An uncountable countable set
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From: mueckenh on 11 Aug 2006 13:00 Dik T. Winter schrieb: > > I am using pi but I know that it is not a number but only an idea. > > False mathematics results only from the assumption that pi was a number > > and that all of its digits could be enumerated by natural numbers. > > Ah, you are opposing mathematical terminology. Not the mathematical ideas. > Why do you know it is not a number? What is your definition of "number"? A number must be in trichotomy with other numbers. A detailed explanation would require too much space. Look into my paper "Physical constraints of numbers" on my home page. > > It is not in the set, but the number of the elements of the set is > > aleph_0. Under this aspect aleph_0 should have some existence in the > > set, namely as the number of elements. Of course it doesn't. Therefore > > it cannot be at all. > > An idea, just like pi. But less consistent. Sqrt(2) is a consistent idea, and its square is even a number. Aleph_0 is an inconsistent idea. > > > 2) If we have only potential infinity oo, then we cannot even consider > > the whole set N and there is no equivalence class of N neither of > > larger sets. > > Possibly. But when we assume the axiom of infinity, there is. But it entails infinite natural numbers which are not, hence there is no omega. > > > > When I talk about a natural number n, I > > > am talking about a potential infinity, it will never become infinity > > > itself. On the other hand, when I talk about number of elements of > > > the set of natural numbers, it is actual infinity I am talking about. > > > > Either N is actually infinite or not. How should an actually infinite > > set exist if the elements are only potentially infinite? > > Yes, that is what the axiom of infinity asserts. And removing philosophycal > ramblings... > > > |{2, 4, 6, ..., 2n}| < 2n > > > > lim [n --> oo] {2, 4, 6, ..., 2n} = G = set of all positive even > > numbers > > lim [n --> oo] |{2, 4, 6, ..., 2n}| = aleph_0 > > lim [n --> oo] n < aleph_0 (because n is finite) > > Now, again, you fail to define what you *mean* with n --> oo. The same is meant as in sequences and series like: lim [n --> oo] (1/n) = 0 n becoming arbitrarily large, running through all natural numbers, but always remaining a finite number < aleph_0. > I would > state that the first two limits exist and are both aleph-0, The first is only a set, not a number. > because we > are talking about ordinal numbers. I would assert that the third does > not exist, because there we are talking about natural numbers. But > unless you provide proper definitions of those notations I can not say > anything to it. I do not provide a definition of aleph_0. Cantor provided that definition: A number larger than any natural number. > > > Nope, in that case you get a different picture. I did show you two > pictures originally, one with terminating paths and one with non-terminating > paths. You elected to discuss the former, you can not now change to the > latter. > I did not change. All the paths are infinite. Every edge starts from a node and ends in a node. > > > > The number of edges is twice the number of paths. > > > > > > Yup, and each path is terminated. > > > > ? I thought you had understood: No path is terminated. There is no edge > > without appending node and no node without appending edge. > > In the first picture, there is no edge without terminating node. In the > second there is no edge with terminating node. In the naturals there is no even number without a following odd number and there is no odd number without a following even number. In the tree there is no edge without a node from which it starts and to which it leads. But no edge and no node is terminating a path. > > > > So you did show, quite nicely, that the number of terminating paths is > > > countable. But the terminating paths do not include the reals. > > > > No path is terminating. Otherwise the path 0.1 would be present as 0.10 > > and 0.100 too and so on. > > Yes, that is true in the first picture. No problem. > > > Further each terminating path would carry only less than two edges. Non > > terminating paths carry exactly two edges. There is no doubt that each > > path carries 1 + 1/2 + 1/4 + ... edges. Or do you doubt the geometric > > series? > > What is 1/2 edge? Think about it. What is a half cake? What s a half horse? When our ancestors tried the fractions they may have asked those questions. Do you really have problems? > > Why? I tell you that with a particular definition I get a number that is not > on the list. And I can even show (without knowing the particulars of your > list) that the number I get is not the n'th element on your list, because > it is different in the n'th digit. No process at all. But the list is infinite. Therefore it is not sufficient to prove that a finite segment of the list does not contain the diagonal number. Regards, WM
From: Virgil on 11 Aug 2006 16:15 In article <1155314190.919903.129280(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Do you believe in infinite natural numbers? Or is it true that you > never really thought about that topic, i.e., how infinitely many finite > numbers can exist without one or the other being infinite? I believe that in ZF, ZFC and NBG one can, and must, have infinitely many finite natural numbers. The axioms require it. They do not require any infinite natural numbers. What is "mueckenh"'s axiom system which prohibits what ZF, et al, require?
From: Virgil on 11 Aug 2006 16:21 In article <1155314314.212493.69980(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > The "Hilbert Hotel" method allows one always to insert one more and > > still have rooms for all. > > Just this argument shows that the diagonal number of Cantor's list has > always reserved a place in the list. Wrong. It may have a place in a new list, but none in the original list. What Cantor does say that is that no list can be complete, but does not say that there is any number which cannot be listed. > Because arguing in set theory is > arbitrary and misleading? It is the way "Mueckenh" does it. WE have a concrete set of axioms, say ZF, from which we proceed. "Mueckenh" apparently has nothing but his intuitions as foundation, but intuition is notoriously unreliable in mathematics.
From: Virgil on 11 Aug 2006 16:31 In article <1155314605.714007.167990(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > The line number is infinite by the axiom of infinity. Where in the axiom of infinity are infinite number lines mentioned? > If no line is numbered as omega, where can we find the line omega + 1? Who says there is one? > > Thus "mueckenh" is claiming to be able to inject the power set of the > > naturals into the set of naturals. > > > > We should be interested in seeing his attempts to perform this > > impossibility. > > Where does *my arguing* fail? > Consider the binary tree which has (no finite paths but only) infinite > paths representing the real numbers between 0 and 1. The edges (like a, > b, and c below) connect the nodes, i.e., the binary digits. The set of > edges is countable, because we can enumerate them > > 0. > /a\ > 0 1 > /b\c /\ > 0 1 0 1 > ............. > > Now we set up a relation between paths and edges. Relate edge a to all > paths which begin with 0.0. Relate edge b to all paths which begin with > 0.00 and relate edge c to all paths which begin with 0.01. Half of edge > a is inherited by all paths which begin with 0.00, the other half of > edge a is inherited by all paths which begin with 0.01. Continuing in > this manner in infinity, we see that every single infinite path is > related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any > other path. >The set of paths is uncountable, but as we have seen, it > contains less elements than the set of edges. "Mueckenh" conflates the finite cases with the infinite case. What is true only for finite cases need not be true for infinite ones. And until "Mueckenh"can find some fault with my constructions, based on specific axioms, I will continue to have more faith in them that in anything of "Mueckenh"'s intuitionals.
From: Virgil on 11 Aug 2006 16:34
In article <1155314675.845888.174190(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > So we have infinitely many finite triangles, but without that "final > > edge", no infinite triangle. > > Cantor's list has no final line. Is it finite? Since a triangle requires 3 edges, without all 3 it is not a triangle. An endless list does not require an end, so is "complete" without one. |