An uncountable countable set
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From: Dik T. Winter on 23 Aug 2006 19:41 In article <1156363838.403378.231610(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1156149324.184708.191200(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > Now, if we complete the tree (and the list above) the number > > > > of edges is still countable, but *none* of the edges terminates at 1/3. > > > > You want "a last line" for completion, but there is none. > > > > > > I don't want a last line, nevertheless the set of all edges is > > > countable. > > > > So 1/3 is not in your tree. > > 1/3 is that path which turns left, right, left, right, left, right, ... > and never ceases to change its direction. > Which edge do you miss? Which dge could not be enumerated? As in your tree, by definition of the tree, each edge terminates at a node, what is the number of the edge that terminates at 1/3? What is the number of the edge that terminates at 1/5? As the set of all edges is countable, you should be able to state that. The problem with your counting method of edges (count first level 1, next count level 2, etc.) you will already have exhausted all natural numbers by counting all edges that are at the end of finite paths. > Do you believe that the set of edges of 1/3 is uncountable? That is not the set of all edges in the tree. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: MoeBlee on 23 Aug 2006 19:58 Tony Orlow wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> Set theory contradicts with: > >> > >> (1) E y e N, A x>y, x< 2*x < x^2 < 2^x (y=2) > >> > >> because: > >> > >> (2) A y e N, aleph_0>y > > > > I don't know what you intend '<' to stand for. For the domination > > relation? The less than relation on ordinals? > > The standard "less than" operator, commonly used for finite reals. This is in set theory, right? w is not in the field of the less than relation on the reals. Therefore, your (2) is incoherent. Also, if were in set theory, 'finite' adds nothing in description of real numbers. > > I don't know what is meant by '(y=2)' in the larger formula. > > I mean that for x>2, 2*x < x^2 < 2^2. For x=2 they're equal. Is aleph_0>2? So you mean that y=2 is a counterexample to the claim that ~Ey...etc. ? (But then you mean 2^x, not 2^2 just above, I guess.) And w is not in the less than ordering for real numbers, so it doesn't exactly make a lot of sense to ask whether w is greater than 2 in that ordering. > >> and > >> > >> (3) aleph_0/2 = aleph_0 = aleph_0^2 < 2^aleph_0 > >> > >> (1) is trivially inductively provable. > > > > Do you mean (1) is a theorem of set theory, or do you mean it is > > provable that (1) is the negation of a theorem of set theory? > > That (1) contradicts set theory. It's certainly not a standard theorem, > in the infinite case, but it should be. However, its incompatible with > aleph_0 and the system of limit ordinals. If you made (1) a rendering of a formula in set theory (without ambiguity as to exactly what ordering '<' is to stand for), then we'd be in a better position to evaluate whether or not it is a theorem of set theory. > >> (2)and (3) are from transfinitology. > > > > What is transfinitology? > > You know, Cantorian religion. That Hollywood stuff. ;) Right. Just like a Mel Gibson movie. > > What is the definition (and in what theory is > > this definition?) of '/' where w (omega) is in the numerator? > Well, I was calling it Bigulosity. The definition of x/y is "how many > intervals of length y fit into an interval of length x?". This is > constructible using straightedge and compass. Where it's not integral, > is as approximable as it is with symbolic reals. I asked you what theory this is in. So, since you now say that (3) is part of set theory, I take it that you intend for the above to be something defined in set theory. There are some basic definitions of 'interval', 'length', and 'integer' in set theoy, and in certain contexts in set theory, but I don't know what 'symbolic reals' are in set theory, nor can I fathom how what you claim to be a definition works for omega as x. Maybe you're forgetting that you're mixing set theory up with your own informal notions again. MoeBlee
From: Dik T. Winter on 23 Aug 2006 19:56 In article <1156364007.547996.270100(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > I do not think so. With blocks of height 1/2^n and width 1 - 2^n, after > > k steps the total height is 1 - 2^k and the total width 1 - 2^k. When > > we complete we get the limiting case. But there is still no block > > at either height 1 or with width 1. > > > > And there is no smaller container that contains the complete stair. > > According to Cantor: The infinite set of finite numbers. aleph_0 is > actually infinite, no natural number is actually infinite. The > staircase has width [0, 1] and height [0, 1). It is a difference. I do not see a proof in your statements. The staircase is within [0, 1) and [0, 1) after completion. How is that in contradiction with what Cantor states? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 23 Aug 2006 19:51 In article <1156363950.351759.83230(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Indexing is covering due to the form of the numbers of my list. > > > If position n can be indexed, then 1 to n are covered. > > > > In repeatal mode again? That is no proof for either of my statements. > > Each position (say n) of my number can be indexed, so each segment > > from positions 1 through n can be covered. This does *not* mean > > either that my number can not be indexed or that my number can be covered. > > Then prove your assertion. Show us how your number can be completely > indexed while it cannot be covered. Again, definition: K is the number with K[p] = 1 for all p in N and no other digits. Do you agree that is a valid definition? If not, why not? K can be completely indexed because each digit position is a natural number. Do you agree with this? If not, why not? K can not be covered because there is no natural number n such that all digit positions are less than or equal to n. Do you agree with this? If not, why not? > No, don't tell us that this was > true or that you defined that, but show us *how* you manage this trick > which, in my eyes, is impossible. With the axiom of infinity it is dead easy, see above. And I thought we were arguing with the axiom of infinity in mind. > Give us at least one example how you > index a number without covering all the preceding numbers. I have no idea what you are stating here. > > It is your lack of a proper proof that if each digit of a number can > > be indexed that number can be covered. And such a proof does not > > exist. > > I do not see how I could avoid my conclusion. But if you are so sure > then give us at least one example how you completely index a number > without covering all the preceding numbers. I have done so many times, and am doing it here again. You still have *not* provided any proof supporting your conclusion. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 23 Aug 2006 20:13
In article <1156364184.155913.12090(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > What is the sum of infinitely many finite numbers? What is the sum of > > > all finite numbers? > > > > I state that it holds for infinitely many 'n'. Not that it holds for > > infinite 'n' (whatever that may be). > > Infinitely many have infinitely many differences. If you sum them up > then you get an infinite sum. Well, if you give me a proper definition of the sum of infinitely many numbers, I may have some idea about your meaning. In mathematics such a thing is not defined. Your statement above is not a definition. > > > > But I am doing so in order to show that his arguing concerns > > > impredicative definitions and is inconclusive.f is the mapping, n is a > > > natural number, M_f(n) is a set which contains all nongenerators, > > > including n if not including n which is mapped on M. > > > > Yes. Such triples do not exist. And that precisely shows why Hessenberg's > > proof was right. > > That it is false. An argument which makes use of the last digit of pi > is void, because the last digit of pi does not exist. Again avoiding the issue by going on with completely different things. > > If there is a surjective mapping f from N to P(N) it is > > a requirement (of surjectivity) that such a triple *does* exist. > > I gave an example that this set cannot exist, independent of the > surjectivity, independent of the cardinalities of the sets involved in > the mapping. But it is trivial that such a triple can not exist. But if there is a surjective mapping from N to P(N), such a triple *must* exist. So there is no surjective mapping from N to P(N). In a mapping from N to the set S(N) of finite subsets of N the triple also does not exist. But there it is not needed, because M(f) is not necessarily a finite subset. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |