The Definition of Points
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From: Lester Zick on 24 Apr 2007 14:44 On Mon, 23 Apr 2007 21:47:51 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> Truth tables and logical statements involving variables are >>> just that. If I say, 3x+3=15, is that true? No, we say that IF that's >>> true, THEN we can deduce that x=4. >> >> But here you're just appealing to syllogistic inference and truisms >> because your statement is incomplete. You can't say what the "truth" >> of the statements is or isn't until x is specified. So you abate the >> issue until x is specified and denote the statement as problematic. > >I don't consider it a problem, but a state of affairs. If one doesn't >know what the state of affairs IS, they can say, IF it's THIS then THIS, >and IF THAT then THAT. I don't know if THIS or THAT, but once I know, >then I'll know. That how logic is. This comes back to axioms. Surely (or >probably, or maybe) you see that. Well certainly it's a problematic state of affairs, Tony, but look up the meaning of "problematic". It means "questionable" not necessarily causing problems. If one doesn't know certain factors circumstances are certainly problematic regarding the truth of the proposition. ~v~~
From: Lester Zick on 24 Apr 2007 14:53 On Mon, 23 Apr 2007 21:47:51 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >> The difficulty with syllogistic inference and the truism is Aristotle >> never got beyond it by being able to demonstrate what if anything >> conceptual was actually true. > >May I address that? When I say that true() and false() are "zero-place" >logical operators, that really means that they take no "logical >parameters". That is, what is passed to the function for evaluation >includes neither "true" nor "false" but is rather a statement which is >to be evaluated, in the context of whatever "theory" we are considering, >as either "true(x)" or "false(x)". Wouldn't "true(1)" represent such a statement? > The truth of x or y depends on what >rules we consider valid in out language. But just "considering rules valid" doesn't make them or their results true, Tony, and that's what I'm trying to ascertain. You claim to have proven certain things about the nature of truth true but those things rest on assumptions of truth regarding the nature of truth in specific contexts which aren't necessarily true in general. Really all you seem to have done is describe rules you consider true versus proving them true. ~v~~
From: Lester Zick on 24 Apr 2007 15:13 On Tue, 24 Apr 2007 09:27:05 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> The "truth" of the two truth values is that I've declared them a priori >>> as the two alternative evaluations for the truth of a statement. >> >> Well there are two difficulties here, Tony. You declare two a priori >> alternatives but how do you know they are in fact alternatives? In >> other words what mechanism causes them to alternate from one to the >> other? It looks to me like what you actually mean is that you declare >> two values 1 and 0 having nothing in particular to do with "truth" or >> anything except 1 and 0. > >1=not(0) >0=not(1) >x<>not(x) > >That says it all. "not" defines the relationship between 0 and 1. In >arithmetic terms, not(x) can be equated with 1-x, for obvious intuitive >reasons. If 1 is the universe, and x is some portion of that universe, >everything that is not(x) is everything in the universe, minus what's in x. > >"not not" then equates to 1-(1-x), which is just x, which doesn't have >any truth value until you assign it one. Okay, Tony. Basically you're agreeing with what I've been saying. But I'm still bothered by this binary arithmetic approach to the mechanics of "not" and "not" compounded in terms of itself. I think you really need a symbolic rather than arithmetic approach to the problem. In other words you've got the idea of differences but the differences are not between binary logic values. They're taken between real things in logical terms which can result in any and all kinds of differences any and all of which can be regurgitated tautologically in terms of subsequent differences. The 0 stays but the binary 1 goes. We can't just take the truth of one set of circumstances and compare it to the truth of another set of circumstances. Both sets may be true but may be true in different ways which preclude analyzing the truth of both together in exactly common terms. This is why I analyze the problem in terms of predicate combinations and not in strictly binary terms. We might have a "large apple" and a "green apple" both of which predicate combinations could be true of one specific real object without having any common "truth value" in terms of each other. This is why I analyze the problem in symbolic terms rather than strictly binary terms of common truth values. ~v~~
