Xerxes

As to quantum fields I know literally nothing, but otherwise yes. As to my fudging by using the term "space", yes I was well aware I was doing something slightly worse than hand-waving. In mitigation, I did not want to use the "m-word" without explaining what a manifold is, which in turn I didn't feel I could do justice to without first going through topological spaces. A big ask in a single post!! As a matter of interest - suppose in my last post one assumes my "space" a manifold (which my notation clearly telegraphed to the initiated), would it not be sufficient that the manifold is a least [math]C^2[/math] i.e. not necessarily smooth?

Of course, hence my somewhat sarcastic appeal. But lookee here. I quite myself Although this is not be best definition of a (say) a vector field, it tells us all we need to know. We can still have fun with it, though. Suppose I denote our "space" by [math]M[/math]. (Some of you you will know why I make this choice, but for now it is unimportant). Suppose that I denote the collection of all vector fields on [math]M[/math] as [math]\mathcal{X}[/math]. By construction there may be any number of possible fields on [math]M[/math]i.e there may be any number of "choices" of the single vector [math]v[/math] at [math]p \in M[/math]. Now since, by the definition of a vector space restricted to [math]p \in M[/math]I will have that 1. [math]v = u+w[/math] in this vector space 2. [math]\alpha v = z [/math] in this space it follows that, for the vector fields [math]X,\,\,Y \in \mathcal{X}[/math], then (say)[math]\alpha X + \beta Y[/math] for scalars [math] \alpha,\,,\,\beta[/math]is also a vector field. BUT this is nothing more nor less than the definition of a vector space. Which leads to the curious (but true) conclusion that [math]\mathcal{X}[/math] is a vector space. And since by construction [math]X,\,\,Y \in \mathcal{X} [/math] and since elements in a vector space are vectorswe are left with the curious conclusion that a vector field is a vector!! Same is true for scalar and tensor fields, so next time you hear a mathematician tutting (tsk, tsk)when a physicist talks about the metric tensor (which is both a field and a vector) or the curvature tensor, bear this in mind.

You are welcome. Just remember - everybody - this facility is open to abuse. As in DO NOT DO IT

Yeah, it took me a while to figure it. Bottom right click "quote". You will likely get a pretty useless box that will not allow you to split the text you want to quote In the quote box look for the top left icon, click, and you then have plain ASCI text. Following the portion of the text you wish to respond to enter [ /quote]. Enter your reply. Delete the rest, or continue as above, but this time remember to start the relevant portion of the remaining text with [ quote]

If you know of one (or more) rings with no additive identity, I implore you to share them with us.

So, you truly believe that I learned my mathematics from some "dictionary" ? I can assure you this is not the case. I can also assure you that the definition I gave of rings vs, fields is correct. If you want a lesson on integral domains, rings and fields just ask. Otherwise, do the other thing......

Conway, first note that in mathematics the term "field"is context-dependent. In abstract algebra a ring is defined as a set together with a pair of closed binary operations (typically + and x), each with an identity (respectively here 0 and 1) and an inverse for the operation + but none for x. A field in abstract algebra is defined as a ring with the additional property of an inverse for the operation x. Rings and fields in this sense comprise numbers of various sorts (mathematicians call them scalars) which, in applications, are generally taken to be Real or Complex. In linear algebra, on the other hand, a field assigns to each and every point in some chosen space a value, generally scalar-, vector- or tensor-valued. Here relativistic effects do NOT apply to scalar fields. So to your question: In the sense of abstract algebra, relativistic considerations are out of context - they simply do not apply. Moreover, the assertion that every set contains the empty set as a predecessor set can be used in the set-theoretic construction of the Ordinals and not, as far as I am aware, in the construction of a field of any description. The convention that, in the construction of the Ordinals, the empty set be re-labelled as 0 and its successor set as 1 etc is carried over to the Real numbers (with 0 as a field additive identity) seems to depend only on the simple observation that for any set [math]S[/math] that [math]S \cup \emptyset = S[/math]. Confused? You should be! This is a HUGE area of mathematics, which some of us have spent (wasted?) years studying

