The Definition of Points
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From: Brian Chandler on 26 Apr 2007 03:26 stephen(a)nomail.com wrote: > In sci.math Tony Orlow <tony(a)lightlink.com> wrote: > > Well, if the axiom systems we develop produce the results we expect > > mathematically, then we can be satisfied with them as starting > > assumptions upon which to build. My issue with transfinite set theory is > > that it produces a notion of infinite "size" which I find > > unsatisfactory. I accept that bijection alone can define equivalence > > classes of sets, but I do not accept that this is anything like an > > infinite "number". So, that's why I question the axioms of set theory. > > Why would you question the axioms? The axioms do not contain the > words "number", "size", or "infinite". As you have been told over > and over again, the results do not depend on the names we use. Only the other day, Tony _appeared_ to agree that there was nothing at all "wrong" with cardinality, just as long as it wasn't called "size". Perhaps he really means he questions the names of the axioms of set theory... Well, who knows. He is spending his energies conversing with Liza at present, so give him a break. Brian Chandler http://imaginatorium.org
From: Lester Zick on 26 Apr 2007 13:46 On Thu, 26 Apr 2007 02:21:09 +0100, Ben newsam <ben.newsam.remove.this(a)gmail.com> wrote: >On Wed, 25 Apr 2007 11:27:00 -0700, Lester Zick ><dontbother(a)nowhere.net> wrote: > >>The fact is two of the predicates are true and one false. Does that >>mean t=0.000 or t=0.667? The same would apply to combinations of >>propositions. Are we supposed to be taking an arithmetic average or >>exercising some kind of intuitional insight? Not even to mention the >>weighting of predicates. I just can't imagine that all predicates have >>the same significance in terms of probablistic truth. Are we supposed >>to just adopt someones weighting opinions on the subject of truth? > >Porridge dancing again, Lester Yeah I really wish I knew what that meant, Ben. I wish even more that you knew what that meant. Then instead of merely dancing in oatmeal we might trip the light fantastic for a change. ~v~~
From: Lester Zick on 26 Apr 2007 14:49 On 25 Apr 2007 23:11:47 -0700, Brian Chandler <imaginatorium(a)despammed.com> wrote: >> Why would you question the axioms? The axioms do not contain the >> words "number", "size", or "infinite". As you have been told over >> and over again, the results do not depend on the names we use. > >Only the other day, Tony _appeared_ to agree that there was nothing at >all "wrong" with cardinality, just as long as it wasn't called "size". Cardinality refers to the number of. >Perhaps he really means he questions the names of the axioms of set >theory... What theory? I see no theory. I see axiomatic assumptions of truth. I see SOAP operas. I don't see any theory. What is it this set "theory" is supposed to explain that can't be explained without set "theory". ~v~~
From: Lester Zick on 26 Apr 2007 15:11 On Thu, 26 Apr 2007 02:19:36 +0100, Ben newsam <ben.newsam.remove.this(a)gmail.com> wrote: >>>Absolute truth underlies the universe. Science only confirms a >>>theoretical truth to within some degree of accuracy, or disproves it. >> >>So this absolute truth thingie, Tony. Does it lie out there in space >>with the conjunctions you hypothecate and the dimensions Ben >>hypothecates? > >Er... is this me? What dimensions that I hypothecate? I presume that >you no more know the meaning of the word "hypothecate" than you do the >meaning of "meretricious", because I have not mortgaged any >dimensions. Oh, Ben, Ben. You're so meretriciously pretentious sometimes. You "pledged" empirical tests of spatial dimensionality "as security" presumably for your standing as an empiric "without delivering over". In other words you mortgaged your standing as an empiric to undelivered empirical tests of spatial dimensionality. There is also a secondary usage for "hypothecate" meaning "hypothecize". But I much prefer the usage which sprang the trap I was confident you'd fall into or I wouldn't have included the observation to begin with (dangling preposition, what?) . I'm just so witty I can hardly stand it. > Assuming that you mean "hypothesised about", I feel it >only right to point out that I have not hypothesised the existence of >any dimensions beyond the standard three, since that number seems to >be sufficient to describe physical space as we percieve it. It was empirical tests of spatial dimensionality I was referring to. ~v~~
From: Chas Brown on 26 Apr 2007 15:11
Tony Orlow wrote: > 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. I point this out in the general spirit of highlighting that you often say things like: "Order is defined by x<y ^ y < z -> x < x". "Successor defines the naturals". "It's actually the set of vertices of a |S|-dimensional cube". which are flat out /wrong/ without further clarification. And what on earth could you mean when S is not a finite set? What is an |N|-dimensional cube? What is an |R|-dimensional cube? (Besides being "a cube with |N| and |R| dimensions, resp.") > 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 S = P(P(N)), then it's hard to see how to apply your logic. Which "bit" corresponds to the set of all sets of odd naturals? >>> 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. > How is that /more/ specific? What is /not/ perfectly specific about the definition of a lattice? >> >> 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. > Sure it is. You claim that a lattice can be a tree, if we create "redundancies". I claim that a pentagon can be a cube, if we create "redundancies". The question I have in both cases is why would we /want/ to do such a thing, not whether it /can/ be done using "redundancies". >>> 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. > I'm still not sure why you want to call such a thing "tree-like". One of the characteristics of a tree is that it has /no/ "loops": if x is incomparable to y, and z <= x, then z is incomparable to y. One of the characteristics of the lattice (P(S), <=) is that it has /tons/ of such "loops". Why would you want to call a thing that which it isn't? >>> >>>> 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. > Let's define a new type of order: a step-wise order. An ordering (T, <=) is