Cantor Confusion
in [Math]
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From: MoeBlee on 9 May 2007 19:54 On May 9, 3:36 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 8 Mai, 19:01, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On May 8, 5:49 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 7 Mai, 22:36, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > (1) Who is the author of that book? I'd like to look it up to see just > > > > what is written there. > > > > Karel Hrbacek and Thomas Jech: "Introduction to Set Theory" > > > Marcel Dekker Inc., New York, 1984, 2nd edition. > > > In which book we find the standard set theoretic definition. Your > > mistake is in conflating their mention of the general non-rigorous > > notion with the actual definition they give. > > Their mention is nevertheless correct. I quoted them including "A > function, therefore, > represents a special type of relation, a relation where every object > a > from the domain is related to precisely one object in the range, > namely, to the value of the function at a." The issue at that juncture was what DEFINITION is in use for a formal proof, in formal Z set theory, of Cantor's theorem. And that DEFINITION is the one Hrbacek & Jech gave as the DEFINITION, and NOT that of a formula, domain, and range. It's plain as day: I said the definition in use for formally proving Cantor's theorem in formal Z set theory is that of a certain kind of relation. And that is the definition given by Hrbacek & Jech. And it is not that of a 'formula, domain, and range'. Axiom: y=z <-> Ax(xez <-> xey) Definition: {r s} = x <-> Ay(yex <-> (y=r v y=s)) Definition: {r} = {r r} Definition: <r s> = {{r} {r s}} Definition: p is an ordered pair <-> Ers p=<r s> Definition: f is a relation <-> Apef p is an ordered pair Definition: f is a function <-> (f is a relation & Axyz((<x y>ef & <x z>ef ) -> y=z)) (Here, as with most authors such as those I mentioned previously, we say 'relation' where Hrbacek & Jech say 'binary relation'.) > In order to define that relation, we need a formula or rule or > whatever, a domain and a range. There are different matters: 1) defining the predicate 'function', 2) defining a PARTICULAR function, 3) stating "f is a function & P", where P is a formula (especially one in which the variable 'f' occurs free). And any function has a unique domain and range, but it is not the case that in every instance we must MENTION that domain and range to carry out 2) and 3). In particular in certain ordinary proofs of Cantor's theorem, we suppose f is a function & dom(f)=x & range(f) subset of Px. There you can see, from the definition of 'function' that the MERE statement 'f is a function' does not require mentioning the domain and range. However, in this proof, we go on to say that x is is the domain and that the range is a subset of Px. So, again, you need to understand the difference between, on the one hand, defining 'function', and on the other hand, speaking of PROPERTIES of a PARTICULAR function (or in the case of the proof of Cantor's theorem, 'f' as a variable to be universally generalized), such as the particular domain and A (not THE) range that f is a subset of. And most importantly, the function f in the proof of Cantor's theorem does NOT require any formula definining it. The proof starts with "Suppose f is a function & dom(f)=x & range(f) subset of Px." There, 'f' is a variable that will be universally generalized at the end of the proof, and thus, by the very NATURE of universal generalization, we assert about f ONLY those suppositions that we will NOT dischage at the end of the proof, thus any specification of anytyhing about f other than that its domain is x and its range is a subset of Px would be INCORRECT for this proof, just as, for example, it would be incorrect to try to prove that all natural numbers n are either even or odd but start supposing that n is DEFINED by some formula that distinguishes n from all other natural numbers. This is just basic predicate logic. > > Those people give the same definition as Hrbacek & Jech, > > then they give a formula or rule and two sets Now you're just being silly after it's been pointed out to you that even Hrbacek & Jech give the explicit DEFINITION (as even they MARK it 'definition') in such a way that a 'function' is defined as the other authors define it, which does not include mentioning that a function has a particular defining formula with it or that a function is some kind of object whose components are a formula, domain, and range. A function is a set of ordered pairs that has a certain property (the property sometimes called 'many-one', i.e. for all x, y, z, if <x y> and <x z> are members of the set of ordered pairs then y=z). That is the ENTIRE DEFINITION. And thus such a set of ordered pairs is not ITSELF some object that is system (or whatever you would call it) of components that are a formula, domain, and range. Yes, every function has a unique domain and a unique range, and sometimes we do specify PARTICULAR functions by a formula that defines that function. But the definition of 'function' ITSELF is NOT that a function has a defining formula. This is understood by the rest of the people in this thread who are talking with you about this. Why don't YOU understand it? Possibly because you WILLFULLY AVOID understanding it? > > you fool. > > You are loosing self control. I am in fine control when I specifically choose to comment that you are a fool. > Or do you think such expressions will > support No, I think such expressions state an auxillary fact, which is that in such discussions of set theory you act the fool. > your wrong claim that H&J would contradict themselves within > few lines? Then you are loosing your intellect too. I never claimed any such thing. And if you persist to claim that I have, I'll call you a liar as well as a fool. Among my recent posts you will find my ample discussion explaining, for YOUR benefit, the difference between, on the one hand, an informal description of an informal notion as a prelude to a rigorous defintion, and on the other hand, the rigorous definition that captures the intuitive sense of the informal notion but avoids the intensionality and vagueness of the informal notion in favor of a purely extensional and rigorous definition. And as Hrbacek & Jech do that, it does NOT harbor a contradiction. So why don't you understand that after it has been explained to you already by so many people other than me? > > > Which university did you attend? > > > I did not say that I have or attended or have not attended any > > university. It's irrelevent. > > According to your performance here, in particular your discussion > about what a function is and your lack of understanding any textbook, > you never saw a university from inside, I suppose. I've given you pinpoint sharp explanations of a bunch of basic stuff that you need to learn, whether at a university or on your own. > So it is in vain to exchange any further argument with you. You have no coherent argument, understanding, or knowledge of this subject to put into exchange. And so it stands: The definition: f is a function <-> (f is a relation & Axyz((<x y>ef & <x z>ef ) -> y=z)) and there is no correct objection to the use of 'function' in the proof of Cantor's theorem as that theorem is proven in ordinary set theory textbooks, including in Hrbacek & Jech. Thus, it stands that your initial comment in that regard are plainly incorrect. > Bye. MoeBlee
From: MoeBlee on 9 May 2007 20:09 On May 9, 3:52 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 8 Mai, 20:32, Virgil <vir...(a)comcast.net> wrote: > > In article <1178636838.282834.114...(a)w5g2000hsg.googlegroups.com>, > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 7 Mai, 23:50, Virgil <vir...(a)comcast.net> wrote: > > > > > >http://books.google.com/books?id=Er1r0n7VoSEC&pg=PA23&ots=1afi1e6mts&... > > > > > k++Jech+set+theory+function&sig=Zx5hPqZZ2icNy3mkguHi9kyrFVA#PPA24,M1 > > > > > /Introduction to Set Theory/, by Karel Hrbacek, Thomas J. Jech > > > > > (1999) [pages 23-4] > > > > > : > > > > > : * 3. Functions * > > > > > : > > > > > : Function, as understood in mathematics, is a procedure, a > > > > > : rule, assigning to any object /a/ from the domain of the > > > > > : function a unique object /b/, the value of the function > > > > > : at /a/. A function, therefore, represents a special type > > > > > : of relation, a relation where every object /a/ from the > > > > > : domain is related to precisely one object in the range, > > > > > : namely, to the value of the function at /a/. > > > > > : > > > > > : * 3.1 Definition * A binary relation /F/ is called a > > > > > : /function/ (or /mapping/, /correspondence/) if /aFb_1/ > > > > > : and /aFb_2/ imply /b_1 = b_2/ for any /a/, /b_1/, and > > > > > : /b_2/. In other words, a binary relation /F/ is a function > > > > > : if and only if for every /a/ from dom /F/ there is exactly > > > > > : one /b/ such that /aFb/. This unique /b/ is called > > > > > : /the