From: Lester Zick on 24 Apr 2007 15:40 On Tue, 24 Apr 2007 09:27:05 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >> The second problem is what makes you think two "truth" alternatives >> you declare are exhaustive? This is related to the first difficulty. >> You can certainly assume one thing or alternative a priori but not >> two. And without some mechanism to produce the second from the first >> and in turn the first from the second exclusively you just wind up >> with an non mechanical dualism where there is no demonstration the two >> are in fact alternatives at all or exhaustive alternatives either. > >See the mechanism for their mutual definition, above. > >As far as the two truth values being "exhaustive", it's simply declared >in Boolean logic that these are the only two values allowed, true() and >false(). Now, these are not really all the possibilities, if one looks >at truth in the context of probability. Where a probability of truth of >x is 100%, the truth value is 1, meaning "definitely true". Where a >statement has a 0% probability of truth, the value is 0, meaning >"definitely false". Between these two lie a continuous set of possible >probabilities. So, in this context, {0,1} is NOT an exhaustive set of >all alternatives. But, this requires that you accept partial truth, >which doesn't seem to be acceptable for you. What other alternatives do >you see besides true(x) and false(x) for a statement x? Well as noted in the preceeding reply segment there are certain subtle complexities in relations between true predicate combinations which obscure the idea of binary truth values but have nothing to do with probabilities whatsoever. I don't reject the idea of probability but I think the idea of truth in general has to be straightened out before we can get to any mechanically exact understanding of probability. I agree the common boolean notion of truth only allows for two truth values but I don't agree that actually reflects what truth means in strict mechanical terms. Taking two symbolic predicate combinations such as "green apple" and "large apple" both of which are true of one specific real thing we are then left to consider the truth of each in relation to the other. And what we find are that there are complex subordinated differences between them not represented in simple binary 1's. It's a sequential combinatorial logic problem where the truth of each predicate has to be considered individually and in combination. Let's suppose for example we have two predicates "animal" and "fox" both of which are true with respect to one specific real thing. And we represent the truth of each with a binary 1. Then we further consider the truth of the predicates in combination and we can't tell whether the thing is an "animal fox" or a "fox animal" from the representation of truth in each instance. Thus the binary arithmetic representation is completely misleading even though it correctly identifies the truth in each instance. It just can't tell us how the true but different predicates "fox" and "animal" relate to each other. And if they're related incorrectly the resulting predicate combination will not be true even though each predicate in the combination is. ~v~~
From: Tony Orlow on 24 Apr 2007 16:06
cbrown(a)cbrownsystems.com wrote: > On Apr 23, 10:18 am, Tony Orlow <t...(a)lightlink.com> wrote: >> cbr...(a)cbrownsystems.com wrote: >> >> <snip> > >>> So the "tree" criss-crosses like a chain link fence. This type of >>> partial order is usually called a lattice. >> Yes, it becomes a sort of a lattice-looking thing from one level to the >> next. It's actually the set of vertices of a |S|-dimensional cube, if >> the same subset may only occur once. > > More is required: to consider the ordering, we also need to include > the edges /between/ the vertices; and they must be /directed/ edges; > and there must be no cycles. That's a very "special" kind of cube; not > just /any/ cube will do. > The edges between the vertices represent the relationship between proper superset and subset, and are indeed directed. It's not a particularly "special" cube. The choice of a vertex on a n-dimensional cube can be considered a n-bit string, indicating which of the two directions afforded by each additional dimension one will choose to place the next point in that specification. Where we turn a bit from 1 into 0, which means we removed an element from a set to get a proper subset, the '<' relationship holds, whether in terms of binary string or subset, no? >> If you allow the redundancies, so >> that S-[x,y] appears both as a child of S-{x} and of S-{y}, then you get >> a tree, but not every element is unique. > > And that's why most would not consider it a tree, but instead a > lattice. It's more specific than a lattice, otherwise, but sure. > > Similarly, most would not consider a pentagon to be a cube; even > though one can (by replicating points to create "redundancies"), label > the points of a cube variously with the five points of a pentagon. > That's not particularly similar. >> I guess on each level n, where >> S is on level 0, one gets each unique subset n times, and the number of >> unique elements generated at each level is 2^n-n? Something like that. >> > > I'm not sure