Cantor Confusion
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From: David Marcus on 10 Jan 2007 14:43 Andy Smith wrote: > Or the not very elegant Peano IIb: > > 1) All natural numbers form an ordered set bounded > by alpha (aka 0) and omega (aka infinity). If we are talking ordinals. > 2) All natural numbers have a successor and a > a predecessor; Except zero. It doesn't have a predecessor. > 3) alpha has no predecessor, omega has no > successor, and neither are natural numbers. Zero most certainly is a natural number. > 4) For any natural number n there exists another > natural number strictly bounded by n and (omega-n). You haven't defined omega - n, and I doubt you can in any useful way. > This has some dodgy consequences I think. > > not least that this does now define the infinite integers. > > omega = -1 > > (because omega*2 = omega -1 (in binary ...1111*10 = ...1110 > =omega - 1). The universe operates in 2's complement > arithmetic! > > I think I might wait a decade or two before mentioning this > elsewhere. Or, here. -- David Marcus
From: Virgil on 10 Jan 2007 14:56 In article <1168429849.191051.275370(a)k58g2000hse.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > The diagonal consists of line ends, of finished lines. > > > > But this does not make the diagonal a line, so the question > > of whether the diagonal exists is not relevant. > > Wrong. On the contrary, WM is wrong and WH is right. WM has this perpetual hitch in his logical giddyup which makes him jump illogically from "for all x there is y such that f(x,y)" to conclude also "there is y such that for all x f(x,y)". > > > > > Every end of a line belongs to a line. However, these lines > > > > can be different for different line ends > > > > > > Your assertion is wrong. There are never different lines required for > > > different line ends. So which different lines have the same ends, WM? > > > Different line ends are always elements of one > > > single line. In one sense yes, for every pair of ends there is a line containing both but there is no line containinig every pair of ends. There is that hitch is WM's giddyup again. > > > > The assertion was that the lines can be different. This > > is trivially true. Your statement that "Different line ends are > > always > > elements of one single line" is equivalent to the statement that > > L_D exists. We know this is false. > > Then you should re-investigate your "knowledge". We have done so several times and found it valid every time. We have also seen WM's quantifier dyslexia at work (the hitch in his giddyup). It is obvious for any > given pair of natural numbers, that they belong (as indexes) to a > single line. Induction supplies the proof for all natural numbers. That says that for every pair (or indices) there is a line containing both, which is quite true. It does not say (and it is not true) that there is a line containing every pair of indices. There is that hitch in WM's giddyup again! > > Name two line ends which do not belong to one single line, or stop > claiming you could do so. We have never claimed that. There is that hitch in WM's giddyup again! > My claim was that L_D exists as a fixed line IF the complete diagonal > exists actually. And in ZFC and NBG, you are flat out wrong. > > > > No. It is clear that given any two natural numbers, or indeed any > > set of natural numbers you can write down, there exists a single > > line which contains all the natural numbers you wrote down. But > > this line *can* depend on which set of natural numbers you write > > down. This is only true for finite sets and is patently false for the infinite sets of ZFC or NBG. > > Of course. And if you write down all natural numbers, then you write > down L_D. Not in ZFC nor NBG. > If the set of all natual numbers is complete, then L_D is complete. False in ZFC and NBG. > If the set is potentially infinite, then L_D can change. It depends on > which set of natural numbers you just have. I always have at least on which is not any L_D.
From: Virgil on 10 Jan 2007 15:00 In article <1168430561.744398.116400(a)i56g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > Since there is no largest element in "potentially" infinite sets (in > > "actual/complete/finished", too) this sentence makes no sence at all. > > A potentially infinite quantity (set or not) is always finite. > Therefore in a linearly ordered set here is a last element. Contrary to > the claim of set theorists, a set is not fixed in reality. Sets do not exist in "reality" but in imagination. And in the imaginations of mathematicians, sets are fixed. What happens in the imaginations of anti-mathematicians like WM is irrelevant in mathematics. > > > In any case, a line containig all elements of the diagonal does exist. > > > > Proof? > > The existence of the diagonal (if existing) and the fact that a given > set of natural numbers is always a subset of a natural number. Not in ZFC or NBG! The set of all natural numbers in either of those is not itself a natural number.
From: Virgil on 10 Jan 2007 15:17 In article <1168431141.572934.249490(a)o58g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > >> mueckenh(a)rz.fh-augsburg.de wrote: > > >> > > >> > Franziska Neugebauer schrieb: > > >> >> mueckenh(a)rz.fh-augsburg.de wrote: > > >> >> > > >> >> > The finite union of two or more finite trees is a finite tree. > > >> >> > An infinite union of finite trees is the infinite tree. > > >> >> > > >> >> What is an infinite union? (Please give a definition) > > >> > > > >> > An infinite union of trees is the union of all (or nearly all) > > >> > finite trees which reach from level 0 to level n where n eps N. > > >> > > >> 1. Please give a definition of "union of all finite trees". > > > > ==================================================== > > > Definition: Denote the nodes of the tree by > > > > > > (0,0) > > > (1,0) (1,1) > > > (2,0a) (2,1a) (2,0b) (2,1b) > > > ... > > > (n,0a) (n,1a) ... > > > > > > The union of all trees up to the n-levels tree is > > > > > > {(0,0)} U {(1,0), (1,1)} U .. U {(n,0a) (n,1a) ...} > > which is obviously the same as > > {(0,0)} U {(0,0), (1,0), (1,1)} U .. U {(0,0), (1,0), (1,1),..., > (n,0a) (n,1a) ...} > > End of definition. This presumes falsely, that every tree has the same nodes and edges and thus that all trees are subtrees of some ur-tree. But suppose I have a family of trees in which no two trees share any nodes or edges, what is the union of such a family? > ================================================= > > > > 1. "Definition By Example" is considered Bad Pratice. > > The upper part contains the definition. To support a definition by an > exampe is good practice in most text books. > > > 2. The union operators are not "evaluated". How do they evaluate? > > Read a book on set theory. Books on set theory would not require the union of a set of disjoint trees to be a single tree. > > > 3. Why don't you take the standard graph theoretical approach? > > (Hint: graph G = (V, E), V: set of vertices, E: set of edges). > > Because it veils the conradiction raising from the tree. Actually, it reveals the contradiction in WM's representations, which is why WM avoids it. If S is your set of trees, then unless you can show that V_union(S) = union{V:V e G and G e S} and E_union(S) = union{E: E e G and G e S} S will not be a tree at all. > > > 4. What is "infinite" in "infinite union of trees which reach from > > level 0 to level n e N"? I can only spot finitely many trees having > > finitely many levels. > > Then you think that there is no infinite union of naural numbers? Spot > a natural number which is not, as a level, in the tree. Since WM claims that are no infinite sets, at least outside the likes of ZFC, how does he justify infinite trees? And once he allows ZFC, or its like, how does he claim no actual infinities?
From: Virgil on 10 Jan 2007 15:21
In article <1168442463.926610.193620(a)o58g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > No. It is clear that given any two natural numbers, or indeed any > > > > set of natural numbers you can write down, there exists a single > > > > line which contains all the natural numbers you wrote down. But this > > > > line > > > > *can* depend on which set of natural numbers you write > > > > down. > > > > > > Of course. And if you write down all natural numbers, then you write > > > down L_D. > > > If the set of all natual numbers is complete, then L_D is complete. > > > If the set is potentially infinite, then L_D can change. It depends on > > > which set of natural numbers you just have. > > > > L_D cannot change. L_D is a line. A line cannot > > change. > > No. But the property of being the greatest line can and does change. Then there isn't any one L_D, there are endlessly many of them, in fact an infinite sequence of them, for no sooner is one created than it is eclipsed by its successor. > > > You are saying that for every maximum integer, n_m, > > that can be shown to exist, a line L_M(n_m) > > which contains all integers up to and including > > this maximum (i.e. all integers that have been shown to > > exist at this point) can be shown to exist. > > No one is disuputing this. However, you cannot find > > a single line L_D that contains every L_M(n_m) > > that can be shown to exist. > > Of course you cannot find that line. But in ZFC and NBG, the completed diagonal exists and your L_D cannot. |