value of F at a/ and is denoted /F(a)/ or /F_a/. > > > > > : [F(a) is not defined if /a [not in] dom F/.] If /F/ is > > > > > : a function with /dom F = A/ and /ran /F/ [subset] B/, > > > > > : it is customary to use the notations /F: A -> B/, > > > > > : /<F(a)| a [in] A>/, /<F_a>_a[in]A/ for the function /F/. > > > > > : The range of the function /F/ can then be denoted > > > > > : /{F(a)| a [in] A}/ or /{F_a}_a[in]A/. > > > > > : > > > > > As usual, WM includes only the irrelevant bits > > > > I only wanted to avoid typing infinite definitions. > > > > > and excludes the part > > > > that gives the formal definition, and, incidentally, proves him wrong > > > > The second paragraph proves the first one wrong, in your opinion? > > > A formal definition always REPLACES any informal ones, and governs the > > meaning and usage of the thing defined. > > But the formal definition does not specify how a and b are related > other than by mentioning F. This F however is undefined unless you > know from the first paragraph that it is a procedure or rule. WRONG. F is a SET. > Further definition 3.1 contains: It is customary to use the notations / > F: A -> B/. Why do you think A and B were not two sets and F was not a > formula, as I said (and as anybody says who ever studied some > mathematics)? What are you talking about? F, A, and B are all SETS. In Z set theories, every object is a set (and in ordinary class theory, every object is either a set or a proper class). 'F' is NOT a variable of the meta-lanaguage ranging over formulas. 'F' is a variable of the OBJECT language, thus F is a set (or more precisely, upon an appropriate definition of 'set' (such as 'x is a set <-> Ey xey) we have, in Z set theories, the theorem 'Ax x is a set', thus we have, as a derivable line in a proof,. the formula 'F is a set'. > > And the formal definition requires a function to be a set of ordered > > pairs (a relation), so that it is first of all a SET, which is quite > > different from being merely a rule.- > > Of course it is not merely a rule. It is a rule and two sets, which > are connected by this rule. No, here it is: 3.1 Definition * A binary relation F is called a function (or mapping, correspondence) if aFb_1 and aFb_2 imply b_1 = b_2 for any a, b_1 and b_2. There it is in EXPLICIT mathematical English. F is a relation. And a relation is a set (it is a set of ordered pairs). F is not required to be a formula. No matter that is has been EXPLICITLY shown to you that you are incorrect, you still persist to confuse the informal notion (and you even get THAT wrong) with the rigorous definition. MoeBlee
From: MoeBlee on 9 May 2007 20:31 On May 9, 8:10 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 15:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > But that is not the definition, their formal definition is: > > "A binary relation F is called a function (or mapping, correspondence), if > > aFb_1 and aFb_2 imply b_1 = b_2 for any a, b_1, and b_2." > > And how do you think the a's and b's are selected? And from what sets > do you think > are they selected? In many instances we use formulas to prove that certain a's, b's, etc. are in the domains and ranges of certain functions, but that is a DIFFERENT matter from the MERE DEFINITION of 'function' which to specify the property that makes a set a function, as that property is satisfied NO MATTER what PARTICULAR way or by what formula we may prove that certain a's and b's are in the domain or range of any PARTICULAR function. Why oh why don't you UNDERSTAND this? > And why do you think H&J give that informal > statement? Just in order to confuse the students? No, they give it because it is a common informal non-rigorous notion of 'function' that students may be familiar with from high school algebra and even college level calculus, as that notion may suffice for such limited purposes; but for the purposes of certain more advanced mathematical studies such as in axiomatic set theory, we require a rigorous EXTENSIONAL definition; so the informal notion is presented so that the student can see how the informal notion motivates the actual rigorous set theoretic definition. > It is ridiculous to > follow this discussion. You're not following it! > I can assure you, if one of my students would > not know that a function is a formula (or rule or whatever) together > with a domain where it is defined and a range, then he or she would > not pass the exame. And this is the same in the better math courses in > Germany. If the student doesn't