why you want a tree to be a proper representation of a > lattice. I was simply pointing out that it is completely at odds with > the definition of a tree to consider the partial ordering by inclusion > as being a tree when we consider vertices as representing distinct > elements, and the directed edges to represent the "<=" relation > (which seems perectly natural to me), > The recursive generation of the subsets acts as a tree. The fact that you can reach the same subset by different branches makes it redundant, but the process is treelike. Like I said, if you restrict nodes to the unique, then you necessarily create a binary hypercube. >> >>> In order to get a "tree", we need to make sure that we don't do this >>> kind of criss-crossing; which is more than you've stated in 1. and 2. >>> It's easy to see how to do this is S is finite. It's harder to see how >>> to do this if S is not finite, without some sort of choice function. >>> And then we get to the whole axiom of choice implies well-ordering >>> thing. >> I was assuming redundancy. >> > > To me, that means you were (implicitly) assuming, e.g., x = {a,b} and > y = {a,b}, but not x = y. Seems a bit convoluted, don't you think? > > <snip> > Not from a process perspective. Such a recursive definition may only produce new elements, or may produce redundant elements. We may consider each rational value to be unique, but the definition of rationals produces both 1/2 and 2/4, and is redundant. >>> A neN E meN : m>n >>> doesn't imply that m is the /smallest/ m satisfying m > n. Yes, we >>> know from the properties of the naturals that there /is/ a smallest >>> such m which would then be the m that "comes right after" n (thus >>> satisfying your allusion to a successor); but that isn't always the >>> case. >> Right. IF we define '>' to mean "successor" we get an inductive set, and >> if we define "successor" to mean "increment", then it seems to me we >> really get the naturals. >> > > Again, that is simply not enough. > > Yes, the naturals have the property that 1 = 0 + 1, 2 = 1 + 1, 3 = 2 + > 1 and so on. But that is not a /definition/ of the naturals; it is an > observation /about/ the naturals. > > This is just like your statement "Order is defined by x<y ^ y<z -> x < > z". Order is /not/ defined that way; you need more. And you need more > to define the naturals. > Like? >> >>> I'll formalize "m is the smallest natural which is larger than the >>> natural n" by: >>> (m,n in N) and (m > n) and (for all s in N\{m}, if s > n, then s > m) >>> But if we switch to the rationals, >>> (m,n in Q) and (m > n) and (for all s in Q\{m}, if s > n, then s > m) >>> then there is no such m for any n in Q. >> Well, once you have ordered the rationals so as to appear countable, ... > > "Appear"? 6 doesn't "appear" to be even; it /is/ even. > > Similarly, the rationals can be bijected with naturals. Therefore they > don't simply "appear" countable, they /are/ countable. > > They don't "appear" equinumerous with the naturals to me... >> ... then there is such an m for any n, in that order. That's changing the >> meaning of '>' from the normal quantitative interpretation of a rational >> to one of a different order. > > I wonder why, then, most people think that 3/2 > 5/6. What on earth do > you think they mean? Pardon me, but in Cantor's diagonal bijection between the rationals and the naturals, doesn't 3/2 come before 5/6? Doesn't that mean that, in countable rationals, 3/2 < 5/6? Isn't 1/1 the "smallest" rational, in that order? > >> Personally, I think comparing subsets of >> the reals out of quantitative order is one of the methods that lead to >> screwy results. I mean, what you just demonstrated is that, in the >> normal quantitative order, N is sparse and Q dense, which to most naive >> intuitions says |Q|>|N|. > > It all depends on what "most naive intuitions" mean by "|Q| > |N|". By > some definitions of ">" and "|X|" this would be true, and by others, > it would be false. This is similar to "3/2 > 5/6" being true for some > definitions of ">", and false by other definitions. It's true, quantitatively. > > <snip> > >>>>>>> Let's order the subsets of {a,b,c} by inclusion. Then: {a,b} < >>>>>>> {a,b,c}. {a,c} < {a,b,c}. {a} < {a,b}. {a} < {a,c}. But not ({a,b} < >>>>>>> {a,c}); and not ({a,c} < {a,b}). Does that mean that {a,b} = {a,c} >>>>>>> "for the purposes of that order"? >>>>>> Where b<c, {a,b}<{a,c}. >>>>> Since "<" indicates subset inclusion, it is the case that not b < c. >>>> If a, b and c are members of a set such that xeS ^ yeS -> x<y v y<x v >>>> x=y, then either b<c v b<c v b=c. >>> Yes, if a set is totally ordered (i.e., the ordered pair (S, <) >>> defines a /total order/ on the /elements/ of S), then (trivially) it >>> can be totally ordered. >>> But I was talking about the ordered pair (P(S), <) where < is subset >>> inclusion; which defines a /partial/ order on the /subsets/ of S. >>> In that case there are distinct /subsets/ X, Y of S with not (X < Y or >>> Y < X). Which is to say that there are distinct /elements/ X, Y of >>> P(S) with not (X < Y or Y < X) >> Yes, in that partial order. But, if you partition a set... > > what do you mean by "partition a set"? > Divide into mutually exclusive subsets of which the union is the set. >> ... such that there >> is a total order on the subsets, and a total order within each subset, >> then you can establish a total order on the entire set... > > Trivially, if you establish a total order on the subsets of S, then > you have a total order on the subsets of S. > > On the other hand, if you select some limited class of subsets of S > upon which we have a total order ">", then it says nothing about > subsets which are /not/ in that limited class. > > For example, if you establish a total order on the finite subsets of > some infinite set S, then it doesn't say /anything/ about how that > ordering should be extended to /infinite/ subsets of S. > I would imagine it does, but I might need a counterexample to see what you're saying. >> ... using something >> like a choice function. I just don't see how an uncountable set can be >> partitioned into a countable set of subsets, each countable, to produce >> a well ordering. >> > > Equally, I find it difficult to imagine how you could have a countably > infinite collection of non-empty subsets {X_1, X_2, ..., X_n, ...} and > yet somehow /not/ be able to select one element from each of these > sets. With a countably infinite collection of sets, one can choose one from each. Then one can repeat an infinite number of times to get a well order, selecting remaining elements. But, if any of the sets is itself uncountable, then we end up with an uncountable number of "next set", or limit, elements. If that is allowed, such that there exist limit elements after an infinite number of other limit elements, then we can define a well order on an uncountable set, but not otherwise, that I can see. > > Thus the joke: "The axiom of choice is obviously true, the well- > ordering principle is obviously false, and who can say about Zorn's > lemma?". The question is whether a well order can occur on an uncountable set. Do the limit elements themselves need to be well orderable? If so, the problem regresses...uncountably. > > <snip> > >> Consider Some countable set S, and a set of binary strings representing >> subsets, with one bit for each element of S, which is a 1 if that subset >> contains that element and 0 otherwise. Don't we have the set of all >> binary reals in [0,1)? Given any two, can't we determine which is greater? >> > > No (at least not in the way you state); because there are two > different subsets of S which correspond to the same real number. Not to the same string, and those two representations can easily be ordered based on the most significant bit difference. > Therefore, by representing the same real number, it is not the case > that one is "greater" than the other using R's usual ordering. They > are different subsets; but we have identified them with the same real > number; so we cannot determine which is "greater" in this way. > > <snip> > >>> Imposing the ordering of R as you propose is a /pre/-order (or a pre- >>> total-order) on the sequences you describe. >> Huh? Isn't that a total ordering on P(S)? >> > > No. In a pre-order, it is not required that > > x <= y and y <= x implies y = x. > > The other rules regarding a partial order /do/ apply: > > x <= x > x <= y and y <= z implies x <= z > > and that is why it is called a "pre-order". > Well, I am considering .1000... to be greater than .01111... based on the most significant bit of difference. >> >>> I think you would agree that the "sequence of bits": >>> 0000111... >>> is not the /same/ sequence of bits as: >>> 0001000... >> No, but of course you consider them the same value. I'm not sure I do, >> outside of standard reals. >> > > We start with subsets of N. We apply a /bijective/ function that maps > each subset of N to a sequence of "bits". Next, we map a sequence of > "bits" to lim sum (b_n*2^-n) of those bits ("the quantitative order of > the reals in [0,1)"). It turns out that the latter function is not a > bijection. > > "Outside the standard reals" a sequence of bits could mean anything > you wish it to. What did you intend? > The most significant bit difference determines order. >> >>> Equivalently, the set X = {n : n > 3} and Y = {3} are not equal. >>> And yet, with the ordering you describe, (0000111... <= 0001000...) >>> and (0001000... <= 0000111...); which is to say X <= Y and Y <= X and >>> yet, not (X = Y). >> If they are ordered such that x<y iff the first bit where they differ >> has a 0 in x and a 1 in y, then 00001111.. < 0001000... > > And if they are ordered in such a way X <= Y such that every bit which > is a 1 in X is also a 1 in Y, then we get get partial order by > inclusion. We can define many different orderings to sequences of > bits, depending on our interests. So what is your point? > 0 >> That is, if the >> elements of the set are well ordered, then there can be a total order on >> the subsets. > > But the order you describe is /not/ a well-order on the elements of > P(S). Do you recall the definition of a well-order? > >> I think you are equating those two strings according to the >> standard conception of the reals, but those are two different subsets, >> one of which includes an element which comes before any element in the >> other, in the well ordering on S, so that one comes first. >> > > You are the one who said: > >>>> if We have >>>> created a binary string, and if we consider all bits to be to the right >>>> of the binary point, with the first being the first bit, >>>> then we have >>>> ordered the subsets of N according to the quantitative order of the >>>> reals in [0,1]. > > If instead of the quantitative order of the /reals/, you meant some > other thing, then perhaps your statement is true; or perhaps not. I > can't say. As we agreed in the section I snip, it is surely the case > that there is a total order on the set of all bit sequences > (lexicographic order). That corresponds to the quantitative order, except for those questionable pairs of strings, so sure, the lexicographic order. > > <snip> > >>> But I wasn't talking about ordering the /elements/ of P(N). I was >>> talking about ordering the /subsets/ of P(N); which is to say ordering >>> the /elements/ of P(P(N)). E.g., one element of P(P(N)) is the set X >>> of /all/ subsets of N having prime order; and another is the set Y of / >>> all/ subsets of N consisting of a finite number of primes. How should >>> I proceed to determine whether X <= Y or Y <= X in such a way that <= >>> is a total order on P(P(N))? >> Well, if you have established the total order on P(N), then for >> xeP(P(N)) and yeP(P(N)), x<y iff the first bit in x which is different >> from y is a 0. > > Think: What is meant by "the /first/ bit in x" in this context? > > Yes, we have /totally/ ordered P(N) using the usual total order (yea; > even, well-order) of N. > > But we have /not/ thereby put a /well-order/ on P(N); which is to say, > we have not (yet) defined an ordering on the /elements/ of P(N) such > that every /subset/ of P(N) has a least element. So we have not > defined what the "first" or smallest "bit" of each possible /subset/ x > of P(N), which is to say element x of P(P(X)), might be. > >> That is, if you can determine which subset is the most >> significant, or first in their union which is not in their intersection, >> the element containing that subset precedes the other. >> > > This is the very nut of the idea behind the definition of "well- > ordered". Can we /always/ do this, given that we start with an agreed > upon well-ordering on N? Maybe, and maybe not. Consider the fact that > this question already has a name. That implies that other people, > besides you, have already wondered about this. > >> Can I apply this to your example? Let's see. No, because of the infinity >> of bits in P(N), one cannot use the normal ordering of binary reals to >> accomplish this, because that order is not a well order suitable for >> addressing the bits. > > Yes. Exactly! See, on one level, you /do/ get it. > >> One might try using the order of the binary >> naturals, but for infinite sets, we would have trouble ordering, say, >> ...1010 and ...011. > > No; that we /can/ do (lexiographic ordering). What is in question is > whether we can relatively totally order /sets/ of bit sequences; not > whether we can relatively totally order /bit sequences/ (the latter we > can easily do by lexicographic ordering of N x {0,1}). > > A better analogy would be: > > Suppose (S, <=) is a well-order on S. Then there is an ordering (P(S), > <=') such that <=' is a total order. > > But as we have just seen, it is not /obviously/ true that (without > outright assuming something like AoC, which makes the first clause > unnecessary): > > Suppose (S, <=) is a total order on S. Then there is an ordering > (P(S), <=') such that <=' is a total order. > Well, didn't we just go through an example where it appeared to be untrue that this is always the case? Why "assume" something is true that's not obvious, and even obviously not always true? >> So, perhaps that doesn't work for P(P(N)), but what >> was the point about a total order on P(P(N)) again? >> > > I brought all of this up to highlight the problems with your > assertion: > >>>>>>>> Defining equality >>>>>>>> where there is no relative order doesn't make sense. > > It is not obvious what the "right" definition, or even "a" definition, > of ordering creates a total order (let alone a well-order) on P(P(N)). > However, it is quite sensible how we "should" define what /equality/ > means for two elements x, y of P(P(N)). To whit, the same as is usual > for sets: x = y iff for all z, z in x iff z in y. > > Cheers - Chas > Yes, equality of sets is defined by membership of elements, each of which may be included or excluded. In determining equality, the ability to detect the difference between inclusion and exclusion is required. Usually, inclusion>inclusion, but that depends. Doesn't "x=y" mean "there is no difference between x and y"? Tony |