Cantor Confusion
in [Math]
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From: MoeBlee on 10 May 2007 13:25 On May 10, 7:14 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 03:59, Virgil <vir...(a)comcast.net> wrote: > > In article <1178738274.909825.208...(a)e65g2000hsc.googlegroups.com>, > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 9 Mai, 20:16, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1178707943.860551.269...(a)h2g2000hsg.googlegroups.com>, > > > > > > 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. > > > > > If it is a set, it is defined as sets are defined, which does not > > > > require anything as undefined as "procedure" or "rule". > > > > How do you define a set, unless you cannot specify the elements > > > belonging to it? > > > I am reasonably certain that that question does not mean in standard > > English, what you intended to ask. > > Should have been: > How do you define a set, unless you can specify the elements belonging > to it? I gave you an answer. > > The axioms of ZF tell how to specify sets in ZF. > > Yes, how to specify, but they do not specify sets except few. The axioms provide for certain theorems of the form: E!x P and thus to define: c = x <-> P (where 'c' is a new constant symbol added to the language; so 'P' is a defining formula). But there is no theorem of set theory that says: If x is a set, then there is a defining formula for x. Do you understand that? > A set can be specified by specifying its elements. This can be done > very well, in particular for finite sets, by a list. Yes, the rule or > formula can be a list listing the elements of a set. A formula is a certain kind of sequence of symbols (so a formula is a certain kind of "list" of symbols). But just listing the names of sets is not a formula. To even begin here, you don't know the definition of 'formula'. > > > Do you know what a relation is? Do you know that there is also the > > > domain and the range required? > > > There are a domain and range to a relation, but they are consequences of > > its definition, not antecedents to it. > Answer only by yes or no: Do you know that there is also the domain > and the range required in a relation and all the more in a function? Your question rides on the meaning of 'required'. It's been explained to you already: Just to DEFINE the predicate 'is a relation' does NOT require mentioning that for every relation there exists the domain and the range of that relation (just look in Hrbacek & Jech, e.g.). However, it follows as a THEOREM that for every relation there exists the domain and the range of that relation. So, in the sense of 'required to mention or specify as to domain and range just to form a definition of 'relation'', it is not required. But in the sense of 'it is entailed that for every relation there exists the domain and the range of that relation', it is required (in the sense of 'it is entailed that') for every relation there exists the domain and the range of that relation. Do you understand now? > Can you imagine a relation without having a domain and range? Of course not. But we DO (Hrbacek & Jech, for example) define the predicate 'is a relation' without having to mention the ADDITIONAL point that for every relation there exists the domain and the range of that relation. Do you understand now? MoeBlee
From: MoeBlee on 10 May 2007 13:58 On May 10, 10:15 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > Be sure, I studied every page very carefully, so carefully that I > typed every letter (some time ago). And typing letters does not an understanding make! > > 2.  RELATIONS > > Mathematicians often study relations between mathematical objects. > Relations between objects of two sorts occur most frequently; we call > them binary relations. > ... > A binary relation is, therefore, determined by giving all ordered > pairs of objects in that relation; it dos not matter by what property > the set of these ordered pairs is described. > > !!!!!!!!!!!! But it obviously does matter *that* it is > described. !!!!!!!!!!!!!!! Yes, to DEFINE a PARTICULAR relation, we have to have a formula to define that PARTICULAR relation. But the relation ITSELF is NOT the formula that defines it and the formula is not even a component of the relation. Morevover, there do exist relations that are not DEFINED. For example, say it is a theorem: Er(r is a relation & P) (where the variable 'r'' occurs free in P). So the theorem says that there exists at least one relation that has the property that is specified by P. But it might be the case that there is not a defining formula to DEFINE a PARTICULAR relation that has the property P. Now, if you don't like that, then that is understandable, and you may reject such a non-constructive theory. However, you can't rationally dispute the plain fact of what do and do not happen to be such theorems of non-constructive theories such as Z set theory, since it is objectively verifiable that there are proofs of such theorems, notwithstanding anyone's acceptance or not of the truth of the axioms or of a non-constructive theory. Moreover, it is not a contradiction to have non-constructive theorems. It is quite understandable that a person might prefer only constructive theories and reject non- constructive ones, but that is different from the issue of consistency, and a theory is not inconsistent merely for being non- constructive. > 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} > > 2.9  Lemma  The inverse image of B under R is equal to the image of > B under R-1. > 2.11  Definition  The membership relation on A is defined by > >                 ïA = {(a, b) | a ï A, b ï A, and a ï b} > > The identity relation on A is defined by > >                 IdA = {(a, b) | a ï A, b ï A, and a = b} > > 2.12  Definition  The set of all ordered pairs whose first > coordinate is from A and whose second coordinate is from B is called > the cartesian product of A and B and denoted A ï´ B. In other words, > >                 A ï´ B = {(a, b) | a ï A, b ï B}. > > ???You see a A and B here???? Yes, to define 'AXB' you have to mention A and B. So what? The definition of the Cartesian product of A and B (AXB) is not the defintion of the predicate 'is a relation'. > >  >                          It is ridiculous to > >  > follow this discussion. 