know that the informal notion of a rule is supplanted by the formal definition that Hrbacek & Jech give, then the student will be lost as to set theory until he or she does learn that definition. You go read your Hrbacek & Jech, and see that through the rest of the book, when they talk about functions it is in the sense of the DEFINITION they gave and all proofs in that book regarding functions are with regard to that DEFINITION. In PARTICULAR, as we were discussing, the proof in that book of Cantor's theorem is in regard to functions as in the DEFINITION in that book. Get it through your skull that is as thick as prison wall concrete: When we prove Cantor's theorem, just as Hrbacek & Jech prove Cantor's theorem, we do it with regard to functions in EXACTLY the sense that Jech & Hrbacek give in the DEFINITION and NOT in the informal non- rigorous sense that the rigorous definition SUPPLANTS. MoeBlee
From: MoeBlee on 9 May 2007 20:33 On May 9, 10:30 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 8 Mai, 20:00, Virgil <vir...(a)comcast.net> wrote: > > > And perhaps, as a non-mathematician, one might even say an > > anti-mathematician, WM is incompetent to judge the value of mathematical > > definitions to mathematics. > > If you call yourself a mathematician, then I will agree to the honour > of being an anti-mathematician. > > I attended Harvard and Oxford, among others. > > Not studying mathematics there, I hope. At least not learning it. Or > did they teach you infinite definitions there? Probably they did not > get ready yet? > > > You need it taught by anyone at all with any more mathematical > > competence than you have yourself. Only the ignorant laugh at their own > > ignorance. > > Then try to stop it. Do you know the definition of a binary relation > by H&J? > > 2.1 Definition A set R is a binary relation if all elements of R > are ordered pairs, i.e., if for any z Î R there exist x and y such > that z = (x, y). > > 2.3 Definition Let R be a binary relation. > (a) The set of all x which are in relation R with some y is called the > domain of R and denoted by dom R = {x | there exists y such that xRy}. > dom R is the set of all first coordinates of ordered pairs in R. > (b) The set of all y such that, for some x, x is in relation R with y > is called the range of R, denoted by ran R. So ran R = {y | there > exists x such that xRy}. > > Can you find the words domain and range in this text? What is your point? MoeBlee
From: Dik T. Winter on 9 May 2007 21:09
In article <1178720551.560480.296960(a)y80g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 9 Mai, 04:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178463202.150771.47...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 6 Mai, 04:33, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > Your conclusion is invalid. Indeed, each node is passed by a > > > > > > mutlitude of paths. But the path "0.10101010..." is nevertheless > > > > > > separated from all other paths. > > > > > > > > > > No. Only from those few you can ask for. > > > > > > > > And those are (in your opinion) only finitely many. Or can I ask for > > > > infinitely many paths? > > > > > > No. You may ask for a set of infinitely many paths, but not for > > > infinitely many paths. > > > > So in your opinion there are only finitely many paths in the infinite tree. > > I did not say that. Quote: "Only from those few you can ask for". I can ask only for finitely many paths. > > Great. You reject the axiom of infinity, as I wrote already many times > > before. (Note: reject, not refute.) > > I did not reject anything. I ask you whether you can put infinitely > many questions. Please answer. If the answer is no, then you should > recognize that you cannot ask whether all paths which accompany > 0.101010... will get separated from it. I can not ask infinitely many questions, but that question is a single question. > > > > > No. Only from those few you can ask for. > > > > > > > > Opinion again. > > > > > > It is proven by Cantor that there are uncountably many real numbers. > > > It is also clear that there are only countably many questions, no? > > > > I would think there are only finitely many questions. But all this is not > > mathematics but philosphy. > > No. That is mathematics, because you imply to be able to ask > infinitely many question. Not at all. I ask the single question: "are all other paths different from a particlar path?" In the same way: "what is sum{i = 1..n} i" a single question. You want to split the first