The Definition of Points
in [Logic]
|
Prev: On Ultrafinitism
Next: Modal logic example
From: Lester Zick on 30 Apr 2007 13:50 On Sun, 29 Apr 2007 19:51:51 -0400, Bob Kolker <nowhere(a)nowhere.com> wrote: >Lester Zick wrote:> >> So lines are definitely and demonstrably not composed of points? > >He didn't say that. He said it is possible to describe manifolds by >means of function spaces. What he said in part, Bob, was: >"In modern mathematics, lines and other manifolds are NOT defined as >point sets, but rather the opposite way from top down in terms of >their function spaces . . ." And I think my question is perfectly reasonable in that context. In other words I just requested clarification as to whether lines are or can be defined as "the set of all points . . ." or not. Saying one can define manifolds by means of function spaces doesn't address that particular point that I can see. It's my understanding that you believe lines are constituted of the "set of all points . . ." and integration via the calculus applies to such points for the purpose of constituting lines. So I suppose what I'm really asking is whether what Rock is talking about represents an issue in mathematical definitions considered as pure formalisms in terms of manifolds and their function spaces or whether his comments directly concern relations between manifolds. If his comments are just directed at definitional relations between manifolds and their function spaces I don't see that it addresses relations between manifolds and that was the purpose of my question. On the other hand if his comment regarding the top-down nature of mathematical definition was just intended to refer to function spaces and manifolds alone without reference to relations between manifolds then I don't see how it addresses relations between lines and points. And simply commenting on the mathematical definition of manifolds in terms of function spaces doesn't really contribute much to resolution of that issue which was the purpose of the thread to begin with. ~v~~
From: Tony Orlow on 1 May 2007 12:17 Chas Brown wrote: > 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.") > 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? :) >> 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)). >>>> 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. >>> >>> 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". > Whatever. >>>> 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? > Just to annoy you I guess. >>>> >>>>> 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". > 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? >>>>> 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. Integers aren't totally ordered? And that if succ(x) = succ(y), then x = > y. Order doesn't necessarily require successor. Are the reals totally ordered? And the inductive principle. The inductive principle is important, but utilizes order with successor, rather than defining it. Those are all wonderful things, but x<y ^ y<z -> x<z is the root of what order is. > > 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. 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. >>>> >>>>> 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. >>>> ... 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. > >> 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, and you're referring to the quantitative order of the rationals, within which there is no successor relation for your bijection with the naturals. >>> >>>> 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.) > It seems obvious to me here that one would WANT to have a single order on ALL subsets, if possible. You can't select a largest element of an infinite subset of N, so for all sets, start with the smallest. Each subset will have a smallest element. If N starts with 1, and each element indexes a bit to the right of the binary point, then each subset corresponds to a bit string between 0 and 1 in value, and all subsets are totally ordered. Of course, in this ordering, the infinite set of evens will come "before" the set {3}, so it's not ordered by the subset relation, but it's totally ordered. I agree the subset relation creates only a partial order, largely because there are multiple sets of any given size. >>>> ... 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. > 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. >>>> >>> >>> 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. >>> >>> 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? >> If so, the problem regresses...uncountably. >> > > I think instead it prospers... vegetatively. > In other words, you don't know what I'm saying. Okay. >>> >>> <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. > If the bit string represents set membership, then the dual representations within the binary reals can be totally ordered by most significant bit difference. I'm not arguing the quantitative difference between 1 and .999... right now. >>> 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". Lexicographic, which corresponds to quantitative for all but those cases. My oops, then. > > <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> > snippy snippy >>>> 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". Which two of those subsets are equal, given your definition? > > 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. >>>> 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 Sure. Sorry for the lack of clarity. Tony
From: Lester Zick on 1 May 2007 14:44 On Mon, 30 Apr 2007 23:43:28 +0100, Ben newsam <ben.newsam.remove.this(a)gmail.com> wrote: >I googled. There aren't any, and you know it. Thanks for the opine, opie. ~v~~
From: briggs on 1 May 2007 15:33 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. In this cube B and C are both 3 edges away from both F and G. So this graph is not isomorphic to the graph-theoretic representation of a classical cube. So this definition is too weak to capture what we conventionally mean by "cube".
From: Tony Orlow on 1 May 2007 17:27
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". > > In this cube B and C are both 3 edges away from both F and G. > > So this graph is not isomorphic to the graph-theoretic representation of > a classical cube. > > So this definition is too weak to capture what we conventionally mean > by "cube". I should have said "exactly one". Tony |