Cantor Confusion
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From: mueckenh on 24 Dec 2006 04:34 Virgil schrieb: > > As not all natural numbers do exist, the set is potentially infinite, > > i.e., it is finite. It has a maximum. L_D. Taking this maximum and > > adding 1 or takin L_D ^ L_D or so yields another maximum. In any cae > > there is a maximum. > > Which is always exceeded by a larger maximum. Which then is the maximum. > > > > > The proof that L_D does not exist > > > makes no assumption about whether all natural numbers > > > exist. All we need is that the natural numbers are not > > > bounded. This is true whether or not all the natural numbers > > > exist. > > > > Here is anothe proof: > > The set of all even natural numbers contains at least one number larger > > than its cardinality. > > Can WM name that number? If not, how can he prove its existence? Nobody can name it, because the set is not fixed. But the existence can easily be proved (and set theorists usually like to believe in objects which cannot be known but the existence of which can only be proved, like the well-ordering of the real numbers). Here is the proof: |{2,4,6,...,2n}| < 2n for *all* existing natural numbers n. > Which "largest existing natural number" is that? > You name one, and we will find a larger one. That is it, then. > > > > > > Assume it exists. Call it N_L. > > > > > > It is easy to see that the set > > > A={1,2,3,...,N_L} exists. > > > > If we do not take into accoun that there are physical constraints, yes. > > What physical constraints limit non-physical existence? The absence of more than 10^100 bits limits the existence of objects which require more than 10^100 bits to exist, like any irrational number (don't mistake its name like pi with the number as defined, e.g., by Cantor by fundamental series). Regards, WM
From: mueckenh on 24 Dec 2006 07:08 Newberry schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Newberry schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > 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. > > > > > > > > You cannot construct a surjection because irrational numbers do not > > > > exist. > > > > > > Are you saying that x^2 = 2 does not have a solution? > > > > Yes. But that is another topic. (In short: There are less than 10^100 > > bits in the universe. Therefore, no irrational number can be > > approximated better than to about 10^-100. The Cauchy criterion for > > converging series fails.) > > What cannot be approximated? You know the Cauchy criterion. For every epsilon > 0 you must be able to find an n_0 such that for n > n_0 we have |a_n - pi| < epsilon. If you have less than 10^100 bits available, then you cannot realise an a_n which reproduces more than the first 10^100 bits of a number like pi. (Unless pi was a number showing some pattern, like the number 0.111.., which would make it possible to express every a_n using less than 10^100 bits.) You cannot find such a number a_n without having pi already. But this number shall be used to establish the existence of pi. This is impossible. pi does not exist *as a number* (it exists as an idea and in form of close approximations). Regards, WM.
From: mueckenh on 24 Dec 2006 07:12 Dik T. Winter schrieb: > In article <1166845904.426550.122020(a)48g2000cwx.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > > stephen(a)nomail.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. > > No, you do not. In the finite case you do, approximately. In the > infinite case you do not because *all* parts you pass to a path are > reduced to size 0. All parts? The diagonal digits a_nn of Cantor's argument have to be multiplierd by 10^-n in order to yield the diagonal number SUM a_nn * 10^-n. But there is no problem with yielding zero for n --> oo? --- Remarkable. > Consider what we pass on to the paths that go > left only. Look at the following: > level 1: 1 edge at level 1 > level 2: 1/2 edge at level 1 + 1 edge at level 2 > level 3: 1/4 edge at level 1+ 1/2 edge at level 2 + 1 edge at level 3 > etc. for each finite n, but what about the infinite case? > level 'oo': 1/oo edge at level 1 + 1/oo edge at level 2 + ... > Where do you see the 2 coming up? It is certainly *not* the series > 1 + 1/2 + 1/4 + ..., because there is *no* edge that is passed in > full, there is also *no* edge of which 1/2 is passed, etc. There is no edge in the *finite* domain which is "passed in full". I the infinite series there is an edge passed in full. Otherwise the path does not exist at all. The infinite sum 1 + 1/2 + 1/4 + ... yields 2, although there is no smallest term 2^-n. It can be applied to calculate the edges per path. You should know that absolutely converging series can be reordered. =============================== > I now understand what you were trying to do. It lasted rather a while. > But in the complete tree >the parts of edges that are assigned to an infinite path are not 1, >1/2, 1/4 etc. There is *no* edge that is completely assigned to a > path, Then the tree is incomplete. You fail to understand infiniteness already on this low level? > Note that in each finite tree, the complete edge assigned to a path is > the last edge on the path. As there is no last edge in an infinite > path, this does not hold in the infinite tree. Regards, WM So there are different paths without different edges?
From: mueckenh on 24 Dec 2006 07:18 Dik T. Winter schrieb: > In article <1166907306.745536.315650(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Virgil schrieb: > ... > > > > 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? > > I wonder what the correct meaning of your words are. > > > 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, it can be done *when AC is true*. But perhaps we have a different > view on the meaning of the sentence "can be done"? It has been proved that there is *no definable well-ordering* of the reals. So it cannot be *done*. The only thing that can be done is to prove that a well-ordering exists. Zermelo did it. Now we know that this well-odering never can be found. In order to keep contradictions far away from ZFC, we accept that situation. Regards, WM
From: mueckenh on 24 Dec 2006 07:21
cbrown(a)cbrownsystems.com schrieb: > What "authorities" do you claim I am quoting when I state "the function > f(x) = 2*x is a bijection between the set of all naturals, and the set > of all even naturals"? > > I am simply applying specific definitions of the terms "function", > "bijection", "the set of all naturals" and "the set of all even > naturals", using the usual logic. This bijection between sets (initial segments {1,2,3,...n} and {2,4,6,...2n}) is only valid for finite sets. > > > 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 ? > > Of course not. How do you infer that from my statements? >From your statement that the same calculation, in case of edges, yields less than edge per path. The number of edges accumulated by one path up to level n is a function like that above: f(n) = 2 - 1/2^n. Very easy. Regards, WM |