts observation, and necessity.

------------------------------
I don't think the above would make an actual infinit impossible, it just wouldn't be possible in worlds that have real things that meet in actuality.

I know this is an interjection.. and I am falling way behind on my responses.  Reply With Quote

9. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by MindTrap028 An interjection...

Well, there most certainly is an "infinitieth" particle... namely the fist one when counting backwards.

Counting down, as with an infinite past, or counting backwards of the OP's set of particles.
(from infinity).... 10, 9 , 8, 7, 6, 5, 4, 3, 2, 1..

There are an infinity number in the set. (cardinality...is that right?), and 1 is the "infinith" member of that set.
The observable fact is that we can only observe the infinith set by counting down.

So it would seem to me, that any set that has defined limits on either end.. must have an "infinith" member.
Okay, so here we're getting into the distinction between cardinals and ordinals. Consider the difference between {1,2,3,4,...} and {1st, 2nd, 3rd, 4th, 5th,...}. The first is a set of cardinals; they answer how many. The latter are ordinals, which refer not to how many but rather to an ordered collection of objects.

So take the set of objects {CliveStaples, GoldPhoenix, MindTrap}. This is the same collection as the set {MindTrap, CliveStaples, Goldphoenix}.

We can order this set in a number of ways: alphabetically by name, numerically by number of posts, etc. If we order it alphabetically, the first object is CliveStaples, the second GoldPhoenix, etc. So we could write CliveStaples < GoldPhoenix < MindTrap, where we understand x < y to mean x precedes y in alphabetical order.

So the first element of an ordered set if it exists is the element x such that there is no y with y < x.

Not all ordered sets have first elements. Consider the set of numbers 0 < x < 1; this has no "first" element and no "last" element. No element has a successor.

So it's just not at all clear what the "infinitieth" element would be. In the example {..., 3, 2, 1}, is 2 an infinitieth element as well?  Reply With Quote

10. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by CLIVE
So it's just not at all clear what the "infinitieth" element would be. In the example {..., 3, 2, 1}, is 2 an infinitieth element as well?
Well, I would say the confusion is not if they are the "infinith" it is o.k. to say it is, and it is o.k. to say that they all are.
The problem of course is that we can point to the elements that fullfill the idea of "infinith" when counting backwards (I would say because it is against the defined beginning of the set) but not when counting forward (even though it too comes to a defined end ie...from the OP 2)

However, the fact that conceivable all the elements in the set can be called "infinith", is a real problem to any claim that there is no "infininth" element.
I mean, consider.. In the same breath, all the elements are the ininith one, and the other way none of them are.
That would appear to be a problem in and of itself, and should point to a problem with the very concept of infinity, rather than some other explanation that we don't understand infinity.

----
None of that addresses the point about the concept of meeting., as far as I can tell.
Also... thanks.. i could not for the life of me (even though you have said it a few times) remember "ordinals".

I would say that ultimately, in regards to cardinals, and ordinals.. it only matters as it relates to other objects.
Certainly we can understand that if one car lot is next to another car lot, we can point to the last car in one lot, and the first car of the next car lot which directly follows.
(to repeat myself I know).

It is one thing to say that a lot is so big, I am not certain where one begins and another ends.
It is something all together different to say that these two lots are next to each other, but never actually next to each other (Ie there is no last car, and no first car).
That is a contradiction. But as pointed out earlier, we test our math against the observed to see if it matches up, and this is a point where we can observe that the math of infinites does not match up with the very idea of objects meeting.

so should I conclude that I don't understand infinits well enough, or that actual infinites are not possible in a world with meeting (no matter how handy it is for specific calculations and tasks)?

I would also say, that if we can not point to where two objects specifically meet, then they can not be said to meet. Or be adjacent or however one may like to phrase it.

*Final note* Geez, this is like the 3rd rabbit, and I am terribly sorry for it.  Reply With Quote

11. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by MindTrap028 Well, there most certainly is an "infinitieth" particle... namely the fist one when counting backwards.

