Cantor Confusion
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From: MoeBlee on 10 Nov 2006 13:43 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > > > > > > > > Wrong.. You have not understood the meaning. We have: The set of all > > > > > sets does not exist. But if all mathematical entities including all > > > > > sets do exist in a Platonist universe, how can it be that the set of > > > > > all sets does not exist? > > > > > > > > That is NOT Fraenkel, Bar-Hillel, and Levy's point at all! > > > > > > You admitted that you do not understand their philosophical remarks. So > > > let it be. > > > > No, please do not misparaphrase me. I said that I admit that I don't > > fully MASTER their philosophical remarks. But I said that I do know > > basically what they're talking about and what they're saying. > > You may think so, but that is obviously false (see below). > > > > > And they don't advocate it at all. > > > > > > They mention this point of view as a possible one, and, as far as I > > > remember, as the only possibility for a platonist. > > > > WHERE? Where in that book do they mention the point of view that a set > > can grow or have different members at different times and that that is > > the only possible view for a platonist? You've got it completely > > backwards. It is very much a platonist view that sets are NOT as just > > described. > > p. 118, considering that the set of all sets does not exist, they > write: a reasonable way to make this conform to a Platonistic point of > view is to look at the universe of all sets not as a fixed entity but > as an entity capable of "growing". > > Why should they write "reasonable" if they found it unreasonable? Because, as YOU SKIPPED RESPONDING to my explanation for you, and as would be clear had you bothered to read and UNDERSTAND the remarks in that section of the book, theyr'e NOT claiming that sets in general grow or are indeterminate. They are pointing out that if we adopt different axioms, then the entire universe of all sets is something different depending on which axioms we adopt. > I did not say strictly that this is the only possible Platonist point > of view, I said "as far as I remember". But I will not reread this book > only because you did not understand it. You never read this book in the first place with any kind of reasonable understanding of it. You haven't a clue. MoeBlee
From: MoeBlee on 10 Nov 2006 14:05 mueckenh(a)rz.fh-augsburg.de wrote: > Never! I gave a mathematical proof. That has nothing to with set thery. David Marcus listed your exact remarks. It was clear that you represented to give a proof in set theory. As I said, "I am open to seeing your argument as SET THEORETIC if you would just cooperate". [all caps added]. But now you you say your argument "has NOTHING to do with set theory" [all caps added]. I'm DEFINITELY not interested in your informal arguments that have no axioms, no primitives, no formal definitions, and no specified logic that have NOTHING to do with set theory.. So let's be definite here. Do you have a strong belief that your argument can be expressed in the language of set theory and uses only first order logic applied to the axioms of set theory? If so, then, your patience allowing, we can work together as you explain your concepts and terminology to me and I strive to see how to formalize it as a set theoretic proof. But if you really do NOT have a strong belief that your argument can be expressed in the language of set theory and uses only first order logic applied to the axioms of set theory, then I'm not interested in wasting my time. You can let me know which it is. MoeBlee
From: Virgil on 10 Nov 2006 16:33 In article <1163153505.869336.50170(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Your assertion: "The number of lines is omega. This does not mean that > there is > a line omega" contains a self-contradiction. 0 1 2 3 The number of lines is 4 but there is no line numbered 4. The whole problem is easily solved by using ordinals.
From: Virgil on 10 Nov 2006 16:40 In article <1163153634.279931.8900(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > Explain how in a square matrix (squarity proven by the existence of a > > > diagonal) the columns can be longer than the lines for *every* line > > > and every column. > > > > To be square we must have: for every line k, there exists a colmun k. > > This is true. Note that there is no line aleph_0 and no column aleph_0. > > To be square we must have both: For every line, there exists a column, > and there exist as many lines as columns. Not quite. it is enough to have a bijection between lines and columns. If line n ends in column n, the bijection is trivial: each line is paired with its last column. Note that as long as each line is longer than its immediate predecessor this shows that there are at least as many columns as rows, and possibly more.
From: Virgil on 10 Nov 2006 16:41
In article <1163153906.747213.22290(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > > The length is not only a supremum but exists as a maximum (if all > > > natural numbers exist). > > > > > > > No. > > > > Consider the two sets. > > > > A={1,2,3,...} > > B = {1,2,3,...,omega} > > > > Both sets have omega as supremum. Set A does not contain its supremum, > > set B > > does. In the first case the supremem is not a maximum, in the second > > case > > it is. So it is not true that the supremum must also be a maximum. > > That is correct for two independent sets. But it is wrong for two sets > which are connected by a bijection d_nn like the diagonal of a matrix. > Every element d_nn maps a line with n elements to an initial segment of > the column with n elements. > 1 1 > 2 12 > 3 123 > ... > n 123...n > ... > > Continuing we collect less than omega lines. Maybe you do but nobody else does. Just goes to show how WM doesn't count! |