• # Thread: Different kinds of infinity?

1. ## Different kinds of infinity?

 I recently watched a documentary that introduced the Continuum Hypothesis (nothing in depth at all) and George Cantor, the famous mathematician. They discussed the notion of different kinds of infinities, but didn't get into the details or explanations. I assume this is because this aired on public television, and the technicalities would go over the heads of... everyone. I don't expect I'd understand right off the bat either. In any case, the video described a circle in a new way to me. A circle was superimposed on the screen, and the narrator would draw a triangle, a square, and so on inscribed within the circle, and then went on to explain that a circle was essentially a shape with an infinite number of angles if all of the side and angles were the same. After thinking about it for a moment however it also occurred to me that circles have perfect curves, and no sides as can be distinguished be being separate from one another by any angle. A circle has infinite angles, but it also has zero angles? Is something like that theoretically possible? Cantor went insane and was put into an asylum several times over the course of his attempts to prove the Continuum hypothesis. The documentary touched upon contradictory elements in his work.. Again though, there isn't much in terms of specifics. To conclude: I want to know more about different kinds of infinity, and whether or not infinity can be equal to zero in some cases. PS- I found the documentary on page 65 of the Abstruse Goose webcomic, aptly titled, "The Cantor Madness." PPS- I have some understanding of countable and uncountable number sets.

2.  Originally Posted by Invader I recently watched a documentary that introduced the Continuum Hypothesis (nothing in depth at all) and George Cantor, the famous mathematician. They discussed the notion of different kinds of infinities, but didn't get into the details or explanations. I assume this is because this aired on public television, and the technicalities would go over the heads of... everyone. I don't expect I'd understand right off the bat either. In any case, the video described a circle in a new way to me. A circle was superimposed on the screen, and the narrator would draw a triangle, a square, and so on inscribed within the circle, and then went on to explain that a circle was essentially a shape with an infinite number of angles if all of the side and angles were the same. After thinking about it for a moment however it also occurred to me that circles have perfect curves, and no sides as can be distinguished be being separate from one another by any angle. A circle has infinite angles, but it also has zero angles? Is something like that theoretically possible? Cantor went insane and was put into an asylum several times over the course of his attempts to prove the Continuum hypothesis. The documentary touched upon contradictory elements in his work.. Again though, there isn't much in terms of specifics. To conclude: I want to know more about different kinds of infinity, and whether or not infinity can be equal to zero in some cases. PS- I found the documentary on page 65 of the Abstruse Goose webcomic, aptly titled, "The Cantor Madness." PPS- I have some understanding of countable and uncountable number sets. I don't see why not. Infinity is essentially comprised of everything. Positive and negative numbers, which it goes in both direction. A lot of people just view infinity as going in one direction with a line of positive number. If you look at it from a different angle, zero consists of both positive and negative numbers. It is essentially perfect, like the sum total of all positives and negatives explained with one symbol or a symbol explaining there being no positives or negatives. It sounds like the guy was heavy into sacred geometry. What the images essentially show are infinities within infinities within infinities etc.. They are comprised of nothing but circles, which it is like the circles are in and of themselves angles and/or sets.

3.  Just just watched a video of someone breaking down the theory. The Infinities In Between (1 of 2) The Infinities In Between (2 of 2)

5.  Originally Posted by Xei No, he wasn't. He was heavy into proper mathematics. Sacred geometry is a load of nonsense. It doesn't even mean anything. What you described makes sense, which it seems he probably wasn't into sacred geometry. How is it a load of nonsense when mathematics is applied to nature or vice versus? It is like having an anchor point for observation of the world around you.

6.  Originally Posted by ArcanumNoctis How is it a load of nonsense when mathematics is applied to nature or vice versus? It is like having an anchor point for observation of the world around you. Sacred geometry, as a model for the world around you, is as useful and as valid as the idea that everything is made of fire, water, air and earth.