Counting down, as with an infinite past, or counting backwards of the OP's set of particles.
(from infinity).... 10, 9 , 8, 7, 6, 5, 4, 3, 2, 1..

There are an infinity number in the set. (cardinality...is that right?), and 1 is the "infinith" member of that set.
The observable fact is that we can only observe the infinith set by counting down.

So it would seem to me, that any set that has defined limits on either end.. must have an "infinith" member.
Yeah, that's not quite right (As Clive has gotten into in post#69, and I've discussed with Clive in post#62/63/64), but we can continue this conversation when you reply to my post#61/63 or Clive's post#69.

It seems like you are conflating ordinal numbers with cardinal numbers, which is the root of your confusion. It's a valid confusion to have, these concepts are identical in finite cases and all of our counting intuitions stem from conflating the two --but they are different. They have different definitions, they capture different notions, and when you generalize to the case of infinity, they diverge immediately as concepts. Thus, because your intuition tells you to treat these two different concepts as the same, you're interpreting their difference as a contradiction in the nature of infinity (Rather than acknowledging that the two concepts are not interchangeable). That's my understanding of your objection, anyways, so we'll see how that plays out as we continue to discuss this. But the above seems to me to suggest that I'm correct about where your conceptual problem is arising from. Originally Posted by CliveStaples Not all ordered sets have first elements. Consider the set of numbers 0 < x < 1; this has no "first" element and no "last" element. No element has a successor.
I think main point is that there is always a larger (and always a smaller) element than any element in that. But yes, this is a crucial point, which forbids the existence of a "zeroeth" point and an "infinitieth" point.  Reply With Quote

12. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by GoldPhoenix Yeah, that's not quite right (As Clive has gotten into in post#69, and I've discussed with Clive in post#62/63/64), but we can continue this conversation when you reply to my post#61/63 or Clive's post#69.

It seems like you are conflating ordinal numbers with cardinal numbers, which is the root of your confusion. It's a valid confusion to have, these concepts are identical in finite cases and all of our counting intuitions stem from conflating the two --but they are different. They have different definitions, they capture different notions, and when you generalize to the case of infinity, they diverge immediately as concepts. Thus, because your intuition tells you to treat these two different concepts as the same, you're interpreting their difference as a contradiction in the nature of infinity (Rather than acknowledging that the two concepts are not interchangeable). That's my understanding of your objection, anyways, so we'll see how that plays out as we continue to discuss this. But the above seems to me to suggest that I'm correct about where your conceptual problem is arising from.

I think main point is that there is always a larger (and always a smaller) element than any element in that. But yes, this is a crucial point, which forbids the existence of a "zeroeth" point and an "infinitieth" point.
Crucially, since there aren't "successor" elements, there's no notion of a "next" element (as there are with ordinals, each of which have a successor ordinal).  Reply With Quote

13. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by CliveStaples Crucially, since there aren't "successor" elements, there's no notion of a "next" element (as there are with ordinals, each of which have a successor ordinal).
That's true as well, but you could do a (monotonic) sequence embedding. In other words, take a sequence that always lies in the set (0,1), and then choose an ordering 1/n (1 - 1/n) for the first (last) elements. Then it's easy to see that no last element exists (there is no natural number that corresponds to the "first" or the "last" points), without having to delineate the importance of a successor or predecessor function, which I doubt people here have a deep intuition for the Peano axiomitization of arithmetic and it's language. But yes, obviously that's a valid way of demonstrating that there is no least or greatest element in the set.  Reply With Quote

14. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by Squatch
Further, I’m more than willing to agree that infinity isn’t a number, which is what you seem to be getting at above. But then how do you explain the statement in your OP: “Suppose this hypothetical universe starts out with an infinite number of particles?” If there are an actual infinite number of particles, how is there not an “infinitieth” particle?
Because in the given ordering, every particle is the "kth particle" for some natural number k. If there were an "infinitieth" particle (whatever that means in this context), that particle couldn't also be the kth particle for some natural number k since particles in different positions are distinct (no particle is in two different places in the order).  Reply With Quote

15. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by GP
Yeah, that's not quite right (As Clive has gotten into in post#69, and I've discussed with Clive in post#62/63/64), but we can continue this conversation when you reply to my post#61/63 or Clive's post#69.