a step-wise order iff: (i) (T, <=) is a total order. (ii) There exists M in T such that for all x in T, x <= M; i.e., T has a maximal element. (iii) For all x in T, there is a unique y in T such that y < x, and for all z in T, if z < x, then z <= y. Now suppose (S, <=) is a partial order. Suppose T is a subset of S. Then we will call T a step-wise chain of S iff (i) (T, <=) is a step-wise order, and (ii) if not (t in T), then (T union {t}, <=) is not a step-wise order. I.e., T is maximal: there are no "missing" steps. For finite S, the "branches" of your "tree" seem to be the step-wise chains of (P(S), <=) which contain S itself, where <= denotes subset inclusion. But we can prove: There is no subset T of P(N) which contains N, and is a step-wise chain of P(N) with ordering defined by inclusion. So no, I don't think that the resulting lattice is "tree-like" or "recursively generated". >>>> 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? > For example, that there exists an element 0 such that there is no element n in N with succ(n) = 0. And that if succ(x) = succ(y), then x = y. And the inductive principle. Without those constraints, the residues mod 3 satisfy your definition: 0 -> 1 -> 2 -> 0 -> 1 ... >>> >>>> 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... > ? Didn't you refer somewhere to such a bijection previously? >>> ... 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? Did you imagine that that was the /only/ way to create a bijection between the rationals and the naturals? > Doesn't that mean that, in > countable rationals, 3/2 < 5/6? Isn't 1/1 the "smallest" rational, in > that order? > In that order yes; in certain other orders, no. >> >>> 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. > Yup. And false in other orderings. That was my point. <snip> >> 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. > Let S = P(N). Let T be the set of all finite subsets of N. (Totally) order the elements of T by their largest element, then next largest element, and so on. Let U be the set of all infinite subsets of N. (Totally) order the elements of U by their smallest member; and then next smallest member, ad so on. S = T union U. T intersect U is empty. T and U partition S. T is totally ordered. U is totally ordered. How does this imply some /unique/ ordering of elements from T relative to elements of U? (Note, I don't disagree that you /can/ order T union U so that the result is a total order on N; just that the above ordering is /insufficient/ to tell us /exactly/ how to do it in the "obvious" way. For example, we can say that for all u in U, t in T, u < t. Or we could equally say for all u in U, t in T, t < u. The given orderings of T and U don't /relate/ to each other.) >>> ... 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. And given even the weak form of choice you agree to below, it is /provable/ there is no way to partition an uncountable set into a countable number of countable sets to start with; regardless of whether they re well-ordered or not. So I think you are confused as to the nature of a well-ordering on an uncountable set: in any case, it is not a countable collection of countable sets. >>> >> >> 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. So you accept the Axiom of Countable Choice? By this I mean, you accept the implications that arise from the axiom of countable choice? > 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. > Yes; I think it's a reasonable conclusion that if there /is/ a well-order on an uncountable number of elements, that there would be more than any finite number of elements which are limit ordinals. >> >> 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? /Every/ subset of a well-order <= is well-ordered by <= (check the definition). So the answer is "yes" for reasonable interpretations of your statement. > If so, the > problem regresses...uncountably. > I think instead it prospers... vegetatively. >> >> <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. > Again, if you mean "1.0 is not equal to 0.9999...." then say so. That is /not/ the case in the /usual/ "bitwise" representation of the reals. >> 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. > So you are saying that you are /not/ using the quantitative order of the reals. Fine; all you need to do is say so. Instead you said "using the quantitative ordering of the reals". <snip> >> "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. > Then /say/ that. Don't say something else which is false. <snip> >> 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. > I.e., that corresponds to the quantitative order, except it doesn't. Be /assertive/. If you mean the lexicographic order, then say that is what you mean; not "sure whatever; if it makes you happy, I guess it could be the lexicographic order". <snip> >>> 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? > You stated "Defining equality where there is no relative order doesn't make sense." I am providing a set S as a counterexample where: (a) S is an easily described set: it is the set of all subsets of P(N). (b) There is no "obvious" explicit relative (total) order to S. (b) Equality between the elements of S does "make sense". The example depends on the power set axiom, and not on choice. If we /don't/ accept choice (no matter how weak), then we can't use the ordering of N to order the subsets of P(N); and yet (whether such an order can be proven to exist or not) equality is defined. If we /do/ accept choice, then (depending how strong a choice we accept) it may well follow that there exists a relative (total) order on subsets of P(N). And equality is /still/ defined. Therefore, equality does /not/ depend on ordering; it is defined the same way whether we can prove the existence of an ordering or not. Instead the difference between a pre-order and an order depends on equality. >>> 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"? > For sets, most people assume that it means /exactly/ what I said above. What different people might mean by "there is no difference between x and y" is not a fixed thing. They may mean, for example, "x and y are isomorphic", which is to say that there is no /practical/ difference between x and y /in the area of interest/. It's up to them to be /clear/ about what they mean if they want to be understood. Cheers - Chas |