The Definition of Points
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From: Mike Kelly on 21 Apr 2007 06:58 On 20 Apr, 17:13, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 19, 7:06 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> MoeBlee wrote: > > >>> You just need to learn it theorem by theorem from the axioms. All it > >>> takes is an introductory set theory textbook, and all of them will > >>> prove the well ordering theorem for you. > > >> But, a well order cannot have infinite descending chains of any sort, so > >> if there are an uncountable number of limit ordinals within the order, > >> then there will exist an infinite descending chain within that sequence > >> of limit ordinals itself. How does one get around that? > > > Again, you're reasoning from vague ideas you have of what the > > terminology means. If you were to pin down these terms you use such as > > 'infinite descending chain' with the precise definitions, then tried > > to derive a formal contradiction, you wouldn't succeed to prove any > > such contradiction nor evidence of anything that we must "get around" > > here. Ironically, your response is right after I reiterated that you > > need to study this in a systematic and precise way, definition by > > definition, theorem by theorem. It is just more effort down the drain > > for me to untangle your confusions over terms of which you've not > > bothered to learn their precise definitions from primitives and > > axioms. > > > A well ordering doesn't preclude that a member of the field of the > > ordering may have infinitely many predecessors. What a well ordering > > does demand is that in any nonempty set of predeccessors, there is a > > least one, which doesn't preclude that there may be infinitely many > > "on top" (as we're "working downwards") of that least one. You don't > > even need to go to the uncountable; all you need is just one limit > > ordinal in the field of the well ordering to see the point: Just take w > > +1. The well ordering on w+1 by the membership relation has w itself > > as member of the field of the ordering. But to "travel from" w "to" > > the least member of w requires "travelling through" an infinte number > > of members of the domain of the field of the ordering. > > As it was explained to me, because every natural is finitely far from > the lest in w, there is no infinite descending chain of predecessors one > can define. In other words, between any two limit ordinals can only a > countable number of elements. Now, can a well order have an infinite > descending chain of limit elements? If so, then an uncountable set could > be partitioned into an uncountable set of countable partitions, and be > well ordered, without a doubt, and it would already have been done. There's a difference between having a limit ordinal as an element and having every element of a limit ordinal as an element. -- mike.
From: Mike Kelly on 21 Apr 2007 07:14 On 20 Apr, 17:24, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 19, 7:01 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >>>> If you say in the same breath, "there > >>>> are infinitely many rationals for each natural and there are as many > >>>> naturals overall as there are rationals", > >>> And infinitely many naturals for each rational. > >> How do you figure? In each 1-unit real interval, there is exactly one > >> natural, and an infinite number of rationals. Which interval has one > >> rational and an infinite number of naturals? > > > That there are denumerably many rationals but only one natural in a > > unit interval defined by the standard ordering on rationals doesn't > > refute that there exists the kind of correspondence Mike Kelly > > mentioned. Several months ago I defined for you a dense linear > > ordering on the set of natural numbers. You ignored it. > > I did? I was out of touch with this stuff for a while. Maybe that was > then. Sorry > > >>> Or, maybe, other people don't share the same intuitions as you!? Do > >>> you really find this so hard to believe? > >> Most people find transfinite cardinality "counterintuitive". Surely, you > >> don't dispute that. > > > Probably most people who are not familiar with the theorem by theorem > > proofs of set theory would find uncountability unintuitive. But I know > > of no evidence that most people in general can't grasp the idea of an > > infinite set such as the set of natural numbers if the matter is > > presented to them in a clear way. Then, if one accepts that there are > > infinite sets, and one grasps the notion of a power set and some other > > basic concepts about sets, uncountability follows. > > Some people have problems with uncountability, but that's not what I'm > talking about. Most people find it strange that half the naturals are > even, but that half is considered the same "size". Evidence for this? When people are introduced to cardinality in freshman analysis they don't generally seem to have problems with it. Of course, freshman analysis teaches things in a sensible order. You're taught about relations, funtions, injections, surjections and bijections and so on right at the start. When it's time to discuss cardinality, you look at bijecting infinite sets. You discover that the naturals and the evens can be bijected. Naturals and integers. Naturals