It seems like you are conflating ordinal numbers with cardinal numbers, which is the root of your confusion. It's a valid confusion to have, these concepts are identical in finite cases and all of our counting intuitions stem from conflating the two --but they are different. They have different definitions, they capture different notions, and when you generalize to the case of infinity, they diverge immediately as concepts. Thus, because your intuition tells you to treat these two different concepts as the same, you're interpreting their difference as a contradiction in the nature of infinity (Rather than acknowledging that the two concepts are not interchangeable). That's my understanding of your objection, anyways, so we'll see how that plays out as we continue to discuss this. But the above seems to me to suggest that I'm correct about where your conceptual problem is arising from.
Well, I'm not sure the point has misunderstood or confused ordinal and cardinal.

The idea with those concepts, is that one has to do with the size of the total given set, and the other has to do with where an item appears in the set.

So, while a set may be said to be infinite in size, speaking of an infinith object has to do with where it appears in the set.

So, I don't think I am incorrect to say that 1(counting down) is the infinith object in the infinite set of numbers, nor is clive wrong is pointing out that all the objects would also be considred the infinith object.

The intuition you seem to be referring to, is that normally if there are only 4 objects in a set, then the 4th object is inherently the same as the total size of the set.
But that is not my objection here.

My objection has to do with the apparent fact that infinit sets.
1) Can be defined to have an upper boundary(Ie never reaches 2) with no end (Ie infinite members before 2).
2) That this definition precludes objects from touching (see no point of 2, is adjacent to list of 1, who's limit is 2).
3) 1 & 2 can not be true in a world where infinit objects touch.

Seperatly, the idea that all objects are properly defined as "infinith" object in one direction, but not in the other causes me to pause at the plausibility of the idea of infinith.
This maybe where you are refering to my intuition, because with all real objects that I have encountered so far, it has held true that the ordinarily is reversible.  Reply With Quote

16. Re: WLC is Wrong: The Argument for Actual Infinities

The idea with those concepts, is that one has to do with the size of the total given set, and the other has to do with where an item appears in the set.

So, while a set may be said to be infinite in size, speaking of an infinith object has to do with where it appears in the set.
Correct. It's important to note, though, that order doesn't matter so far as the set is concerned. I.e., there's no difference between {0,1,2,3,...} and {1241941,54092376940923,2087606972609209572, ...} so long as both sets contain the same elements.

Now, as sequences {0,1,2,...} and {1,0,2,...} are different, because two sequences are identical if and only if they agree at every position.

This maybe where you are refering to my intuition, because with all real objects that I have encountered so far, it has held true that the ordinarily is reversible.
This is a good observation, as in fact every finite well-ordered set of the same cardinality is order-isomorphic, and reversing the order on a set (often referred to as the dual order) doesn't change its cardinality, so every finite well-ordered set is order-isomorphic to its dual.  Reply With Quote

17. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by CLIVE
This is a good observation, as in fact every finite well-ordered set of the same cardinality is order-isomorphic, and reversing the order on a set (often referred to as the dual order) doesn't change its cardinality, so every finite well-ordered set is order-isomorphic to its dual.
Well, the other points not withstanding(and I think that they really need to be addressed), I guess this fact begs the question... "why".
I mean, is it that sets just so happen to act that way or is there some physical or logical law which makes all real sets act that way?

