An uncountable countable set
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From: Virgil on 21 Aug 2006 15:41 In article <1156148773.420187.122140(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > You are dreaming. It does follow from the form of the numbers of my > list. > Try to come back to reality. Simply show how a digit position n can be > indexed the complete number of which, i.e., the complete sequence of > digit positions 1 to n is not in the list. The Hilbert Hotel method!
From: Virgil on 21 Aug 2006 15:44 In article <1156148773.420187.122140(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > In my list there are all numbers which can index the position n+1 if > they can index the position n. But they cannot index a position larger > than all naturals. > All positions which are not larger than all naturals are in my list. The Hilbert Hotel method allows one to index anything that can be listed in any order. "Mueckenh" may insists that his infinite string cannot be put at the beginning of a list, but Hilbert knew better.
From: mueckenh on 21 Aug 2006 15:45 Dik T. Winter schrieb: > In article <1155999560.728582.29520(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > But I now understand that the series: > > > sum{n = 1 .. oo} (-1)^n/n > > > does not converge according to your logic. > > > > For every natural number n we have n/n = 1. But the notation lim n --> > > oo is not clear. > > You do not know the convention that taking powers goes before division? I overlooked the "(-1)^". You never overlook something? > But I now understand that the series: > sum{n = 1 .. oo} ((-1)^n)/n > does not converge according to your logic. It does to any practicable approximation (not by my logic but in reality). > And why is the notation lim n --> oo not clear. You have defined it just > a few days ago: > > > > 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. That is correct as long as n is a natural number. But often some expression like 0.111... is ivolved where not all digit positions can be indexed by natural numbers and n is understood to approach something I do not know. > > > > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings > > > > where the remainings are |{n+1,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 > > > > > > That makes no sense. As I read it, we have: > > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings = 1 + 1 = 2. > > > > ? > > 1 + remainings = 1 + 1 = 2 > > Which error did you see? > > Read the fifth line above the question mark. The remainings yield 1. That is correct. > > > > But I will allow an error here. But I was thinking about: > > > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > > > which might be written as: > > > lim{n -> oo} |{1, 2, 3, ..., n}|/|{2, 4, 6, ..., 2n}| > > > > = 1 by mathematics and by set theory. > > > > > or as: > > > lim{n -> oo} |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| > > > > = 2 by mathematics but 1 by set theory. > > As set theory does not know about limits, Now we know about limits in set theory. > I fail to see that, but > even if set theory had knowledge about limits, I would state: > No, also 2 by set theory. What you fail to see is that > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > is defined by neither mathematics, nor by set theory. Both do give > *no* definition of aleph-0/aleph-0. The cardinal numbers do not > form a field, only a ring (and not even a division ring). So > arbitrary division is not defined, definition is not even possible. Division was possible and was practised in fact long before rings and fields were known. Do you agree that aleph_0 = aleph_0 ? Do you agree that division by n is a process which leads to a partition of a number (of units) into n equal shares (and possibly a remaining)? Exactly that is done by a bijection. Each share here is one unit. For a process like |{1, 2, 3, ..., 2n}| / |{2, 4, 6, ..., 2n}| we have at first a bijection between the sets {1, 2, 3, ..., n} and {2, 4, 6, ..., 2n} and then we consider the remaining which yields another complete bijection. Therefore the result is 1 + 1 = 2. One must have a very restricted mind to believe that without fields and rings division was impossible. Regards, WM
From: Virgil on 21 Aug 2006 15:47 In article <1156148970.634196.169840(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > In my list holds the following law: > > If a digit position n is covered then there is at least one number not > less than n in the list. > If an infinite number of digit positions is covered than there is at > least one number with infinitely many digit positions in the list. But > in my list there is no number with infinitely many digit positions. > > If you would understand that, then all you question were answered. Apparently "Mueckenh"'s list is so rigid that rearrangements are forbidden, but in counting, rearrangements are always allowed.
From: David R Tribble on 21 Aug 2006 15:48
Tony Orlow wrote: >> If the formula applies to an infinite number of finites, then does the >> sum of the aleph_0 finite naturals equal (aleph_0^2+aleph_0)/2? > Dik T. Winter wrote: >> I think there is a misunderstanding. The formula >> sum{i = 1 .. n} i = n * (n + 1) / 2 >> holds for every finite number n, so it holds for infinitely many finite >> numers n (as there are infinitely many finite numbers n). But we can >> not switch to an infinite sum (that is something different). > David R Tribble wrote: >> Not without proof that the formula works for infinite n, anyway. >> But we're still waiting for Tony to provide that proof. > Tony Orlow wrote: > David, this is not a matter of proof given standard definitions, ... That's right. It's a matter of proof using good old logic, regardless of what new definitions you use. > ... but a > redefinition of the principle of induction such that it applies to the > infinite case. When an inductive proof demonstrates that f(x)>g(x) for > all x greater than some y, all infinite values falls into that category. You can't just declare that and expect it to be true. You must prove it, using your new definitions and axioms, using good old logic. > So, it simply amounts to removing the restriction that induction only > applies to finite n. Once this restriction is lifted, then we can > "prove" that, for instance, for all n>2, n^2>2n, and therefore this also > holds for all infinite n. The implications of this assumption are a very > nice system of ordering infinities which goes far beyond what > cardinality can even hope to attain. Again, you can't simply remove the "restriction" that it doesn't work and expect it to magically work now. You must prove that it works under your new axiomatic scheme. > If you want to see that it applies to infinite n visually, picture the > unary enumeration in question, which forms a staircase, from an infinite > distance. You have some sort of square divided diagonally. whatever > infinite n you have chosen, you have an nxn square divided in half. > That's your n^2/2. The remaining n/2 is a linear measure, being 1 > dimensional, and does not add more than infinitesimally to the area > occupied by the set of units. That may work in your visualization of things, but it directly contradicts my visualization. So now it is incumbent upon you to demonstrate that your geometric interpretation is consistent with your underlying theory. |