and rationals. Then you learn cantor's diagonal argument about set and powerset being unbijectible. And see that the reals and naturals can not be bijected. So, not all infinite sets can be bijected. Interesting. So let's come up with some way to denote these "bijectibility classes". Only then are cardinal numbers introduced. And they aren't controversial. We've already seen that the naturals and the evens are both bijectible, and that bijection aligns with our intuitive notion of size in the finite case. Why is it so horrible to denote their bijectibility with what we call a number, and informally refer to cardinality as size? > I really don't have any problem with cardinality, Took you long enough to admit that, eh? > but I wouldn't call any transfinite cardinalities "quantities" by any means. Nor would I. So what? >I think referring to them as "numbers" is what makes the theory suspicious to people. It's what makes it suspicous to cranks on the internet. They see the word "number" and are so mathematically stunted that they assume this means "number like what I deals with on my calculator". Then they find, horrors, that infinite cardinal numbers are different from finite cardinal numbers. How very confusing! > >>> Well, maybe you'd like to do that. But you have made no progress > >>> whatsoever in two years. Mainly, I think, because you have devoted > >>> rather too much time to very silly critiques of current stuff and > >>> rather too little to humbling yourself and actually learning > >>> something. > >> That may be your assessment, but you really don't pay attention to my > >> points anyway, except to defend the status quo, so I don't take that too > >> seriously. > > > You're ridiculous. Certain people have even OVER-indulged you by > > showing in detail, ad nauseam, exactly what's whack in your various > > proposals. And it doesn't even matter WHO is telling you - that you > > need to study this stuff from a good textbook is just plain, basic, > > good adive that you do need to take seriously. Or continue to be a > > nonsense spouting crank. Your choice. > > There have been lots of objections, and a few valid points, but no major > flaws detected in what I propose. It's just not compatible with ZFC. It's not compatible with formalisation in general. What is your goal in all this? Are you trying to come up with an alternative "geometrical" foundation? Are you trying to build on ZFC to define "bigulosity"? You claim they're incompatible, but mundane stuff like asymptotic density in the naturals is certainly definable in ZFC. Finally, do you think your "proposal" will lead to any kind of novel applied mathematics? Does bigulosity do anything for us other than claim "the bigulosity of the evens is N/2"? > >>> Get this through your head : every relation between objects in set > >>> theory is based on 'e'. It's really pathetic to keep mindlessly > >>> denying this. Set theory doesn't just "try to base everything on 'e'". > >>> It succeeds. > > >> If you say so. > > > No, not just "if he says so". He says so and he's RIGHT. And you can > > VERIFY for yourself just by reading a textbook already. > > > MoeBlee > > So, defining N doesn't involve a successor relation between two > elements, as well as a member relation between an element and a set? > When you define N, doesn't the rule E x e N -> E succ(x) e N define a > relation between two elements, as well as between those elements and N? Successor is defined in terms of 'e'. Or maybe we're lying to you. Crank. -- mike.
From: Mike Kelly on 21 Apr 2007 07:16 On 20 Apr, 20:09, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 20, 9:24 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> MoeBlee wrote: > > >> So, defining N doesn't involve a successor relation between two > >> elements, as well as a member relation between an element and a set? > > > The successor operation is DEFINED in terms of the membership > > relation. EVERYTHING in set theory is defined in terms of the > > membership relation. The only non-logical primitives of set theory are > > '=' and 'e' (and we could even define '=' in terms of 'e' if we want > > to set it up that way). There is NO formula of set theory that doesn't > > revert to a formula in the primitive language with just 'e' and > > '=' (or even just 'e') as the ONLY non-logical symbols. I've been > > telling you this for probably over a year now. Why don't you > > understand this? > > The particular successor relation for the vN ordinals is defined in > terms of 'e', but that's not the only model of successorship. But every relation in set theory is defined in terms of 'e'. I thought we were talking about set theory? > >> When you define N, doesn't the rule E x e N -> E succ(x) e N define a > >> relation between two elements, as well as between those elements and N? > > > "Define a relation". From N of course we can define the successor > > relation on N. So what? That doesn't refute that every definition in > > set theory ultimately reverts to the membership relation. I'll say it > > YET AGAIN: > > Before you go on your rant, let me just say this. As a relation, I > specifically mean that each combination of inputs produces either a 0 or > 1. It's a truth table conception of what a relation is. Oh, we're not talking about set theory. -- mike.