Now, myself I would suspect that it is the logical law of identity which causes this to be the case, and that, that law doesn't simply apply only to finite sets.  Reply With Quote

18. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by MindTrap028 Well, the other points not withstanding(and I think that they really need to be addressed), I guess this fact begs the question... "why".
I mean, is it that sets just so happen to act that way or is there some physical or logical law which makes all real sets act that way?

Now, myself I would suspect that it is the logical law of identity which causes this to be the case, and that, that law doesn't simply apply only to finite sets.
I'll try to unpack this for you.

I. Introduction to Orders on Sets

Given a set S, an order on S, which I'll write as <, is a collection of pairs of S-objects. If (a,b) is in <, we usually write a < b. If we wish to refer to the ordering of S by <, we usually write (S,<), as there may be in general many orderings on S other than the particular one we're looking at.

This is perfectly true for the familiar orders. On the natural numbers, we have the familiar order 0 < 1 < 2 < ... , right? This makes (N,<) an ordered set. Since 0 < 1, we have that (0,1) is in <. Similarly, (0,2) is in <, (0,3) is in <, etc. And likewise (1,2) is in <, (1,3) is in <, etc. Most generally, (a,b) is in < if and only if b = a+k for some positive integer k. Note that just because (0,1) is in < doesn't mean that (1,0) is in <.

Ia. Partial Order

An order < on S is a partial order if it satisfies a few properties:

(i) Reflexive: s < s for all s in S.
(ii) Antisymmetric: If a < b and b < a, then a = b.
(iii) Transitive: If a < b and b < c, then a < c.

You're already familiar with some partial orders. The "less than or equal to" relation is a partial order on N:

(i) Reflexive: n n for all n in N. (0 0, 1 1, etc.)
(ii) Antisymmetric: If a b and b a, then a = b.
(iii) Transitive: If a b and b c, then a c. E.g., 1 2, 2 3, so 1 3.

There's another partial order that you're already a little familiar with. Given a set S, the powerset of S, P(S), is the set of subsets of S. So if S = {0,1}, then P(S) = { {}, {0}, {1}, {0,1} }. P(S) is partially ordered by the subset relation, :

(i) Reflexive: Every set is a subset of itself, so A A.
(ii) Antisymmetric: If A is a subset of B, and B is a subset of A, then A = B.
(iii) Transitive: If every element of A is in B, and every element of B is in C, then every element of A is in C.

Ib. Total Order

A partial order < on S is a total order if every element in S is comparable, i.e. for all x,y in S, either x < y or y < x. Not every partial order is a total order; for example, (P(S), ) has incomparable elements:

S = {0,1}
Subsets of S: {}, {0}, {1}, {0,1}

{0} {0,1}, so {0} < {0,1} by our definition of <. Likewise, {1} {0,1}, so {1} < {0,1}. But {0} isn't a subset of {1}, and {1} isn't a subset of {0}. So neither {0} < {1} nor {1} < {0} are true.

Ic. Well order

A total order < on S is said to be a well-ordering, or S is well-ordered by <, if every non-empty subset of S has a <-least element. "What is a <-least element," you ask?

If A is a subset of S, and if x is an element of A, then x is a <-least element of A if x < a for all the other elements in A.

So for instance, N = {0,1,2,...} is well-ordered by the standard order 0 < 1 < 2 < ...; do you see why?

By contrast, Z = {..., -1, 0, 1, 2, ...} is not well-ordered by the standard order ... < -1 < 0 < 1 < ..., since for example the set of all negative numbers is non-empty but has no <-least element.

II. Dual Order

If (S,<) is an ordered set, then the dual of < is given by >, where a > b if and only if b < a. Note that x and y are <-comparable if and only if they are >-comparable. So < is totally ordered if and only if its dual > is totally ordered.

If you reverse the order on (N,<), you get something like this: ... > 2 > 1 > 0.

Is N well-ordered by >?

Well, think about what a >-least element is. So let's go back to the definition, replacing its instances of < with our order >. Given a non-empty subset A of N, and an element x in A, x is a >-least element if and only if x > a for all a in A. By our definition of >, we can write equivalently that an element x in A is a >-least element if and only if a < x for all a in A. So we see that a >-least element is a <-greatest element.

