Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: stephen on 23 Dec 2006 14:23 Newberry <newberry(a)ureach.com> wrote: > stephen(a)nomail.com wrote: >> Newberry <newberry(a)ureach.com> wrote: >> >> > stephen(a)nomail.com wrote: >> >> Newberry <newberry(a)ureach.com> wrote: >> >> >> >> > stephen(a)nomail.com wrote: >> >> >> Newberry <newberry(a)ureach.com> wrote: >> >> >> >> >> >> > stephen(a)nomail.com wrote: >> >> >> >> Newberry <newberry(a)ureach.com> wrote: >> >> >> >> >> >> >> >> > stephen(a)nomail.com wrote: >> >> >> >> >> Newberry <newberry(a)ureach.com> wrote: >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> > What is wrong with WM's mapping? You divide each edge into two halves >> >> >> >> >> > and pass one half to each branch. Then you divide the passed half again >> >> >> >> >> > and again. >> >> >> >> >> >> >> >> >> >> So which edge is mapped to the path that always goes left? >> >> >> >> >> Remember, you need to uniquely map each path to an edge. >> >> >> >> >> >> >> >> > No, I don't. What I am showing here is that each path will accumulate >> >> >> >> > the weight of two edges as it approaches infinity. >> >> >> >> >> >> >> >> That is not a mapping. >> >> >> > Correct. >> >> >> >> >> >> So that answers your question "what is wrong with WM's mapping?" >> >> >> It is not a mapping. >> >> >> >> >> >> > You need to construct a surjection. >> >> >> > No, I don't. >> >> >> >> >> >> Yes you do, if you want to show that there exists a surjection >> >> >> from edges to paths. >> >> >> >> >> >> >> Showing that each path will accumalate the weight of two edges >> >> >> >> has nothing to do with constructing a mapping. >> >> >> > But it shows that in an infinite tree there are two edges per path. >> >> >> >> >> >> No. It shows that in large finite trees there are approximately >> >> >> two edges per path. It says nothing about infinite trees. >> >> >> >> > Does it converge to two edges per path as we approach infinity? >> >> >> >> Yes. Just as the number of leaf nodes per node converges to 1/2 >> >> as we approach infinity. >> >> > So when we approach infinity there are twice as many edges as paths as >> > this limit shows, >> > lim{n-->oo} (2*2^n - 2)/2^n = 2 >> > correct? >> >> Yes. Just as when we approch infinity there are twice as many nodes >> as there are leaf nodes. > The nodes are countable and there are twice as many of them than paths In an infinite tree there are not twice as many nodes as paths. > ... > But an infinite tree has an infinite number >> of nodes, and zero leaf nodes. >> >> > The cardinality of the indexes n is aleph0, correct? >> >> Yes, but irrelevant. You cannot just claim that something that is true >> "in the limit" is true for the infinite case. > Hmm ... Why not? What is the cardinality of the nodes in an infinite > path? Why not? I just explained why not. >> >> Do you agree that an infinite tree has 0 leaf nodes? > Yes >> Do you agree that for large finite trees the total number of nodes >> is approximately twice the number of leaf nodes? > Yes >> Do you agree that an infinite tree has an infinite number of nodes? > Yes >> Do you think that 2*0 = oo? > No > What is supposed to follow from all this? That what is true "in the limit" is not necessarily true for the infinite case. "In the limit", there are twice as many nodes as leaf nodes. In an infinite tree, there are an infinite number of nodes, but 0 leaf nodes. So something that is true "in the limit", is not true of the infinite tree, unless you think that 2*0 = oo. >> >> Stephen > I think it all boils down to this: something that is true "in the > limit" is NOT true for the infinite case. That is what I said above. More accurately, something that is true "in the limit" is not necessarily true for the infinite case. So you cannot just claim that because something is true "in the limit", it must be true for the infinite case, unless of course you believe 2*0 = oo. Stephen
From: mueckenh on 23 Dec 2006 15:55 Virgil schrieb: > In article <1166737132.026953.46380(a)48g2000cwx.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > Indeed. No axiom is necessarily true. > > > > How then should the truth of AC be proved? > > One does not prove axioms, one either assumes them or does not assume > them. You like to snip in such a way that a wrong meaning of my words becomes manifest? You think it is necessary? It was Dik who insisted that a well ordering of the reals can be done (not only be proved), if AC is true. Here is the discssion: ****************************************** > Indeed, that can not be done, until at some moment there is a proof > (without AC) that that axiom is true. > > > But I explicitly stated how you could construct > > > a well-ordering when AC is true. You can not do it in general, and > > > nobody can, because AC is not necessarily true. > > Are there axioms which are necessarily true? Not according to your > > point of view. > Indeed. No axiom is necessarily true. How then should the truth of AC be proved? ********************************************* ============= >> The sum 1 + 1/2 + 1/4 + ... has no last term. >It is not a sum, but is often used to represent an infinite sequence of >partial sums. Such a sequence may have a defined limit value, but that >limit is not "the sum" except by abuse of language. Whatever you may call it. The value < 1 required by set theory can only be attached to it by total abuse of human brain >> Nevertheless we know, >> even in ZFC, that it is larger than 1 and less than 10. >We do not know yet that it exists at all, until we have defined what it >would mean. And, as that is a matter of delta-epsilonics, it is not as >trivial as WM implies. Whatever delta-epsilontics you may try, the value is greater than 1. ============================== >But in ZFC or NBG, there are infinitely many of those finite digit >positions, one for each of the infinitely many