The Definition of Points
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From: cbrown on 3 May 2007 02:38 On May 1, 9:17 am, Tony Orlow <t...(a)lightlink.com> wrote: > Chas Brown wrote: > > Tony Orlow wrote: > >> cbr...(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.") > > How could anything be "flat out wrong" "without further clarification"? Easily. "Everyone who posts on sci.math has the birthname Charles" is flat-out wrong. "Everyone who posts on sci.math as Chas has the birthname Charles" is not flat-out wrong; and may even be true. > If they aren't wrong WITH further clarification, then they aren't > "flat-out" wrong. Anyway.... > "Every A satisfies property P" is not the same statement as "every A which also satisfies property Q, satisfies property P". The first can be false ("flat-out wrong") while the second may be true. > A "cube" is a form where each element, a point, is connected, though an > edge, to n others out of a total of 2^n, In normal conventions: A "TO-cube" is a graph G = (V,E) where |V| = 2^n, and every vertex has degree n, and ... > such that any one can connect > indirectly to any other through a series of n such connections or fewer, > and there exists one element for each which requires exactly n such > connections. How's that for a definition of a square of n dimensions? :) > Well, since there seems to be some question as regards what you mean by "n (somethings) or fewer" and "exactly n (somethings)" and what everyone else means, I'd say it's not quite clear when n is not finite; e.g., when n = |Q| or |R|. For example, let C = { c : c in 2^N}. C is then the set of all binary sequences (b_0, b_1, b_2, ..., b_n, ...). For any given c in C, there are infinitely many b in C such that they require exactly |N| "such connections", in the usual sense of "exactly as many". > >> 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? > > The power set of the odds? Lessee. It would have to be a bit in an > infinite position. > > {} is bit 0, {1} is bit 1, {2} is bit 2, {1,2} is bit 3, {3} is bit 4, > {1,3} is bit 5. {2,3} is bit 6, {1,2,3} is bit 7, {4} is bit 8, etc.... > > So, we'd have 1's in all the bits corresponding to sets of only odd > numbers. Those bits would have to be indexed by sums of odd powers of 2. > So, we'd have 2, 8, 32, 128, etc, and sums of those, for an infinite > value. Alternatively, if these bit indexes are negative powers of 2 and > summed, we get some real in [0,1] that could be considered to index the > subset and create a total order on P(P(N)). No, that only creates a total order on P(N), and not a well-order on P(N). Thus, we don't instantly get a total order on P(P(N)). > > >>>> 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? > > Nothing. I'm saying it's a special case of a lattice. > Jesus, man! Just admit you were wrong to call a lattice a kind of tree! A tree implies that if a and b are incomparable, and c <= a, then it is /never/ the case that c <= b. A lattice implies that for every a and b, even if they are incomparable, there /always/ exists a c with c <= a and c <= b. <snip> > > 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". > > N-{1}, N-{1,2}, N-{1,2,3}.... has a maximal element and is totally > ordered by inclusion. Why isn't that a stepwise chain? > Because the sequence itself must be /maximal/ (sometimes called: saturated); see clause (ii) above. The empty set {} is an element of P(N); therefore, the sequence S = (N, N-{1}, N-{1,2}, ...) must contain at some point {}. If it doesn't, then it's not maximal: (S union {{}}, <=) is then a stepwise chain (there is a "missing step"). And if it /does/ contain {} (as it must), then it must contain some singleton set {y}, which is the "step that comes before {}". But the sequence S at some point contains an element N-{1,2,...,y}, so we cannot have a stepwise chain including {y}, since {y} is not a subset of N-{1,2,...,y} or any of the elements coming "after" it in (S, <=). <snip> > >>>> 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. > > Integers aren't totally ordered? > What does that have to do with "not exists x such that succ(x) = 0" being part of the definition of the naturals? > > And that if succ(x) = succ(y), then x = y. > > Order doesn't necessarily require successor. Erm? Of course not. Why do you think I said that order necessarily requires a successor? > Are the reals totally ordered? > Obviously. What does that have to do with defining the naturals? > > And the inductive principle. > > The inductive principle is important, but utilizes order with successor, > rather than defining it. It is one /part/ of the /definition/ of what it means to say "this set S is isomorphic to the naturals". > > Those are all wonderful things, but x<y ^ y<z -> x<z is the root of what > order is. > You really are not listening to what I am saying. Yes, x<y ^ y<z -> x<z is a /necessary/ but not /sufficient/ condition for an order. Also, x<=y and y<=x -> x = y is a /necessary/ but not /sufficient/ condition for an order. Also, X <= x is a /necessary/ but not /sufficient/ condition for an order. Each of these rules is /equally/ important when saying "<=" is an order. If /any one/ of them is /not/ satisfied, then what you have is "flat-out" /not/ an order. You are saying something like "A square is defined by being a figure with four sides". Yes, it is /necessary/ that every square has four sides. No, simply having four sides is not /sufficient/ to say that something is a square. > > > > Without those constraints, the residues mod 3 satisfy your definition: 0 > > -> 1 -> 2 -> 0 -> 1 ... > > Yes, that would be an ordered cycle. To define a structure as > non-cyclical, one can state that x<y -> ~y<x. Sigh. Yes, one /can/ say that; but that is not what you /said/. > That's fine, but saying > there is no order to the numbers on the face of a clock is simply wrong. > One doesn't get from 12 to 6 without passing through 3, assuming time is > moving forward. > First: you are conflating the circle group R/Z_12 (think of it as the / reals/, mod 12) with, for example, the cyclic group Z_12 (the / integers/ mod 12). Second: yes, if you specify a /particular/ ordering ("... assuming that time is moving forward"), then obviously, you have specified a / particular/ ordering. That is not the same as saying "there is a uniquely prefered ordering". > > >>>>> 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? > > Yes, so? I've also said I don't agree that bijections without > restrictions actually indicate "equinumerosity" for infinite sets. > Bijections between infinite subsets of reals in quantitative order yield > accurate relative measures of those sets over finite or infinite ranges. > Names, names, names. Do you reject the assertion that there are /many/ different bijections between the naturals and the rationals (whatever we might call such bijections), or not? > >>>> ... 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? > > Please show me how you do it while maintaining the quantitative order of > both sets throughout the bijection. This is an obvious example of where > mangling the quantitative order of a set gives bogus results. > It is not possible in any case, becuase the order type of the naturals is not isomorphic to the order type of the rationals. That is not a "bogus" result. It is simply a fact: there is more than one way (up to isomorphism) to order a set having the same cardinality as the rationals (or the naturals, or the integers). For /finite/ sets we have a different result: every total order on a / finite/ set S is (order-) isomorphic to every other such order; so there is is a basis of talking about "the" total order (up to isomorphism) on S when S is finite. > > > >> 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. > > Okay, but we are talking about "<" as denoting the order on a set... No, no, no. Not "we". /You/ are speaking as if one could describe "the" order on a set, as if there were /one/, and /only/ one, such order for any given set. /I/ and others are consistently pointing out that there is no such thing as "the" order on an arbitrary set; instead there are /many/ orders which can be placed on a set. It is like you are asserting "/the/ integer which, when squared, results in 4, is 2". Yes, "a" such integer is 2; but there are clearly /two/ such integers: 2, and -2. <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.) > > It seems obvious to me here that one would WANT to have a single order > on ALL subsets, if possible. Keyword: "... IF possible". The difficulty, even assuming the well- ordering principle, is /which/ ordering is supposed to be "the single" ordering. > You can't select a largest element of an > infinite subset of N, so for all sets, start with the smallest. You are confusing P(N) with P(P(N)). Yes, every element U of P(N) has a smallest element defined by x in U and for all y in U, x <= y. No, not every element V of P(P(N)) has a smallest element U. <snip> > > 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. > > Each set of successor ordinals following each limit ordinal must be > countable, no? If so, then there must exist an uncountable number of > limit ordinals in any well order of an uncountable set. > It follows from the axiom of countable choice (which you appear to accept) that an uncountable set is /never/ the union of 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? > > I don't see anything wrong with 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. > > It's a step beyond that. If each limit ordinal is only followed by a > countable number of successors, then there must be an uncountable number > of limit ordinals, not a countably infinite number. > Indeed; this follows from the axiom of countable choice. What is the problem? > > > >>> 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. > > Then you can only have at most a countably infinite number of limit > ordinals within the well ordering? > Erm? Why do you think that follows from my statements? > >> If so, the problem regresses...uncountably. > > > I think instead it prospers... vegetatively. > > In other words, you don't know what I'm saying. Okay. > Yes. Less poetry, more rigor. <snippity-snip-snap> > >> 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. > > > (c) Equality between the elements of S does "make sense". > > Which two of those subsets are equal, given your definition? > Let P = {p : p is a prime in N with p > 2} Let S = {x : x = 2^n, n in N} Let T = {y : y in N, y > 0, and not (y divisible by p, p in P)} Let U = {A : A in P(N) and |A| in S}. U in P(P(N)) Let V = {B : B in P(N) and |B| in T}. V in P(P(N)) Then U = V; i.e., W in U iff W in V. Proof: by way of the fundamental theorem of arithmetic, S = T. Thus, | W| in S iff |W| in T; and so W in U iff W in V. > > 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. > > I'm not quite getting your point here. > Axioms of choice (weak or strong) generally are required to allow us to construct various orderings, including your purported "the" ordering on "any" set S. But equality of two sets is not a function of which (or even, any) axiom of choice we choose, and thus perforce some /particular/ ordering; it comes /before/ considerations such as ordering. > > > >>>> 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. > > >> 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. > > Sure. Sorry for the lack of clarity. No problemo. So, does that mean you now see that that your statement "defining equality where there is no relative order doesn't make sense", without further clarification, is flat-out wrong? Cheers - Chas
From: Eckard Blumschein on 4 May 2007 08:15 On 4/28/2007 7:36 PM, Lester Zick wrote: > On Fri, 27 Apr 2007 21:17:55 -0400, Bob Kolker <nowhere(a)nowhere.com> > wrote: > >>Ben newsam wrote: >> >>> >>> A reasonable point in a way, as long as whatever scheme you eventually >>> come up with provides useful results. >> >>Useful and Lester do not coexist. > > Just as Bob and true do not coexist. Shouldn't we suggest to declare Dedekind and Cantor saints because they created the paradise of the impossible? Well, old definitions may still be the best ones. A point is what does not have points. Euclid A continuum is what every part of which has parts. Peirce Obviously, any amount of poinst will never completely fill the ideal continuum. Dedekind's idea of expansion for the body of rational numbers into real numbers would be perfect if his axiom of continuity would not conflict with the Cauchy practice and Weierstrass formalism of epsilon limit. One cannot eat the cake and have it except one thinks as schizophrenic as did Cantor. BTW, in his book Numbers, Ebbinghaus quoted Lessing as to say an obvious mistake may lead to something valuable. He did not reveal what he meant. Did someone find this also in the English edition? I see the problem in the entirely different character of rational numbers and real "numbers". Let's be pragmatic on both sides. Nothing is worse than a rigorously wrong theory with twists without limit. Eckard Blumschein
From: Lester Zick on 4 May 2007 14:50 On Fri, 04 May 2007 14:15:48 +0200, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: >On 4/28/2007 7:36 PM, Lester Zick wrote: >> On Fri, 27 Apr 2007 21:17:55 -0400, Bob Kolker <nowhere(a)nowhere.com> >> wrote: >> >>>Ben newsam wrote: >>> >>>> >>>> A reasonable point in a way, as long as whatever scheme you eventually >>>> come up with provides useful results. >>> >>>Useful and Lester do not coexist. >> >> Just as Bob and true do not coexist. > >Shouldn't we suggest to declare Dedekind and Cantor saints because they >created the paradise of the impossible? > >Well, old definitions may still be the best ones. >A point is what does not have points. Euclid >A continuum is what every part of which has parts. Peirce > >Obviously, any amount of poinst will never completely fill the ideal >continuum. >Dedekind's idea of expansion for the body of rational numbers into real >numbers would be perfect if his axiom of continuity would not conflict >with the Cauchy practice and Weierstrass formalism of epsilon limit. One >cannot eat the cake and have it except one thinks as schizophrenic as >did Cantor. > >BTW, in his book Numbers, Ebbinghaus quoted Lessing as to say an obvious >mistake may lead to something valuable. He did not reveal what he meant. >Did someone find this also in the English edition? > >I see the problem in the entirely different character of rational >numbers and real "numbers". Let's be pragmatic on both sides. >Nothing is worse than a rigorously wrong theory with twists without limit. Well this strikes me as a reasonable overall perspective on the problem. However I'm not sure it solves all the problems we may encounter along the mathematical way. My own take on the classification of numbers is derived from curved surfaces in general and derivatives based on them. Transcendentals I associate with curves of various kinds and rationals and irrationals with straight lines and straight line segments. At least this allows for the combination of both geometry and arithmetic on a common mechanical foundation based on differences. It is also consistent with my general mechanics of "not" as true of everything because "not not" is self contradictory and with boolean conjunctive mechanics and also TvN mechanics based on successive regressions and compoundings of "not" in terms of itself. In other words I don't see geometric figures and arithmetic numbers as existing independent of curves and derivatives based on them through processes of derivation/antiderivation or integration/disintegration. I consider all linear geometric figures with an implicit dr attached to them to specify the lineage and derivation for the figure. Curves are defined through linear velocity and transverse acceleration and non planar curves through curves and transverse acceleration changes. Points are defined through tangency and/or intersection. I think historically we've taken integration of mathematical concepts very much for granted and regardless of how all this plays out it's long since time we developed a common scheme for the classification of geometric figures and numbers on a common mechanical framework. ~v~~
From: Tony Orlow on 5 May 2007 21:53
briggs(a)encompasserve.org wrote: > In article <4637b0c7(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> writes: >> briggs(a)encompasserve.org wrote: >>> In article <46376841(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> writes: >>>> How could anything be "flat out wrong" "without further clarification"? >>>> If they aren't wrong WITH further clarification, then they aren't >>>> "flat-out" wrong. Anyway.... >>>> >>>> A "cube" is a form where each element, a point, is connected, though an >>>> edge, to n others out of a total of 2^n, such that any one can connect >>>> indirectly to any other through a series of n such connections or fewer, >>>> and there exists one element for each which requires exactly n such >>>> connections. How's that for a definition of a square of n dimensions? :) >>> Not good. This appears to be a partial characterization rather than >>> a complete definition. >>> >>> I can come up with a three dimensional "cube" that fits >>> this definition but which is not the same as the standard cube. >>> >>> Nodes A through H >>> Edges AB AC AH BC BD CD DE EF EG FG FH GH >>> >>> Graphically >>> B F >>> /|\ /|\ >>> --A | D--E | H-- >>> \|/ \|/ >>> C G >>> >>> 8 nodes >>> Each node with 3 edges >>> All interconnected using no node-to-node paths longer than 3 edges >>> Each node needing at least one 3 edge path to reach at least one other node >>> >>> Note that in a conventional "cube", each node is a "corner" that has >>> exactly one "opposite corner" that is exactly 3 edges away. >> See above, where n is the number of dimensions: >> >> "there exists one element for each which requires exactly n such >> connections". > > Note that you used "exactly n" but not "exactly one". There exist exactly n elements which require 1 of such connections. There exists exactly one such element that requires exactly n such connections. > > This implies that "one" may be read as "at least one". If you > had meant "exactly one" it's clear that you knew how to phrase it. > > So your definition is bad because it doesn't say what you intended for > it to say. Fine. But, you knew what I was saying, obviously. Do you disagree? TO |