We see immediately that N doesn't have a >-least element, since there's no natural number greater than every other natural number. Since N is a non-empty subset of N, we've found a non-empty subset of N without a >-least element, and thus N is not well-ordered by >.

The key insight here is that a totally-ordered finite set will always have a least element (and thus be well-ordered) and a greatest element (and thus its dual will be well-ordered). Infinite sets can break either one, or possibly both, of these.

III. Well-orders vs. Total Orders

IIIa. Finite Sets

Given a finite set X = {x1, x2, ..., xn}, we have the following:
(Theorem) Every totally-ordered finite set is well-ordered.

To prove this, we'll use the following lemma:

(Lemma) Every totally-ordered non-empty finite set has a least element.

Proof of Lemma: By induction on n.

(Base case) n = 1

We have X = {x1}, which trivially has a least element. So all totally-ordered sets of size 1 have a least element.

(Inductive case) Suppose that every ordered set of size k has a least element, for some k >= 1. Suppose Y = {y1, y2, ..., yk, yk+1} is totally ordered by <. Then {y1, ..., yk} is a totally-ordered set of size k and thus has a least element, say, yn. Likewise, {y2, ..., yk+1} is a totally-ordered set of size k and thus has a least element, say ym. Since Y is totally ordered, either yn < ym or ym < yn.

Case 1: yn < ym. We show that yn is a <-least element in Y. Let y be an element in Y. Then y = yj for some 1 <= j <= k+1. If j < k+1, then y = yj is in {y1, y2, ..., yk}, and thus yn < y by construction of yn. If j = k+1, then y = yj is in {y2, y3, ..., yk+1} and so ym < y. But yn < ym, and < is transitive (since < a total order), so yn < y. Thus yn is a <-least element of Y.

Case 2: ym < yn. This is similar to the above.

Proof of theorem
:

Let A be a non-empty subset of X. Since X is finite, A is finite. Since X is totally-ordered by <, A is totally ordered by <. Thus A is a non-empty totally-ordered finite set, and by (Lemma) has a least element. So X is well-ordered.

IIIb. Infinite Sets

The set (0,1) U {2} is totally-ordered by the standard order on R (or Q) but contains no least element and is thus not well-ordered.

The set {z in Z | z < 0} is totally-ordered by the standard order on Z but has no least element and is thus not well-ordered.

So total orders are not equivalent to well-orders for infinite sets.

IV. Order Isomorphisms

Given two ordered sets (X,<) and (Y,<'), a function f:X->Y is an order isomorphism if, for all a,b in X, a < b if and only if f(a) <' f(b). If an order isomorphism exists between (X,<) and (Y,<'), then we say (X,<) and (Y,<') are order-isomorphic

(Theorem) All totally-ordered finite sets of size k are order isomorphic.

Proof:
Let |X| = |Y| = k, for some natural number k, and let (X,<) and (Y,<') be total orders.

By the earlier theorem, X and Y are well-ordered. We can thus construct finite sequences x1 < x2 < x3 < ... < xk and y1 < y2 < y3 < ... < yk where we label the least element of X as x1, the least element of X - {x1} as x2, etc., and similarly for y. Then the map f(xn) = yn is an order isomorphism, since xa < xb iff a < b iff ya < yb.

This theorem fails for infinite sets, as given by the following tricky example.

IVa. A Tricky Example

Given N = {0,1,2,3,...}, let (N,<) be the standard order on N (0 < 1 < 2 < ..., etc.).

Let X = N U {y}. Define <' on X where <'|X = < (so a <' b iff a < b when a,b are in N), and n < y for all n in N. So (X,<') looks like this: 0 <' 1 <' 2 <' ... <' y

(1) |N| = |X|

Proof:
Define f: N -> X where f(0) = y, and f(n) = n-1 for n>0

(2) (N,<) and (X,<') are not order isomorphic

Proof:

(Lemma) If (A,<A) and (B,<B) are order isomorphic, then |{x in A | x <A a}| = |{b in B | b <B f(a)}| for all a in A.