finite natural numbers. Notwithstanding the fact, that this assumption is purest nonsense, it is without any relevance in this case. (That's why I constructed it.) >Except that the only demonstration of such "inacceptability" would be >the discovery of some internal contradiction within ZFC or NBG. No. The necessary inequality in ZFC 1 + 1/2 + 1/4 < 1 is enough. >And the only conflicts WM can up with are with statements outside of ZFC >and NBG, which are irrelevant to what occurs within them. Everything that can be said against ZFC is irrelevant, by definition. ========================= > >And at last you accept the existence of an actually infinite finite > >number. > What "actually infinite finite number" is that, WM? > ZFC allows, indeed requires something like actually infinite non-natural > "numbers", but only WM claims that any of them are also finite. You agreed yourself to have it: I said: >>You cannot have an actually infinite finite number. You replied: > WM may not be able to, but his inadequacies are personal, not universal. Regards, WM
From: mueckenh on 23 Dec 2006 15:59 Virgil schrieb: > > Are you unable to see that the formal use of fractions like |N|/|R| < 1 > > is completely irrelevant but only a brief way to express |N| < |R| ? > > By definition, |N| < |R| means no more than that there are injections > from set N to set R but none from set R to set N. > > How is E(n) < P(n) relevant to that definition? If you agree that the the inequality aleph_0 < 2^aleph_0 is false or useless, then I have reached one of my aims. But if you do not, then you must agree that the ">" symbol can be used in another framework too. (At least if you applied some logic.) ============================ > There is no line containing every node. Which number is missing in every line of the EIT? Which number is missing in line A which is not missing in line B while line B contains a number that is missing in line A? > Note that WM has so far carefully avoided responding to this refutation. > Any infinite path is a diagonal for some EIT. > The finite subpaths staring at the root nod and ending at > a node of the given path are the edges. > Given any node there is a line containing it > BUT > There is no line containing every node. Which node is missing in every line? =================================== >> It is a number which is at least as large as the largest existing >> natural number. > And must be larger than its successor. No, it is always the successor which is the largest number. Regards, WM
From: mueckenh on 23 Dec 2006 16:04 cbrown(a)cbrownsystems.com schrieb: > > > > > > > I can calculate 1 + 1/2. > > > > > > Good for you! > > > > Bad for you, if you can't. Can't you? > > I know that 2/3 < 1 is a convenient way to express 2 < 3. > > For heaven's sake. "2/3 < 1" is not just "a convenient way" of > expressing "2 < 3". It is *also*, among others, a convenient way of expressing "2 < 3 > > E(n)/P(n) -> 2 is a feature only of finite n. |E(n)| < |P(n)| is a > feature only of infinite n. /Many/ things that can be said in the > finite case are not true in the infinite case; But 1 + 1/2 + 1/4 + ... < 1 is *not* among those. > for example E(n) is > finite for natural n, and not finite for infinite n. For example {1,2,3,...n} <--> {2,4,6,...2n} is a bijection for finite sets. It is obviously false for infinite sets. Never thought about that topic? Presumably not, beause you had to learn that stuff from authorities. Here you see how harmful belief in authorities can be. > It is when you try to argue "lim n-> oo E(n)/P(n) = 2 contradicts > E(n)/P(n) < 1 in the infinite case" that your confusing notation > "looks" like a contradiction (to you; to me it looks like nonsense). Then you really do accept 1 + 1/2 + 1/4 + ... < 1 ? That looks like nonsense to me (and, I hope, to all who are really mathematicians). Regards, WM
From: mueckenh on 23 Dec 2006 16:09
Dik T. Winter schrieb: > > No, you did no show it yet. You referred to some Hamel basis. A > > construction either says: This is the first element and that is the > > second etc. or it gives a prescription how this can be found. > > And I did. No Dik, it has been proved that a definable well-order is impossible to accomplish. > Construct a Hamel basis and do a lexicographic ordering > of the elements R as a vector space over Q (with some caveats). > That one is constructable, but I will not show it here (it takes > too much space). Yes, precisely spoken it would require uncountably many elementary cells, which are not available in the universe. > > > > But apparently you have problems when something can be shown to exist > > > (in the mathematical sense) but can not be computed. That is close > > > to the intuitionistic view. > > > > I stated that there is no possible way to construct a well-ordering. > > You opposed. You cannot sustain your statement. That's all. > > Given AC, there is an Hamel basis and using that a well-ordering can > be defined. No Dik, it has been proved that a definable well-order is impossible to accomplish. > > > > > > But I explicitly stated how you could construct > > > > > a well-ordering when AC is true. You can not do it in general, and > > > > > nobody can, because AC is not necessarily true. > > > > > > > > Are there axioms which are necessarily true? Not according to your > > > > point of view. > > > > > > Indeed. No axiom is necessarily true. > > > > How then should the truth of AC be proved? > > Not. Simply because it can not be proven (at least according to current > mathematical thinking). But I also never stated that it can be proven. > On the other hand, it is quite well possible that AC ultimately will > be proven or disproven; you never know. Cohen did prove that it cannot be proved (there is a model of ZFC without AC. Read my book, there you can find it. I will send you a copy when being back from vacations next year.) Regards, WM |