From: Lester Zick on 21 Apr 2007 14:52 On Fri, 20 Apr 2007 12:00:37 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> What truth have you demonstrated without positing first? >> >> And what truth have you demonstrated at all? >> >> ~v~~ > >Assuming two truth values as 0-place operators, I demonstrated that >not(x) is the only functional 1-place operator, and then developed the >2-place operators mechanically from there. That was truth about truth. And how have you demonstrated the truth of the two "truth values" you assumed? ~v~~
From: stephen on 21 Apr 2007 19:31
In sci.math Tony Orlow <tony(a)lightlink.com> wrote: > stephen(a)nomail.com wrote: >> In sci.math Tony Orlow <tony(a)lightlink.com> wrote: >>> stephen(a)nomail.com wrote: >>>> In sci.math Tony Orlow <tony(a)lightlink.com> wrote: >>>>> stephen(a)nomail.com wrote: >>>>>> In sci.math Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>> Mike Kelly wrote: >>>>>>>> The point that it DOESN'T MATTER whther you take cardinality to mean >>>>>>>> "size". It's ludicrous to respond to that point with "but I don't take >>>>>>>> cardinality to mean 'size'"! >>>>>>>> >>>>>>>> -- >>>>>>>> mike. >>>>>>>> >>>>>>> You may laugh as you like, but numbers represent measure, and measure is >>>>>>> built on "size" or "count". >>>>>> What "measure", "size" or "count" does the imaginary number i represent? Is i a number? >>>>>> The word "number" is used to describe things that do not represent any sort of "size". >>>>>> >>>>>> Stephen >>>>> Start with zero: E 0 >>>>> Define the naturals: Ex -> Ex+1 >>>>> Define the integers: Ex -> Ex-1 >>>>> Define imaginary integers: Ex -> sqrt(x) >>>>> i=sqrt(0-(0+1)), so it's built from 0 and 1, using three operators. It's >>>>> compounded from the naturals. >>>> That does not answer the question of what "measure", "size" or "count" i represents. >>>> And it is wrong on other levels as well. You just pulled "sqrt" out of the >>>> air. You did not define it. Claiming that it is a primitive operator seems >>>> a bit like cheating. And if I understand your odd notation, the sqrt(2) >>>> is an imaginary integer according to you? And sqrt(4) is also an imaginary integer? >> >>> No, but sqrt on the negatives produces imaginary numbers. Besides, sqrt >>> can be defined, like + or -, geometrically, through construction. >> >> You cannot define the sqrt(-1) geometrically. You are never going to draw a line >> with a length of i. >> >>>> You also have to be careful about about claiming that i=sqrt(-1). It is much safer >>>> to say that i*i=-1. If you do not see the difference, maybe you should explore the >>>> implications of i=sqrt(-1). >>>> >>>> So what is wrong with >>>> Start with zero: E 0 >>>> Define the naturals: Ex -> Ex+1 >>>> Define omega: Ax >>>> I did that using only one operator. >>>> >>>> >> >>> Ax? You mean, Ax x<w? That's fine, but it doesn't mean that w-1<w is >>> incorrect. >> >> No, I meant Ax. All of the natural numbers. I know you are incapable of >> actually imagining "all", but others do not have that limitation. >> >> Of course, who knows what your notation is really supposed to mean. >> What is (Ax)-1 supposed to be? How do you subtract one from all the naturals? >> >>>>> A nice picture of i is the length of the leg of a triangle with a >>>>> hypotenuse of 1 and a leg of sqrt(2), if that makes any sense. It's kind >>>>> of like the difference between a duck. :) >>>> That does not make any sense. There is no point in giving a nonsensical >>>> answer, unless you are aiming to emulate Lester. >>>> >>>> Stephen >>>> >>>> >> >>> It's not nonsensical, and may even apply to uses of imaginary numbers in >>> practice, but you can ignore it as I knew you would. That's okay. >> >>> Tony >> >> It is nonsense. Such a triangle does not exist. >> >> A >> # >> # >> # >> # >> # >> C############B >> >> Are you claiming this is a triangle? Are you claiming the distance from A to C >> is i? How exactly is this supposed to be a picture of i? What is i measuring >> in this picture? >> >> If this is not what you meant, please draw your picture of i. >> >> And you still have not answered what "size", "count" or "measure" i represents. >> Is i a number, or not? >> >> Stephen > i is a number produced by extending completeness to the multiplicative > field, after introducing completeness to the additive field to produce > -1, like I said above. > Consider that you have a right triangle so A^2+B^2=C^2. Now, if you swap > the values between the hypotenuse and one of the legs, that formula will > produce a second leg of the original length times i. So, in a way it is > like picturing an impossible triangle. You have to "imagine" the other > leg of length i. :) Swap the values between the hypotenuse and one of the legs? What exactly does that mean? > A > |\ > | \ > | \ 1 > sqrt(2) | \ > | \ > | \ > B i C So you are claiming that the obviously longer line from A to C is shorter than the line from A to B? So you can just assign any old "length" to a line regardless of how long it actually is? There is a way to make sense of this, but it is quite at odds with traditional geometry, which seems to be the basis for all your arguments. Stephen |