Proof:

Since A and B are order isomorphic, let f be an order-isomorphism from A to B. Let c be an element in {x in A | x <A a}. Since c <A a, f(c) <B f(a) so f(c) is in {y in B | y <B f(a)}. Since f is injective on A, it's injective on {x in A | x <A a}. Let d be an element in {b in B | b <B f(a)}. Then f-1(d) <A a, so f-1(d) is in {x in A | x <A a} and thus f is surjective onto {b in B | b <B f(a)}. So f is a bijection from {x in A | x <A a} to {b in B | b <B f(a)}. QED.

Suppose f: N -> X is an order isomorphism. Then | { n in N | n < f-1(0) } | = | { x in X | x <' y} |. But | {n in N | n < f-1(0)} | is finite, and | {x in X | x <' y} | is infinite. Contradiction. QED.

So it isn't until you get up to infinite sets that there are even different (in the sense of non-isomorphic) well-orderings to talk about on a given set. If you cut out infinite sets, you miss out on basically the most exciting developments in set theory and foundations in the last 40 years, which has been active in the area of ordinals and ZFC.  Reply With Quote

19. Re: WLC is Wrong: The Argument for Actual Infinities

The staff has been asked to adjudicate whether or not a claim made in this thread was supported. The claim made in the opening post is that it is physically permissible for particles to be placed infinitely close together. This claim was disputed, then challenged. The staff notes that no objection to the nature of the claim was presented as part of the challenge.

When asked specifically to support that position in post 57:

And, please support or retract that point particles can be placed infinitely close together and at precise locations with a known velocity.

The response was to point back to the opening post in post 59:

I might start with the OP. I would also suggest doing some of the longhand work.

However, when reviewing the OP, no such support can be found, it is only noted that the author sees no objection to this, which is not support.

Until such support is offered, this position is retracted within this thread.

If you have any questions or concerns, please address the staff in an "ask the staff" thread.  Reply With Quote

20. Re: WLC is Wrong: The Argument for Actual Infinities Originally Posted by GP
But I can't see any point in re-repeating what I've already said, so I'll leave it here.
Im sorry you feel this way. It would seem at first blush that if the question I asked was so absurd it would be an easy, trivial thing to answer.

Aside from its clearly condescending nature, your response seems unfortunately to ring a bit hollow. We both know that claims like I can restrict a particles possible location more and more and I can restrict its velocity as well are at odds. Obviously, we could ignore that in the OP because you were restricting it to classical mechanics. But as that scenario played out to have problems and you introduced additional concepts it would seem reasonable to ask about the ramifications of those additions.

Your objection to my questioning of an infinitieth particle seems even a bit stranger given that I was using your language. If repeating back your own language to you makes you upset, how much more frustrating must it be to your opponents?

Regardless, Ill honor your request to leave the thread here and will only include my basic summary.

The initial OP proposed by GP was a stated attempt to provide a counter example that would, if correct, necessitate that at least one central premise of the KCA must be false.

As was initially pointed out however, this conclusion was far too broad of a conclusion for the scope of his argument. Rather, the counterexample, if correct, would only be invalidating one specific defense of one specific argument against a central premise of the KCA.

As detailed in post 3, I showed that GPs counterexample only affects half of the first supporting argument (and doesnt talk about the second supporting argument, nor the evidentiary arguments) against premise 2.

