Cantor Confusion
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From: Lester Zick on 27 Nov 2006 14:02 On Mon, 27 Nov 2006 08:08:55 -0500, Bob Kolker <nowhere(a)nowhere.com> wrote: >Eckard Blumschein wrote: >> On 11/26/2006 3:25 AM, Dik T. Winter wrote: >> >>>Remove all irrational points and in addition those >>>points that have x = 0. You have a disconnected set. >> >> >> A set is thought of elements put into it. Can you really set irreal >> points into it? >> This cannot work because the set has to have an order p/q. Otherwise it >> is not a set but a continuum. > >A continuum is a compact connected metric space. And you're talking in riddles again. >Learn some mathematics. Why? You haven't. ~v~~
From: Bob Kolker on 27 Nov 2006 14:06 Lester Zick wrote: > On Mon, 27 Nov 2006 07:25:21 -0500, Bob Kolker <nowhere(a)nowhere.com> > wrote: > > >>Eckard Blumschein wrote: >> >>> >>>Geometry is not outside mathematics. >> >>True. But Geometry is not all of mathematics either. > > > Nor is arithmetic. > > >>The two historical sources of mathematics has been land measurement and >>counting objects. After which comes tracking the cycles in the heavens. > > > So what? One historical source for chemistry was alchemy. Historical > origins don't have any bearing on the nature of a science. They have some bearing. The earlier fields create a barrier to be overcome. It is a challange. Detaching the notion of measure from length and area was a big jump forward in the theory of real variables. Bob Kolker
From: Bob Kolker on 27 Nov 2006 14:07 Lester Zick wrote: > > > Why? You haven't. I have forgotten more math than you ever learned in your life. What I have was a standard definition of a continuum. And it is not a riddle to someone who knows some topology which you clearly do not. Bob Kolker
From: Lester Zick on 27 Nov 2006 14:15 On Mon, 27 Nov 2006 16:04:52 +0100, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: >On 11/27/2006 2:08 PM, Bob Kolker wrote: >> Eckard Blumschein wrote: >>> On 11/26/2006 3:25 AM, Dik T. Winter wrote: >>> >>>>Remove all irrational points and in addition those >>>>points that have x = 0. You have a disconnected set. >>> >>> >>> A set is thought of elements put into it. Can you really set irreal >>> points into it? >>> This cannot work because the set has to have an order p/q. Otherwise it >>> is not a set but a continuum. >> >> A continuum is a compact connected metric space. >> >> Learn some mathematics. > >I only learned discussion with you is wasted time. Actually Bob can be worth talking to. Not so much for his spin on modern math concepts which are pretty much worthless but he has and can learn to think on occasion. It just takes some prodding. >Continuum is a much older and more comprehensive concept than metric >space and compatification. Less important than it might appear. Modern mathematikers tend to over dramatize their dramaturgy with private definitions drawn in parochial terms which can't be interpreted universally. "Metric space" is a reasonable application of universal ideas because it describes a geometric analog. But "compact" is just a modern mathematical buzzword used to help underemployed mathematikers pretend arithmetic is all of mathematics. ~v~~
From: MoeBlee on 27 Nov 2006 14:18
mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > The countability of the set W of all the words of a finite alphabet is > > > > > proved by other means than the usual countability proof. The latter > > > > > consists in constructing a bijection with N, i.e., in constructing a > > > > > list. But it is impossible to construct a list of all words. It is > > > > > impossible to construct a bijection between W and N. > > > > > > > > WRONG. It is a theorem that from the set of finite sequences of members > > > > of a finite alphabet there is an injection into N. > > > > > > Nevertheless it is impossible to construct such an injection > > > > WRONG. > > Do it. I already had. Look at just about any set theory text book for such proofs. > > > as it s > > > impossible to construct a well order of the reals). > > > > Showing an injection from the set of finite sequences of members of a > > finite alphabet into N is not analogous to well ordering the reals. > > I know that. Why do you stress it? > > The > > former can be explicit; > > Do it. I already had. Look at just about any set theory text book for such proofs. > while the later cannot be. > > > > > > > Why should it be > > > > > possible to construct a bijection between N and N? > > > > > > > > Because ANY x is 1-1 with x, by the identity function on x. > > > > > > It is an unfounded assumption to believe that all x of the set appear > > > in this bijection, only by writing x = x. Ridiculous! > > > > We don't prove the existence of an identity function just by