7.  Yes, good point there. Sacred geometry makes no predictions, uses no logical inference, in fact has no logical meaning whatsoever. It is literally a bunch of pretty pictures with pretty names with no explanation whatsoever. Anybody could make up such a thing.

8.  I'd like to see how many of those sets people can count... They're roughly in difficultly order, though perhaps the last two should be swapped. Most of them are linked to the previous one. Z = {..., -2, -1, 0, 1, 2, ...}, the integers. N^2 = {(a,b) such that a and b are in N}, the set of all pairs of numbers. Q^2 = {(a,b) such that a and b are in Q}, the set of all pairs of fractions. {non-overlapping discs in a plane}; picture a dalmatian with circular spots. Remember there could be infinitely many spots (the dalmatian is grotesquely obese). {non-intersecting figures of 8}; this is hard but awesome. {set of all subsets of N}

9.  ∞ ≠ ∞ That's a little something I learned in Calculus during infinite series. infinity.jpg This is where infinity gets fun.

10.  Originally Posted by Xei Z = {..., -2, -1, 0, 1, 2, ...}, the integers. N^2 = {(a,b) such that a and b are in N}, the set of all pairs of numbers. Q^2 = {(a,b) such that a and b are in Q}, the set of all pairs of fractions. {non-overlapping discs in a plane}; picture a dalmatian with circular spots. Remember there could be infinitely many spots (the dalmatian is grotesquely obese). {non-intersecting figures of 8}; this is hard but awesome. {set of all subsets of N} I think I know how to count the first three, I don't understand the next two and I have no idea how one would possibly count the last one (is it even possible?).

11.  Originally Posted by illidan I think I know how to count the first three, I don't understand the next two and I have no idea how one would possibly count the last one (is it even possible?). Give us the quick lowdown on how you did the first three. The next two are slightly more abstract. Basically with the first one (for example), you're on the infinite 2D plane (like a normal coordinate system) and circles of random sizes are dotted all over it in random positions. There could be a finite number or an infinite number (although if there were a finite number it's obvious they're countable), and you have to find some way of counting them; in other words, in pairing them up with the natural numbers so that no circle is unpaired. A big clue for all of the first five is that they all follow from the previous one. You're correct that the last one is bigger than N, though of course the point is in proving it. The clue here is to try to adopt the method that was used for proving the real numbers are uncountable. Originally Posted by The Invisible Man ∞ ≠ ∞ That's a little something I learned in Calculus during infinite series. infinity.jpg This is where infinity gets fun. Well... infinity doesn't equal infinity because it isn't a number. Though using it informally there shouldn't really be a problem in equating them. A ≠ A can never make any sense if A is well defined. Dunno what's going on with the sum... I don't think it makes any sense. If you mean the sum from 1 to infinity... I don't really know why the ns haven't been cancelled. I think 1/n^2 sums to pi^2 / 6.

12.  I haven't totally been able to wrap my head around this. Non-overlapping discs in a plane is a bunch of visual objects that are distinguished from one another by the space in between them. We designate a number to each one. 1 assigned to this spot, 2 assigned to that spot, and so on for infinite (if there are an infinite number of spots). That doesn't mean that any spot has a value of 1 or 2 or n, the number only signifies the amount of spots. Is that correct? And two questions: What do you mean by intersecting figures of eight? What designation of Aleph do we give to a number set that includes all numbers, like the set N such that N goes from -∞ to +∞ where an infinite number of numbers exists between any two numbers (as in there are an infinite number of irrational and non irrational numbers in between 1 and 2, or between 4.0005 and 4.0005001)? Also, your explanations have been helpful and I thank you profusely for them.

14.  Originally Posted by Xei Give us the quick lowdown on how you did the first three. Oh, ok. I wasn't sure if it's ok to post the solution already. Here's how I did the first three (not sure if they're correct though): Counting Z: z : N → Z and z(1) = 0 z(2) = 1 z(3) = -1 z(4) = 2 z(5) = -2 z(6) = 3 z(7) = -3 etc. Counting N^2: n : N → N^2 and n(1) = (1,1) n(2) = (1,2) n(3) = (2,1) n(4) = (1,3) n(5) = (2,2) n(6) = (3,1) n(7) = (1,4) n(8) = (2,3) n(9) = (3,2) n(10) = (4,1) etc. Counting Q^2: Let q : N → Q be the function that counts Q, let n : N → N^2 be the fuction that counts N^2 and f : N^2 → Q^2 with f( (x,y) ) = (q(x), q(y)). Then q2 : N → Q^2 with q(x) = f(n(x)) would count Q^2. Now I got to think about the other ones.

15.  View of the Two-Element metaphysics. Definition: A thing is any material difference in any shape, form, limit. I.e. shape is not a thing. material difference is not a thing. One cannot predicate of either material or form, they are first principles, elements. One cannot define either matieral difference or form, for the above reason. Definition is the social convention which preserves the naming convention which equates the names of a thing to the names of that things various forms and material differences in those forms. I.e. predication is the inverse function of abstraction. Description is a set of directions leading to a thing by which an abstraction of either material difference or form may be made. One can use the above principles of logic to determine that a great deal of Cantor's work was insane and that the questions you have asked, most are basically non-linguistic--not in compliance with the principles of grammar at all. If one does not know the foundation of langauge, they do not know when they are thinking gibberish or not, they have no standard by which to understand that words, just like arthithmetic, has principles by which words can be assembled into basic sentences, basic sentences into the complex sentences we use now. It is an effect long ago noticed, but ignored. This effect is also a major theme in the Judeo-Christian Scripture and that one day, a point in history would be made signaling a psychological change in man--because it is a linguistic change. Man is still proto-linguistic. Example: "Set of all numbers". A number is no more than a name derived from an order naming convention. Thus one has "Set of all names." Is the material in a boundary the boundary which bounds itself? Can one have the set of all names? How does that effect relation to self? If you do not understand, what is the name of the set of all names? The liars paradox in another form. A set is a thing, the material in a set is not, in reference to the set. One cannot have a thing of a not thing. All of man's so call higher logics and math are gibberish. Plato's Parmenides was suppose to exercise the mind to demonstrate what happens when one uses the same name for a thing, a things form, and the material difference in that form. The principles of predication. Demonstration: If I say a set of marbles. "a set" is a variable, one can substitute "house", A house of marbles. A house made of marbles, to clarify. Which is not really good grammar, we should say a house made with marbles. Without understanding we do not even know when to use "of" or "with". Once more: There is a difference between "naming" which produces "names" and the "names" produced. Naming is a material difference--i.e. it is not a thing. Names, if language were at all possible, must be set in a one-to-one correspondece with that which is named. A name can have NO characteristics of its own in order for it to be logically manipulated as that which is named. Which means, there is no common characteristic of a name qua name by which to even construct a set. Now, if you have studied grammar as taught today, you will quickly realize that what you are taught is contradictory to language itself.

16.  Originally Posted by illidan Oh, ok. I wasn't sure if it's ok to post the solution already. Here's how I did the first three (not sure if they're correct though): Counting Z: z : N → Z and z(1) = 0 z(2) = 1 z(3) = -1 z(4) = 2 z(5) = -2 z(6) = 3 z(7) = -3 etc. Counting N^2: n : N → N^2 and n(1) = (1,1) n(2) = (1,2) n(3) = (2,1) n(4) = (1,3) n(5) = (2,2) n(6) = (3,1) n(7) = (1,4) n(8) = (2,3) n(9) = (3,2) n(10) = (4,1) etc. Counting Q^2: Let q : N → Q be the function that counts Q, let n : N → N^2 be the fuction that counts N^2 and f : N^2 → Q^2 with f( (x,y) ) = (q(x), q(y)). Then q2 : N → Q^2 with q(x) = f(n(x)) would count Q^2. Now I got to think about the other ones. Yeah, perfect. I kind of left a clue for the second one with the picture for counting the rationals. You did pretty much the same thing it looks like... the trick of course is counting them diagonally; if you count them horizontally or vertically you'll never get past the first line. And yeah, with the third you just gotta realise that you can combine the previous functions. Are you in education at the mo? You could probably handle maths at Cambridge, especially if you manage the next questions.

17.  Originally Posted by Xei Are you in education at the mo? Yeah, I'm a university student, studying Software Engineering. So, here's my idea for counting non-overlapping discs in a plane. I've already shown that one can count Q^2. So I have an (infinite) list of all pairs of rational numbers. I go down that list, interpret each pair as coordinates on the plane, and check if the point described by the coordinates is in one of the discs. If I get a hit, I assign the next number to that disc – I'm starting at 1, of course. The only thing I'm not sure about is how would I know if I've hit a disc. In what form are the discs given. I assume they're just given visually (somehow). If they were in a some sort of a list I would already have a way to count them, wouldn't I. I'm probably thinking too much like a programmer here.

18.  Yeah, that's pretty much perfect. :V I dunno about your question about what form they're in... in maths we'd think of a collection of discs as a well-defined objective entity. You could think of them as a set of coordinates of centres paired with radii or something, but we'd view that as a description really, rather than the actual object itself. A couple of subtleties about the answer; firstly each disc is actually counted an infinite number of times. Of course, that's not actually a problem. If you show that each disc can be mapped onto N an infinite number of times you can obviously do it a single time. Secondly the proof relies on what we call the 'density of the rationals'. What this means is that, between any two non-equal values, you can find a rational number. This is necessary for the subtle step that you actually can find a point in Q^2 inside the disc; if the question were about points, it wouldn't work.

19.  Originally Posted by Xei I dunno about your question about what form they're in... in maths we'd think of a collection of discs as a well-defined objective entity. You could think of them as a set of coordinates of centres paired with radii or something, but we'd view that as a description really, rather than the actual object itself. Maybe I am looking at this too much from a practical standpoint, but I still want to explain what bugs me about this. If the discs are given as a well-defined set and I want to check if a given point is in one of the discs, I imagine I would start looking through the set, trying to find a disc that contains my point. If there was such a disc, I would eventually find it (in finite time). If there wasn't, however, I would look for it forever and I wouldn't be able to count that set, i.e. not all discs would eventually get a number assigned to it because I would be stuck trying to find a disc for a point that is not contained in any disc. I also just realised that, to be consistent, I would have to raise the same issue for the counting of Z (or N^2, Q, or Q^2). I find this all a little confusing. It almost seems like a philosophical issue. Maybe all I have to do is imagine how such a bijection would look like, but I'm a little sceptical of proofs that I am unable to express formaly. How would you show that the set of non-overlapping discs (let's call it D) is countable in a rigorously formal way? Such a proof should have the following basic structure: 1. Explicitly give a function f : N → D. 2. Formally show that f is injective, i.e. show that f(a) = f(b) → a = b. 3. Formally show that f is surjective, i.e. show that ∀d∈D: ∃n∈N: f(n) = d. I imagine that you could do that for Z and probably for Q, N^2, and Q^2, too, but I am unsure how this could be achieved with D. I think the difference between Z and D is that for Z, even though it is an infinite set, I have a finite description. For D I don't have a finite description, do I? But if I don't have a finite description of D, there is no practical way to check if a given point is in any of discs. You said that D is given as a "well-defined set". Does "well-defined" imply that the definition (= description) is finite? If the definition was finite, I imagine it would be possible to check if a given point is in any of discs. If the definition was not finite, I can't really imagine how it would be done.

20.  Originally Posted by illidan Maybe I am looking at this too much from a practical standpoint, but I still want to explain what bugs me about this. If the discs are given as a well-defined set and I want to check if a given point is in one of the discs, I imagine I would start looking through the set, trying to find a disc that contains my point. If there was such a disc, I would eventually find it (in finite time). If there wasn't, however, I would look for it forever and I wouldn't be able to count that set, i.e. not all discs would eventually get a number assigned to it because I would be stuck trying to find a disc for a point that is not contained in any disc. I also just realised that, to be consistent, I would have to raise the same issue for the counting of Z (or N^2, Q, or Q^2). I had a room mate that raised a similar objection to Cantors work. His exact words: "Stick N in one bucket and R in another and start pulling one item out of each bucket. When one bucket runs out, you know the other one is bigger." The problem with this is that it's not a computational thing. think of there being a set S = {n in N: there exists d in D s.t. f(n) in d} where f is the chosen bijection N -> Q2. We don't need to actually construct this set to work with it.We just need to know that it's well defined. Essentially, you can think of a set S as being well defined if for every element e in the universe of discourse, either e is or is not in S. An example of set lacking well-definition. The canonical example of this is the set A of all sets which are not members of themselves. Is A a member of itself or isn't it? So S defined above is well-defined if d is a well defined set for all d in D and if D is well defined which we'll assume for now. but we can just represent d as the set d = {(x, y) : sqrt((x - c1) + (y - c2)) < r), where c is the center of the disc and r is the radius. This is clearly well defined. Then proving that card(D) = card(N) is equivalent to proving that the map S -> d in D induced by S is surjective.* This is technically trivial but may or may not be challenging for an intelligent person that's not used to this sort of stuff. Technically, we would have to prove a surjection of D onto N as well but in this case, I think were safe to assume that one exists. * this is the map θ: N -> D, that takes n to the disc that contains f(n). This is defined for every element of S by construction. To see that it is well-defined, note that if f(n) lies in d1 and d2 then d1 intersect d2 is non-trivial contradicting our assumptions on D.

21.  Originally Posted by PhilosopherStoned […] and if D is well defined which we'll assume for now. […] I think that's the crux here. How would the (or a) definition of D look like? The definition must be finite, right? If the definition is of infinite length, D lacks well-definition, doesn't it?

22.  xei left the definition of D open because we can just assume that it's well defined and a particular definition of it just obscures the issue. If some particular D happens to lack well-definition, then the proof doesn't apply to that D. I'll give you an example though. For simplicity of notation that's define D(x, r) as the disc of radius r centered at x. D = {D(x, 1/2): for all x with positive integer coordinates} If we take this D, then the proof is utterly trivial, because we can just apply the same argument that you used to prove the denumerability of N2. D = {D(x, 1): for x = (r, θ) with r = eθ and θ in N} This one is super trivial. I'm not really in the mood too cook up an esoteric example, but those are two definitions that yield well defined sets. They also illustrate the value of assuming that something you want is well defined: it lets you construct general proofs. with either of those sets, the particular proof is trivial. As for as for your question about well-definition requiring a finite definition, I don't know how to answer that because I don't know what an infinite definition would look like. Both of the examples above 'yield' an infinite set. You mention that you're a computer programmer. If you've done any functional programming (and CS profs love love love functional languages) then you are probably familiar with the filter function. think of a set definition like the filter function. I toss stuff into it and if it passes through the definition, then it's part of the set. I also reread my last post and am going to edit it a bit to make it clearer and fix one ugly typo.