Defense of the Kalam Cosmological Argument
Second Premiss

First Supporting Argument
Now (2.11) maintains, not that a potentially infinite number of things cannot exist, but that an actually infinite number of things cannot exist. For if an actually infinite number of things could exist, this would spawn all sorts of absurdities.
Perhaps the best way to bring home the truth of (2.11) is by means of an illustration. Let me use one of my favorites, Hilbert's Hotel

But Hilbert's Hotel is even stranger than the German mathematician gave it out to be. For suppose some of the guests start to check out

That takes us to (2.12). The truth of this premiss seems fairly obvious

Second Supporting Argument
This argument is distinct from the first in that it does not deny the possibility of the existence of an actual infinite, but the possibility of its being formed by successive addition.

Thus we can at least reasonably conclude that GPs argument would, at best, invalidate one defense of a premise out of four offered defenses.
___

Now, a bit should be said about GPs strategy. GP explicitly states that he will be arguing via counter-example. An excellent technique given Craigs argument here.

However, there is something that needs to be noted about arguments via counter example, as I highlighted earlier in thread.

Counter examples, in order to be valid, must:
1) Be logically coherent.
2) Must directly invalidate the rule being cited.
3) Must be permissible in the context of the argument that brought about the rule being argued against.

In GPs case, 2 and 3 means that his counter example must show that an actual infinity can have non-contradictory results when existing within this universe (or at least given rules consistent with this universe given that Craigs argument is about this universe).

I think GPs argument fails both 2 and 3 above (I think it also fails 1, but that discussion seems to be still underway).

Below are a list and very brief description of the objections I raised against GPs counter example and which, in my opinion, he did not rebut, either successfully, or in some cases, at all.

Infinite density problem

GPs argument presupposes that particles can be placed infinitely closely together. This is stated in his OP by his discussion of particles being placed according to a descending geometric operator. Essentially, each particle is half the distance from the next most left particle as that particle was from its left neighbor. Clearly, if you continue this halving process on indefinitely, the distance between the particles approaches zero. IE that they are infinitely close together.

GP recognizes this, which is why he tries to make his particles point particles, otherwise there would be a lower bound on how tight he could put them together.

But when asked to support that such a density was possible within our universe, GP declined to detail why that would be the case.

I think it is because generally these kinds of infinite densities are seen as breaks in our understanding of physical law, not as features of that law. It is the reason we say that the infinite density of a black hole represents a breakdown of our understanding, not a real, infinitely dense object.

Thus this objection undermines the feasibility of his counter example in this universe.

Point particle objection

Related to this is GPs use of point particles (as described above, a necessary condition if particles are to be placed infinitely close together).

Let me first point out the consequence of point particles not being invoked and why GP uses them here. If the particles in the setup he describes occupy any spatial extent, the region he described either becomes contradictory or infinitely large. Point particles are the method he is using to allow for the infinite density issue above without resorting to multiple particles occupying the same spatial region, indeed, an infinite number of particles occupying the same spatial region.

So why should we object to this? Because, as I noted in the thread, Point Particles are useful idealizations, not actual things. They are used when the actual structure or size of the particle in spatial dimension is irrelevant to the question asked (as it certainly isnt here) or when the underlying particle is really more a set of fields than a classical particle. Similar to when you say Oh New York is 5,000 miles from my hometown. Now your hometown is not a single point, nor is New York, so you might be calculating between single points on the borders of each of them, or center points in each locale, etc. You are doing so because the varying distances between all the different points in your home town and in New York isnt really relevant to the point you are making.

The same is true here, as I pointed out earlier in thread.

In summary, extended particles have a fixed size, although they may have a fuzzy edge; point-like particles are mathematical abstractions with zero size. But even zero-size particles have an extended effect, due to the effect of the field surrounding them.

A point particle (ideal particle or point-like particle, often spelled pointlike particle) is an idealization of particles heavily used in physicsA point particle is an appropriate representation of any object whose size, shape, and structure is irrelevant in a given context
https://en.wikipedia.org/wiki/Point_particle

A point particle, also known as a point mass, is an idealized object which has mass but no extent in space.
http://www.anselm.edu/internet/physi...ula1/point.htm

You can find a great discussion of their use here: http://physics.stackexchange.com/que...ntum-mechanics

Now, one of the responses the OP had was that there are no problems if we extend this to the quantum mechanical realm, which it was forced to resort to in order to respond to an objection dealing with particle interactions, (because particles that dont occupy the spatial dimension are essentially non-existent unless they have some other properties) which I detail a bit more below in the boundary problem section. This invocation led to another problem with the proposed model.