question into infinitely many questions about all paths. In that case you should also split the second question into infinitely many questions about all n. Using your logic the split questions can only be answered for finitely many cases, so the formula "sum{i = 1..n} i = (n + 1) * n / 2" is only valid for those cases that you did ask. > > > > There is a subtle difference. With the diagonal proof we always > > > > remain in the finite. > > > > > > That's why it fails for numbers with infinitely many digits. > > > > No. > > Assertion, no argument. Yours was also an assertion without argument. > > > No. Only for those p' you can ask for. There must be others, because > > > there is no node which belongs only to p alone. > > > > I do not ask for each individual path. I show it for all. Or do you > > reject the formula sum{i = 1..n} = (n + 1) * n / 2? You can not ask > > the validity for all n, but only for finitely many n, according to your > > thinking. So for each and every n you have to prove it again. > > This formula is obviously only correct for finite sets f nubers, Yes, so what? > becasue for the whole set the result aleph_0. Pray give a proof of this statement. > Do you think it would be > possible to sum an infinite set by (n + 1) * n / 2? No, as such has not been defined in mathematics. But you reject the formula above for any 'n' for which you did not ask. So each time you use the formula with a different 'n' you should prove it for that particular 'n'. > > > > > "completed" means nothing remains. > > > > > > > > Perhaps. > > > > > > Sure. > > > > Oh. A mathematical proof, please. > > The set of natural numbers is completed such that there is no natural > number outside of that set. Do you disagree? Ah, now you define completed, I think. But it is not true that nothing remains. We have still the rational numbers, the real numbers, and whatever. > > > > You can do countably many tests? I thought you could only do finitely > > > > many tests. > > > > > > Please read carefully. We cannot do more than countably many. And in > > > finite time, we cannot do more than finitely many. > > > > Perhaps. You are getting more and more philosophical each turn. > > You refuse to talk and think about what mathematicians can do? I refuse to talk philosophy. The rules of mathematics are pretty simple. You start with a set of axioms, throw in a few definitions, and deduce properties using logic. > > > > But again, in mathematics it is *not* necessary to test > > > > each and every individual. Otherwise you could not even prove that > > > > sum{i = 1..n} i = (n + 1) * n / 2 > > > > for all n, but only for finitely many n. > > > > > > Of course this is only true for finitely many n. For infinitely many > > > natural numbers, we get aleph_0. > > > > You are wrong. We do not get that. If so, pray show a proof. > > See, for instance, Hrbacek and Jech, p. 188, if you do not believe me. Again you misread what I state. I do not state "for finitely many i", I state "for finitely many n". The formula is a statement about a single n. > (There are more books which treat that topic, but this in my > possession: It will not before Friday that I have again access to that book, but context is *very* important. > "It is not very difficult to evaluate infinite sums. For example, > consider > 1 + 2 + 3 + ... + n + ... + ... (n in N). > It is easy to see that this sum is equal to aleph_0." Off-hand I would state that they are wrong, because infinite sums are not even defined in mathematics. To wit: sum{i = 1..oo} is shorthand for: lim{n -> oo} sum{i = 1..n} so there is a limit lurking behind. > What else should this sum be? It cannot be finite and it cannot be > uncountable. In analysis it is undefined. As it is in algebra. And also in many instances of set theory. But as I have stated already before, Hrbacek and Jech give their particular definition of limits in ordinal arithmetic, and with those definitions it makes sense. I think the statement is in the context of that definition. > > What is the relevance? But try to find the first occurrence where pi(x) > > and Li(x) switch place. Nevertheless, it *is* a natural number > > Are you sure? Well, it should be a natural number, otherwise we could not even talk about pi(x). > What evidence is there that 2^65536 is a natural number? Why not? > http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ndjfl/1093634481&abstract= Not mathematics, but logic. There is another newsgroup next-door. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |