Cantor Confusion
in [Math]
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From: Virgil on 21 May 2007 14:26 In article <1179747854.168331.52890(a)x35g2000prf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM > > You do *not* give an injection from paths to nodes. What node does > > the path 0.0101010101... inject to? How do you define the injection? > > I define it in the same way as you define the injection for definable > numbers, namely by a catalogue. Map, for instance, node no. 17 on the > path 0.010101... > > Obviously, the catalogue can contain all nodes. It is not obvious at all, until a rule is given by which a path is associated with a node, whether all paths OR all nodes are represented in your "catalog", much less whether that cataloging injects either paths to nodes or nodes to paths. It is easy enough to inject the set of nodes to the set of paths. For example, associate the root with the always left path, and each left child node with the path through it that thereafter always branches right and each right child node with the path through it that thereafter always branches left. But this mapping, and every other injection from nodes to paths, will leave most paths unassociated with any node. > So let's wait whether > it also can contain all paths. As several people have have shown here several times, no injection from nodes to paths can be a surjection and no surjection from paths to nodes can be an injection in a CIBT. > > > > > > > An injection is also possible for the set of all paths into the set > > > > > of > > > > > all nodes. (There are two nodes per path.) > > > > > > > > *Give* that injection. > > > > > > Map every node onto the path which leaves it to the left-hand side. > > > > I would think that that is *not* an injection from paths to nodes. > > Moreover, > > at each node there are many paths that leave it on the left-hand side, so > > it > > is not even an injection. > > Here is another mapping: Map the node on one of the many paths, call > it A. The others will get their nodes when they separate from A. How? And even if one maps each node to a path, WM would need to show that every path must then be the image of some node, which his vague hand waving does not even come close to establishing. > > > > > > Map every node onto the path which leaves it to the left-hand side. > > > > Which of the paths that leaves it on the left-hand side must I chose? > > Choose the path bunch. If the paths which you assume to be present in > the bunch will become separated then choose the due nodes by which > this happens. Deliberate ambiguity like this proves nothing except the weakness of the arguer's position. > > > > > > > > The number of paths is the same from the root node, because every > > > > > > path > > > > > > starts at the root. Or are you suggesting that there are paths > > > > > > *not* > > > > > > starting at the root? > > > > > > > > > > Every path starts at the root node. But in order to count the > > > > > paths, > > > > > they must be distinguishable, i.e. separated. > > > > > > > > Makes no sense. > > > > > > How would you count inseparated paths? > > > > You are trying to do the counting. When I count I find at every node > > uncountably many paths. > > You do not find any path. You find at most path bunches. WM just isn't looking hard enough. Those who look hard enough, see through the deliberate obfuscation of "bunches" and see at every node uncountably many paths through it. > > > > > > > Every bunch starts at the root node. But in order to count the > > > > > bunches, they must be distinguishable, i.e. thy must be separated > > > > > bunches. The number of separated bunches is doubled at every level. > > > > > > > > You were talking about bunches going in and out of nodes. What you > > > > are > > > > doing is counting edges, not bunches, and the number of edges is > > > > countable. > > > > > > The number of paths cannot be larger than the number of edges. > > > > Why not? > > Because every path contains more edges than it has in common with any > other path. But it is only when each path contains one edge not in ANY other path that that limit applies, and this is NOT the case in a CIBT. In a CIBT, every edge is shared by uncountably many paths.
From: Virgil on 21 May 2007 14:38 In article <1179750646.698997.275410(a)b40g2000prd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > And n fact, if there were all the natural numbers actually existing, > then an infinite one must be among them. (a) 1 is in N and (b) if n is in N, so is n+1, and (c) if 1 is in S and if n in S means n+1 is also in S then N is a subset of S Given any S such that 1 in S and if n in S then n+1 in S let T be the set of finite elements in S. Since 1 in T and if n in T is finite then n+1, also being finite, is also in T, so that N is a subset of T by (c) above Which proves that, while there may be sets of form S which contain infinite members, N is not one of them.