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 that is true, it tells us a lot about mathematics education in Germany. > > Functions in set theory are *not* the same as functions in analysis. > > I talked about mathematics, not about set theory. Nevertheless > functions *in set theory* consist of a prescription (formula, rule, > whatever), domain, and range. No, that is not the DEFINITION of 'function'; and there exist functions that have no defining formula; and every function has its domain and its range, but those are not required to mention to DEFINE 'function' and a function is NOT and does not "consist of" a formula, domain, and range, as the function ITSELF is: A set of ordered pairs that is many-one. THAT is what the function IS. Nothing more and nothing less. It is a certain kind of set of ordered pairs and it is not a formula with a domain and range. > Exercises > > 2.1 Let R be a binary relation; show that dom R c U(UR), ran R c > U(UR). Conclude from this that dom R and ran R exist. > > The last exercise is a good homework for such "mathematicians" who > studied in Oxford or Harvard or at home and, therefore, do not know > that dom R and ran R do exist for any binary relation, or in other > words: without domain and range R would not be a binary relation. We've done that exercise and many similar exercises. And we've done the "double drill down" into the union of the union of a relation MANY times, especially to talk about elements of the domain, range, or field of a relation. And for every new operation symbol (as 'domain of' and 'range of' are locutions representing operation symbols) we prove the existence and uniqueness clauses that support the definition. Doing the above exercise is just one (two, actually) of the many proofs we perform to prove the existence of sets having certain properties. MoeBlee
From: Virgil on 10 May 2007 14:03 In article <1178791459.103475.178210(a)l77g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 20:31, Virgil <vir...(a)comcast.net> wrote: > > > > In order to avoid the clumsy expression of path bunches, I call a path > > > bunch now a finite path. A finite path ends at some node. > > > > > The infinite path {0.000...} is the union of all finite paths with > > > only 0's, namely {0.} U { 0.0} U {0.00} U ... > > > The union is one infinite path and, therefore, has not more paths than > > > were unified. > > > > The union of infinitely many path bunches is a single path? > > As the union of infinitely many initial segments of N is N. According to at least one of WM's definitions, a path "bunch" will, in general, contain more than one path, so that the union of any number of such path bunches will contain as many paths as one of them does. At least if 'union' is to mean what it means in mathematics. > > > > > > > > > All nodes of the tree are last nodes of finite paths. > > > The union over all finite paths is the set of all infinite paths. This > > > set has not more elements than were unified. > c > > Then it is NOT the same 'tree' as a CIBT in ZF or NBG. > > > > In any CIBT of ZF or NBG, the set of paths is easily shown to > > equinumerous with P(N). > > > > And WM cannot refute this bijection. > > And Virgil cannot refute that this is a contradiction. That it contradicts WM's beliefs, I concede, but that it contradicts anything actually mathematical, I deny, and as WM cannot prove any such contradiction, his claim that it contradicts anything of mathematical interest fails.
From: Virgil on 10 May 2007 14:09 In article <1178791668.851141.297820(a)y80g2000hsf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 20:47, Virgil <vir...(a)comcast.net> wrote: > > > In fact, to establish the difference between the diagonal and any one > > other only requires finitely many digits, which WM knows but chooses to > > ignore. > > This is also true for the numbers 1.000... and 0.999... > > Why must this case be excluded from the proof? Because they are both excluded from appearing after the point in the diagonal. The diagonal rule excludes both 0's and 9's after the decimal point, so automatically cannot be equal to any number whose decimal expansion can have them there.
From: MoeBlee on 10 May 2007 14:11
On May 10, 10:15 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > The last exercise is a good homework for such "mathematicians" who > studied in Oxford or Harvard or at home and, therefore, do not know > that dom R and ran R do exist for any binary relation, or in other > words: without domain and range R would not be a binary relation. P.S. As has been explained to you now a hundredfold, yes, every relation has its domain and its range, but that doesn't entail that is required to mention that in the mere definition of 'relation' and not even just to prove that a set is a relation. Take the simplest example possible: Theorem: {<0 0>} is a relation. Proof: Ax(x e {<0 0>} -> x is an ordered pair). And in that proof it was not required to mention that {<0 0>} has a domain and a range; and it was not required to mention that {0} happens to be both the domain and range of {<0 0>}. Do you understand now? MoeBlee |