In quantum mechanics, the concept of a point particle is complicated by the Heisenberg uncertainty principle, because even an elementary particle, with no internal structure, occupies a nonzero volume.
https://en.wikipedia.org/wiki/Point_particle

Which is what led to my response in the last post concerning momentum and position, which was, in itself, a large objection to the model the OP proposed.

Boundary Question

In parallel discussion to the point particle question was a hypothetical I raised about what would happen in that model if a particle approached the OPs set from the infinite density side (ie from point 3 towards point 1). Essentially it was a what would happen next question, what are the consequences of adding a relatively simply addition to the model? Where would the new particle begin interacting with the OPs particle set?

The answer, essentially, was it wouldnt.

There were two possible routes to take in that explanation, neither of which was firmly committed to by the defender of the OP. If we ignore adding fields to resolve the above question we run into an odd scenario where a particle can both be blocked by the set of particles (in the sense that it certainly cannot reach the other side at point 1) and not actually encounter any of them (since there is no initial particle to interact with, just an endless sequence). Which leads us to a scenario where the hypothetical can answer a question both yes and no, a problem.

The second route is to invoke fields. Since the particles of the OPs set are simply contributing to the field the new particle encounters, this problem would seem to be resolved a bit. (Ill note that there were some additional questions that arose from invoking fields, which GP declined to answer and led to his withdrawal from the thread). After all, the particle simply encounters one field (assuming the unsupported position that such a field was coherent is accepted) rather than an endless boundary condition.

But that invocation led to another problem

Particle velocity/position problem

By invoking the fields created by the particles above, we create a much richer description of these particles than had initially existed in the OP. The OPs particles were, essentially, non-existent. They were at certain points, but they didnt exist in space, didnt produce effects on each other, etc. It is hard to describing them as anything real.

So in reply, GP offered us a specific interpretation, de Broglie/Bohm. The problem is, that that interpretation means that if you confine a particle to a small space, its momentum must be proportionally large, thus invalidating the model he laid forward in the OP of particles being confined to an increasingly small space with an increasingly small momentum:

This interpretation you are using also seems to run into another, related problem. Given that you have localized the particles by confining their locations to arbitrarily small points, you would seem to be restricting the de Broglie wavelength (the possible range of locations the particle could show up in, https://en.wikiversity.org/wiki/De_Broglie_wavelength).

Now, given that de Broglie wavelength and momentum are inversely related, it would seem that as we move to a more and more restricted location for each successive particle, we would need increase the particles momentum. That would seem to be a problem with your OP, given that the momentum of each successive particle must be lower in your system.

We could approach this from the opposite angle as well.

You define the mass of each particle in post 10 as: mn = m0 (1/n)2

In the OP you define the velocity of each particle as: vn = (1/n)2

Meaning the momentum (m*v) of each particle is: m0 * 1/(n4)

Given that de Broglie wavelength is: lambda = hbar/momentum we would see that your wavelength is increasing for each successive particle (http://www-physics.ucsd.edu/students...reofMatter.pdf
).

Therefore at some point your wavelength exceeds the range of possible positions defined in your OP for that particle, which is a problem given your statement that  each particle is also never lying on top of one another; for each n, each particle is at a distinct point.

Summary

This was a fun model to think about and to play with. However, as a counter example, it unfortunately fails. The OPs original model suffered from some problems that when additional complexity was added in an attempt to resolve, only added problems of its own.

And such complexity is inherent when infinity goes from being treated as a limit to a value actually attained.

I want to thank GP for constructing it and being willing to engage in the conversation.  Reply With Quote Posting Permissions

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