From: Virgil on 21 May 2007 14:49 In article <1179751344.202204.88880(a)y18g2000prd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 21 Mai, 05:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179663816.282116.232...(a)h2g2000hsg.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > But, again, that is *not* the diagonal proof of Cantor. > > > > > > The following wm-proof certainly even in your opinion belongs to the > > > diagonal proofs considered by Cantor: > > > > > > 0) mmm... > > > 1) wmmm... > > > 2) wwmmm... > > > 3) wwwmmm... > > > 4) wwwwmmm... > > > ... .......... > > > > > > And if the list can be considered as a completed entity, then there > > > must be all natural numbers in the first column. And there must be a > > > line with all natural indexes mapped on w's, i.e., no w must be > > > missing (as would be the case if one m was present). > > > > Why? Show a proof. You are again assuming that there is a last natural > > number. > > If the diagonal is complete, then the list must also be complete, > because the diagonal cannot be longer or broader than the list. If the > list is complete, then there must be a line with only w's, because > otherwise the list is not complete. Circular reasoning! Again! WM assumes that there must be a "last" member in order to justify his need for a last member, which is specifically not a requirement in ZF or NBG or most other set theories. WM's definition of "completeness" requires all ordered sets to have last members. Requiring "completeness" of all sets in this sense is not a part of any standard set theory nor a part of any standard mathematics It is an axiom of WM's MathUnRealism which directly contradicts standard set theories and standard mathematics.
From: Virgil on 21 May 2007 15:09 In article <1179756663.678085.237680(a)y2g2000prf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 21 Mai, 13:47, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > > > > > > > > > >On 20 Mai, 17:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> WM says... > > > > >> >> > It is. Set theory is simply biased. Consider the list > > > > >> >> > 0.666... > > >> >> > 0.3666... > > >> >> > 0.33666... > > >> >> > 0.333666... > > >> >> > ... > > > > >> >> > If the diagonal number is defined by "replace 6 by 3", then we > > >> >> > have > > >> >> > two answers none of which can be preferred by logic, but the > > >> >> > second of > > >> >> > which is suppressed by convention. > > > > >> The diagonal number is 0.333... which is not on the list. > > > > >The diagonal number cannot have more 3's than every list number has > > >3's. > > > > Of course it can. > > Consider the following infinite matrix M: > > xooo... > xxooo... > xxxooo... > ... > > You claim the (unchanged) diagonal can have more x's than any line? Each line contains a finite number of x's. The diagonal contains more that any finite number of x's. So "yes"! > Than is wrong because the diagonal consists of line elements. WM assumes, contrary to fact, the there is a last line and a last character in each line in each line and a last character in the diagonal. > > > With your rule for the diagonal: replace 3 by 6, then > > if we start with the following list: > > > > 0.6000 > > 0.06000 > > 0.006000 > > 0.0006000 > > 0.00006000 > > ... > > > > then the diagonal will be the same: 0.333... > > There is one "3" for each number in the list. > > What do you think why I chose the special list? In that list the > diagonal number cannot have more 3's than at least one line of the > list (well, one more). x0 = 0.666... x1 = 0.3666... x2 = 0.33666... x4 = 0.333666... xn = 1/3 + 1/(3*10^n) for n in N u {0} y = 1/3 So that xn - y = 1/(3*10^n) =\= 0 for all n in N u {0} > > > > >If the diagonal number is complete in the sense that at every > > >position indexed by a natural number there is a 3, then there must be > > >a complete sequence of 3's among the list entries too. > > > > No, that's false. It's *provably* false. > > It is also provably right. See matrix M above. WM has yet to construct a valid proof of anything. > > > > I say in this case because not every sequence has a limit. For example, > > the sequence > > > > 0.600... > > 0.330... > > 0.6660... > > 0.33300... > > 0.666600... > > 0.3333000... > > > > has no limit. But it still has a diagonal, namely > > 0.363636... > > But if this diagonal is to be considered a real number, then we need > the factors 10^-i. Factoring decimals is a muggs game. > > >Forall n e N we have 0 < 1 - 0.999...9 (with n 9's). > > > > Yes, that's true. 1 does not appear on the list > > > > 0.9 > > 0.99 > > 0.999 > > 0.9999 > > 0.99999 > > etc. > > > > and 0.33333 does not appear on the list > > > > 0.6666... > > 0.36666 > > 0.336666 > > 0.3336666 > > etc. > > > > >> So r does not appear on the list. > > > > >So the problem of replacing 0 by 9 is not existing. > > > > What are you talking about? Of course the number 0.9999... > > (all 9s) exists. And it is equal to 1. But it is not on the > > list > > > > 0.9 > > 0.99 > > 0.999 > > 0.9999 > > 0.99999 > > etc. > > > > >But we have in the limit j --> oo: D_j = 0. > > > > Yes. That's true, but irrelevant. For the diagonal to be > > on the list, it must be that D_j = 0 for some finite j. > > That is only required for a finite list. WRONG! > > > It's not good enough that limit of D_j = 0. > > Consider Dave L. Renfro's example: > > 0.49999999... > 0.49111111... > 0.44911111... > 0.44491111... > 0.44449111... > . . . . . . . > > Replacing 4 by 5 and 9 by 0 gives the diagonal number > 0.5000...0999... =/= 0.5 for any finite j. Only in the limit j --> oo > the diagonal number is 0.5000... = 0.499..., the first entry. Here it > is good enough to consider the limit to make the proof fail? That one can construct a rule that fails does not mean all rules fail. It is enough that there is one rule which does NOT fail. Such as: If the nth digit of the nth number in the list is not 2, put a 2 in that place in the diagonal, otherwise put a 3 in that place in the diagonal. This rule works for all decimals.
From: WM on 21 May 2007 16:00
On 20 Mai, 21:09, Virgil <vir...(a)comcast.net> wrote: > In article <1179668114.743217.104...(a)p47g2000hsd.googlegroups.com>, > There cannot be in any set theory of repute any infinite set which is > not actually infinite. > > Whatever WM insists on as being potentially but not actually infinite, > it cannot be a set. Call it a pre-set or a quasi-set of a pseudo-set, > but not a set. So the union U(T(n)) of all finite trees T(n), formed in an obvious way, is an actually infinite binary tree AIBTwith actually countable sets of paths and nodes and edges. Changing to the CIBT does not add any node and any edge, but extends the set of infinite paths already present in the U(T(n)) to the uncountable set of infinite paths. Paths consist of nodes and edges. So nothing is actually changed. That is a theory of repute? Rather it is a theory to refute! ================================ >> (2) ==> p is nothing but a union of finite paths. > Not neccessarily. That is the case only if one requires that a > path be no more than a set of nodes. But every path has a soul in addition? ================================= On 20 Mai, 22:11, Virgil <vir...(a)comcast.net> wrote: > In article <1179663816.282116.232...(a)h2g2000hsg.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Set theory is simply biased. Consider the list > > > > > 0.666... > > > > 0.3666... > > > > 0.33666... > > > > 0.333666... > > > > ... > > > > > If the diagonal number is defined by "replace 6 by 3", then we have > > > > two answers none of which can be preferred by logic, but the second of > > > > which is suppressed by convention. > > > > But, again, that is *not* the diagonal proof of Cantor. > > > The following wm-proof certainly even in your opinion belongs to the > > diagonal proofs considered by Cantor: > > > 0) mmm... > > 1) wmmm... > > 2) wwmmm... > > 3) wwwmmm... > > 4) wwwwmmm... > > ... .......... > > > And if the list can be considered as a completed entity, then there > > must be all natural numbers in the first column. > > As each "column" can contain only w's or m's, there are no natural > numbers in any column. The first column consists of natural numbers only. > > The diagonal, of all w's for the given list, differs from the first > string in place 1 and in the second in place 2, and so on, differing > from the nth in place n, and thus differing enough to be different from > EVERY member of the list. > > Thus it is not in the list. This is correct for a finite list. An infinite list is completely different. Compare the infinite tree. While every path covers all its subpaths, an infinite path p cannot be covered by a single path p' =/= p, but it can be covered by an infinite set of paths p' =/= p. Infinity contains more miracles of that kind. There can even be an entry equal to the diagonal while the diagonal differes from every entry. Regards, WM |