23.  I see now. I guess the problem was my insufficient grasp of the concept of well-definition, as well as my bringing abstract notions like infinite definitions into the discussions (I meant a definition that takes infinite time to write down – then again, maybe that's just nonsensical). I appreciate your patience and your explanations --- On topic again: I just stumbled upon Cantor's theorem, which seems to be pretty much all you need to show that the set of all subsets of N is not countable. Here's my rephrasing it for the given problem (the actual theorem shows that any set has as a strictly lesser cardinality then its power set): Let's assume that we can count P(N), the set of all subsets of N. This would mean that we have a bijective map f : N → P(N), i.e. a map f that provides a one-to-one correspondence between N and P(N). P(N) contains at least as many elements as N, because for any n ∈ N, P(N) contains the set {n}. So, constructing an injective map is quite easy. However, we get a problem with surjectivity, i.e. f cannot map to ALL elements in P(N). Here's why it doesn't work: Let's examine one particular subset of N, that is to say M = {n ∈ N | n ∉ f(n)},or verbally, the set of natural numbers that are not contained in the set that f maps them to. This particular subset of N will give us the desired contradiction. Here's how: M is clearly in P(N) (since it's a subset of N) and because we require that f maps to ALL elements of P(N), there must be some m ∈ N so that f(m) = M. Given any n ∈ N, is it in f(m) = M? Answer: n ∈ f(m) ⇔ n ∈ M ⇔ n ∉ f(n).Now, what if n = m? Answer (just replace all n with m): m ∈ f(m) ⇔ m ∈ M ⇔ m ∉ f(m).So, ultimately we have: m ∈ f(m) ⇔ m ∉ f(m). This, of course, is a contradiction, therefore our assumption must have been wrong and there is no such map f that provides a one-to-one correspondence between N and P(N). It follows that you cannot count P(N), the set of all subsets of N.

24.  Hmm, I think being a computer scientist may have given you a strange slant on this. I don't see why checking if a point is in a disc is something you'd even want to do... we're talking about infinite sets here so computability doesn't really come into it. When I visualise counting discs, I just think about running through N, hence running through Q^2, and hence hitting every disc in finite time. The formal proof would go something like... Construct a function from N to Q^2 [previously justified]. Due to the density of Q, we can always find a point in Q^2 inside any given disc. Hence we have a function that assigns some n to each disc. Note that it is not, as you said (I imagine you said it offhandedly), actually necessary to show that each disc has a unique n (in other words to show injectivity); in fact, with my mapping, it's pretty much as uninjective as can be; each disc has infinite elements in N which map to it. However, what we have done is shown that the discs will fit inside N; if the collection of discs is infinite, it also follows that the size of the set of discs is the same as the size of the set of N (as there is no infinity less than the size of N). You don't need to create a bijection. I wouldn't say that an infinite set is not well-defined. All that was required for the question was to know that these discs existed. I have an inkling that this may be related to the axiom of choice, but I haven't really studied that formally yet. The point is, there's really no problem in assuming that 'somebody' has created an infinite 'list' of discs. They might have used some kind of rule (like the Apollonian gasket) or have done it 'randomly', it's inconsequential. Again I think we're probably weirded out by each other's differences here due to our subtle conditioning by our different educations. You're always concerned with whether you can practically do something or not, but I guess in maths we create infinite objects all the time without worrying, where a computer scientist would probably say 'wurgh!' or something similar. I think most of what PS has said is correct (except the thing about his room mate's 'objection' to Cantor's work, which was less of an objection and more just completely ignoring it). Your proof for the uncountability of P(N) is perfect. It's essentially the diagonal slash method; create an object which differs from the nth listed object by something with something 'n'ish about it.

25.  Originally Posted by Xei;1518637s I don't see why checking if a point is in a disc is something you'd even want to do... we're talking about infinite sets here so computability doesn't really come into it. When I visualise counting discs, I just think about running through N, hence running through Q^2, and hence hitting every disc in finite time. Much clear than mine. Other than bringing 'time' into it of course. This is math not physics Originally Posted by Xei;1518637s I think most of what PS has said is correct (except the thing about his room mate's 'objection' to Cantor's work, which was less of an objection and more just completely ignoring it). No, that was correct. My room mate did in fact say that. In fact we almost came to blows over it a few times. If there's something that's not correct, please call me out on it. I am a little rusty.

Page 1 of 2 1 2 Last

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•