This is getting nuts!! Since I seem unable to cut and reply to quotes, let mesimply enumerate. 1. Orthogonality is a property of vectors. Specifically, 2 vectors are said to be orthogonal if their inner product is zero. So this is the incorrect term to use for planes that are mutually perpendicular 2. The mutualy perpendicular planes that Fred refers to are themselves 2-dimensional vector spaces. So that any space they enclose will be 6-dimensional. Such spaces do exist - they are called configuration spaces - but clearly this is not what is intended here 3. There is absolutely nothing special about the common use of the word "volume". I for one am quite happy to consider the area enclosed by the 1-sphere (i.e. circle) as a 2-volume (since it is a region of the 2-plane). Likewise the "area" enlcosed by any n-sphere (being a region of the embeddiding n+1 plane) as an n+1 volume 4. The universe, according to most people, is modelled as a 4-manifold. Being the "universe" there cannot be assumed to be an embedding space, so all statements must be intrinsic to this 4-manifold. Hence the term "volume", at least in the above sense, has no meaning. 5. Guess what..... we can aslo do intrinsic gemeotry on the 1-sphere, the 2-sphere etc

This is soooo disappointing. Look, nobody here (as far as I know) disputes the truth of the continuum hypothesis, but the contributors to this thread have tried to show you that your "proof" is not valid. Some of us have worked quite hard to explain why, and to give you guidance on fundamental mathematical topics. Your response? Change a word here and there in your original presentation, often leading you even further astray (e.g. the set of Real numbers is NOT compact). I bear no ill will towards you, but if you refuse to learn mathematics either here or elsewhere, then I am done (even though I have enjoyed contributing to this thread). Good luck

Interesting point, and one that persuades me that studiot and uncool are correct - point set topology only can answer pengkuan's issues. Given the set of Real numbers [math]\mathbb{R}[/math], how can one be sure that any 2 members are the same or different? Obviously, given the Euclidean metric one can say that if [math]|x-y|=0[/math] then [math]x=y[/math] and not otherwise. But suppose no such metric exists? Let us suppose the Real Line [math]R^1[/math] (as it is commonly understood) and accept that this has what is called the "standard topology" on [math]\mathbb{R}[/math].That is, it is the union of all open intervals (open sets) of the form [math](a,b)[/math] with [math]a<x,y,z,....<b \in (a,b)[/math]. Let us define the Hausdorff property of any topological space as follows: If there exist open sets [math]U,\,\,V[/math] containing [math]x,\,\,y[/math] respectively and if these sets are disjoint i.e. [math]U \cap V= \emptyset[/math], however "small" they might be, we may say that [math]x \ne y[/math] (and not otherwise). Clearly in [math]R^1[/math] they are "small". How small? Well, the smallest set that contains [math]x[/math] as an element is the singleton set [math]\{x\}[/math], but this is not an open set in any topology - it is both open and closed which, for reasons I can't be bothered to explain, means they are elements in a non-connected topological space. So what is the point of this ramble? Well, if we agree that [math]R^1[/math] is "continuous", that it is connected and that it has the Hausdorff property, then, for any [math]x,\,\,y \in R^1[/math] it is always possible to find disjoint open sets not singletons that contain them Phew! If you read all of that, bravo indeed

First a word on notation and terminolgy. The term "interval" is reserved for a segment of the Real Line i.e all the Real numbers arranged according to their ordering. The notation [math]]0,1[[/math] or equivalently [math](0,1)[/math] refers to a segment - a subset - of the Real line that includes all the Real numberbs rhat lie betweenn 0 and 1 but does not include 0 or 1. Anyway, let's try to work with what you have given us, taken a face value. So you take a 2-element set [math]\{0,1\}[/math] and then remove 0 and 1 to form [math]]0,1[[/math] which is by any argument empty. Now to this empty set you add [math]\frac{1}{2}[/math] and form the 2sets [math]]0,\frac{1}{2}[[/math] and [math]]\frac{1}{2},1[[/math] both of which are empty. But there can be only one empty set. So what have you done? You have effectively iteratively added a number to the empty set and promptly taken it away and then found you are left with the empty set!! I'm not about to write home to mother about that BTW I don't agree that all sets are discrete in the strict sense - they may have a non-empty intersction, neither do I agree that it is improper to define a set in terms of how it is created, butthese are seperate issues, perhaps