writing 'x > > = x'. > > Then demonstrate what you think is a proof. Last time I showed you a proof in set theory, your ultimate response was to point out that I had used the axiom of infinity (which is an axiom of set theory), so that was a pointless dialogue with you. Now, as to showing the existence of the identity function on any set S, we may first prove the existence of the Cartesian product of any x and y (which applies to S and S in particular), then the identity function on S is a subset of the Cartesian prodcuct SxS. See just about any text book of set theory for such routine proofs. > > > > > You are wrong. If a set cannot be well-ordered, it cannot be assigned a > > > > > definite cardinal number. In order to assign a cardinal, at least one > > > > > bijection to an ordinal is required. This requires at least one > > > > > well-ordering. > > > > > > > > WRONG. By using Scott's method (which requires the axiom of > > > > reqularity), we don't need the well ordering theorem to prove that > > > > every set has a cardinality. > > > > > > No. But we need it to determine this cardinality. Again you think > > > erroneously that an existence proof is of any value. > > > > Constructivity is a separate matter. My point stands that we do not > > need the well ordering theorem for every set to have a cardinality. > > Your point is false. In order to have a cardinality, you must know the > number class. In order to know the number class, you must have a > bijection to an ordinal. In order to have a bijection to an ordinal, > you need a well order. > It was Cantor's most important assertion hat all sets can be well > ordered, *in order to ascribe them a cardinal number.* Maybe that you > can ascribe some symbol to some set and call this symbol a cardinal. > That is the same deceit as your bijections to N which are none. > > I think if one swells to explosion about his knowledge of set theory, > he should at least know the very foundation. But I know, that you do > not even understand the simple texts of Fraenkel et al. No, YOU radically MISunderstand what Fraenkel, Bar-Hillel, and Levy wrote. Meanwhile, you are ignorant of even such basic set theory as proving the existence of an identity function. As to cardinality, See Levy's 'Basic Set Theory' for a brief discussion of Scott's method. > > > PS: What about the binary tree in ZFC? Any results? Any idea why it > > > can't be done? > > > > I already mentioned what I would require to discuss your tree argument. > > If you want me to talk about your tree, then please answer my previous > > posts from about a week ago on the subject of even discussing your > > argument. I put it to you very clearly just what you need to tell me > > before we proceed. If you won't do so, then I'm not inclined to waste > > my time on your argument. > > What I can tell you is the following: > > The binary tree > > Consider a binary tree which has (no finite paths but only) infinite > paths representing the real numbers between 0 and 1 as binary strings. > The edges (like a, b, and c below) connect the nodes, i.e., the binary > digits 0 or 1. > > 0. > /a \ > 0 1 > /b \c / \ > 0 1 0 1 > .......................... > > The set of edges is countable, because we can enumerate them. 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 by the infinite recursion > > f(n+1) = 1 + f(n)/2 > > with f(1) = 1 that for n --> oo > > 1 + 1/2 + 1/ 4 + ... = 2 > > edges are related to every single infinite path which are not related > to any other path. (By the way, the recursion would yield the limit > value 2 for any starting value f(1).) The load of 2 edges is only > related to infinite paths because any finite segment of a path with n > edges will carry a load of > > (1 - 1/2^n)/(1 - 1/2) < 2 > > edges. The set of paths is uncountable, but as we have seen, it > contains less elements than the set of edges. Cantor's diagonal > argument does not apply in this case, because the tree contains all > binary representations of real numbers within [0, 1], some of them even > twice, like 1.000... and 0.111... . Therefore we have a contradiction: > > |IR| > |IN| > || || > |{paths}| =< |{edges}| > > If you don't understand this simple and clear exposition, then there is > no hope that you will be able to think any further. > > I really don't know what further information could be required. The information I require about the very basis of this discussion was specified in my previous posts - the ones I wrote a couple of week ago and which I mentioned yet again, but which you yet again ignore. I took the effort to compose those questions for you at that time. If you want me to consider your argument, then please respond to those posts. I'm not inclined to recompose my questions for you again after I already carefully composed them. MoeBlee |