Let me try to stop this getting ugly. Pengkuan is confused but at least he is honest - as uncool rightly says he is using terms which properly belong in point set topology. Hell, even the Real line [math]R^1[/math] is a topological construction, but pengkan admits he knows nothing of these matters. That's fine - it is not a crime. Some of us may consider abusing termiology as a crime, but hey, I think now I see what he is saying. Let's use pengkuan's definitions of a "continuous set" versus a "discrete set". So the assertion is that they have different cardinalities. Fine. Let us take my last post as a (semi)proof that the Real numbers [math]\mathbb{R}[/math] are complete. Let's further assume that this is what pengkuan means by "continuous". Here is another denifintion of a complete ordered field [math]\mathbb{F}[/math]. If for every monotonically decreasing sequence in [math]\mathbb{F}[/math] there is a limit in [math]\mathbb{F}[/math] then this field is complete. I assert that [math] \mathbb{R}[/math] is complete is this sense as well as in my last sense i.e. has no "gaps". I call on Cantor to tell me that the cardinality of [math]\mathbb{R}[/math] is uncountable. I now consider the Rationals [math]\mathbb{Q}[/math] as a subset of [math]\mathbb{R}[/math], and consider the sequence of partial sums [math]S_n=\sum \nolimits_{n=0}^{\infty}\frac{1}{n!}[/math] which converges on [math]e[/math] which is not rational. Therefore [math]\mathbb{Q}[/math] is not complete. Let's assume that is what pengkuan means by "discrete" And since any element in the set of rational numbers can be written as the quotient of two natural numbers (excluding 0), and since the natural numbers are essentially by definition countable, then [math]\mathbb{Q}[/math] must also be a countable set. So yes, using pengkuan's clumsy (and wrong) notation, continuous set is uncountable whereas a discrete set is countable. This applies to the Real numbers and the Rational numbers as a subset - whether it applies more genarally, I cannot be bothered to think about just now

As a late-comer to the party, here's my bottle.... The OP (and indeed the title) sems to confuse the noun "continuum" with the adjective "continuous". They are not related - continuity refers to functions whereas a continuum is taken as an ordered domain that is complete, very loosely indeed this means no "gaps". Try this. Suppose the real numbers [math]\mathbb{R}[/math] ordered in the usual sense. Define subsets [math]U,\,L[/math] such that 1. every [math]u \in U[/math] is an upper bound for the set [math]L[/math]. 2. every [math]l \in L[/math] is a lower bound for [math]U[/math] Since [math]U[/math] inherits the order from [math]\mathbb{R}[/math], there must be a least element (say) [math]x \in U[/math]. So [math]x[/math] is a lower bound for [math]U[/math], and therefore, by construction must also be in [math]L[/math] That is, the "cut" that defines [math]U,\,L[/math] passes "through" [math]x[/math]. And since the "cut point" is entirely arbitrary, one may say there are no "gaps" in [math]\mathbb{R}[/math] so the Real numbers are a complete ordered domain (field in fact) and can be referred to as a continuum

Try induction on [math]x^2-1=0[/math] which has roots [math]\sqrt{1}[/math]. Clearly [math]-1[/math] is primitive since it is not a root for [math]x^1-1=0[/math] i.e. assume for your induction hypothesis that the roots for [math]x^{n-1}-1=0[/math] are [math]^{n-1}\sqrt{1}[/math] BUT beware of multiplicities!

Well I am not sure what you mean by a "process" - let's be conventional and call it a morphism. Assuming I was correct and we are working in the category of fields, then our morphisms are field homomorphisms. Now I will say that 2 such morphisms are equal if and only if they have the same image and the same pre-image. Now consider the "central" object in your diagram [math]R^2 \equiv R \times R[/math]. The projections [math]p_1((x,y)) = x \in R[/math] and [math]p_2((x,y)) = y \in R[/math] always exist. And if [math]f_1=f_2[/math] then their images are equal, say to [math]x[/math]. This mandates that [math]p_1=p_2[/math] (else your diagram makes no sense in any category) in which case your diagram is completly trivial. Otherwise we must have that [math]f_1 \ne f_2[/math] Hmm....is this clear?

Why do you think [math]f_1=f_2[/math] in this case?,. Yes - you rather missed the point. We require your diagram to commute. That is the composite, say, [math]g_1:C \to R^2[/math] exists such that [math]p_1 \circ g_1 = f_1[/math]. Likewise for the "bottom half" of your diagram. Of course the morphism [math](g_1,g_2):C \to R^2[/math] makes no sense notationally, so call it anything you want. In this case it is, of course an isomorphism, though I am not sure this is always the case

In diagrams of this sort, it is importat to "start" in the right place, in this cas it is [math]R^2[/math] Assume we are working in the category of fields Fld.where [math]R[/math] dentes the Real field Then the projections (morphisma) [math]p_1 \to R[/math] and [math]p_2 \to R[/math] define the morphisms [math]R^2 \equiv R \times R \to R[/math]. And if there exist another pair of morphisms [math]f_1:X \to R[/math] and [math]f_2: X \to R[/math] then this induces the morphism [math]X \to R^2[/math].(Personally in this circumstance I orefer to use a dashed arrow for the induced morkisms - it's a matter of taste) In Fld an example would be [math]X = C[/math], the complex field

John, you getting close to insulting me, which I (obviously) do not like For the resord, I understand the Arhimedean perperty of the Reals perfectly, and I know exactly what is meant by a cyclic group of infinite order. So stop it. Moreover, my comment about there being "a kargest Real number" was part of a logical argument to show a contradiction in what I assumed to be your point, that no Real number can be written as a non-terminating string of Natural numbers. If that assumption was wrong, I apologize. Let me ask you this, then. Is it your position that "all Real numbers are finite"? So my question was, and is, this. What do you mean by "a finite number"? Is it that no Real number can be written as a non-terminating string of Natural numbers? Or do you mean that, viewed as a set, the "string representation" of a Real number must have finite cardinality? Or do you mean something else that I have missed?

OK, I assume as you objected to trailing ellipses in a real number, this implies you think every real number can be written as a terminating string of natural numbers. Since the field [math]\mathbb{R}[/math] is a total order, this in turn implies there is a largest real number. Call it [math]x[/math]. Then the Archimedean property claims that there exist some integer [math]n[/math] and some [math]y >0 \in \mathbb{R}[/math] such that [math]ny >x[/math]. Contradiction Moreover, the ingeters [math]\mathbb{Z}[/math] form (under addition) a cyclic group of infinite order, that is, given a generator = 1, then 1 + 1 + 1 +.......= 0 only if we allow this sum to go out to infinity

I disagree - in fact you have not said what exactly you mean by a "finite number" or "infiinite number" Try this. Let [math]S[/math] be set. For every set element [math]x[/math] define the succesor [math]x^+[/math] by [math]x^+ = x \cup \{x\}[/math] Thus, say [math]\O^+ = \O \cup \{\O\}= \{\O\}[/math] I re-label [math]\O = 0,\,\, \{\O\}=1[/math] etc to obtain the natural numbers so [math]0^+=1,\,\,1^+=2[/math] and "so on". In order to give meaning to the "and son on" I need the axiom of infinity which states There exists a set containing 0 and all of its succesors. The first transfinite number so defined is usually written as [math]\omega [/math]. So one may have the seqquence [math]1,2,3,4,5,........,\omega[/math] and then [math]\omega+1,\omega+2,\omega+3,.........,2\omega[/math] and then [math]2\omega+1, 2 \omega+2,.......3 \omega,.........,\omega^2,.......[/math]. Since the Real numbers contain the Natural numbers as a proper subset, there is no difficulty having a Real number the set of whose predecessors has infinite cardinality i.e which can be written as a non-terminating string of natural numbers

Please don't patronize me Look every natural number is an integer is a rational number is real number. The reverse "chain" is false however No non-terminating decimal without a repeating pattern can possibly be rational

So you multipy by 2 then divide by 2 and get back the same number. Not very surprisimg, is it? So in order to assert that 23571113 is rational, you start by assumiing that it is ratiomal. Likewise for 2.357,,,,,, being irratiomal This we call asssuming the truth of the proposition to be proved

Then maybe I misunderstood the OP. Was I wrong to assume it was saying, for example, that if [math]p_1=2,\,p_2=3,\, p_3=5[/math] etc then we are talking about a string of the form [math]p_1p_2p_3,p_4........[/math] which is clearly a natural number, and as such is a subset of the Real numbers and therefore real. Is it your contention that this number is, say, complex? Do you doubt that a natural number can be as "large" as we like? And is it your contention that, given the OP, we may have [math]p_1. p_2 p_3.....p_2 p_3.....p_2 p_3.....[/math] as a rational real number? If so I have misunderstood the OP and I apologize. If not, the above is gibberish

This is silly. If 2 X 23571113..... 2 proves that 23571113..... is rational then why is it NOT the case that 2 X 0.23571113..... 2 is also rational?

No it is not irrational. Since your "list" is a natural nuber, it cannot possibly be irrational. The fact that the ellipses imply it "goes on forever" is neither here nor there, since the natural numbers have a countably infinite number of elements. Interestingly, it is one of the quirks of modulo 9 arithmetic i.e. base 10, that once you introduce the decimal point, everything changes. So no real number, with trailing ellipses to right of the point can be rational (unless of course the ellipses denotes a repeating pattern)