Obections to Cantor's Theory (Wikipedia article)
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From: Robert Low on 20 Jul 2005 16:00 Tony Orlow (aeo6) wrote: > Daryl McCullough said: >>Not quite. "Uncountable set" means "set with a larger cardinality >>than the set of naturals". > Yes, that seems to be the Cantorian definition. Does it actually follow that an > infinite set larger than the naturals can't be enumerated? Personally, I don't > see the connection at all, The connection is fundamental. Set A is bigger than set B if there is no surjection from B to A. If set B is the integers, then saying that set A is bigger than set B is saying that there is no surjection from the integers to A, which implies that there is no bijection between them, and hence that there is no enumeration of A (by definition of enumeration). That's how the game works. We agree on the basic definitions of words like 'bigger', 'enumeration' and so on, and then work out logical consequences of them. Now, until you give a definition of 'bigger', and explain why my first paragraph above is rendered irrelevant, we'll all be very grateful. Or at the very least, very surprised.
From: Helene.Boucher on 20 Jul 2005 16:04 Daryl McCullough wrote: >All of mathematics would > go through just as well without *ever* using the word "larger". > You could just as well use the word "more bloppity": > > By definition, a set S is said to be more bloppity than a set R > if there is a 1-1 function from R to S, but there is no 1-1 function > from S to R. > > Instead of using the term "size" to refer to sets, we could > refer to the "bloppitude". > > Instead of using the words "infinite", we could use the term > "mega-bloppity". > > Nothing of any importance about mathematics would change > if we substituted different words for the basic concepts. > Shouldn't there be one "p" in "blopptitude?" Anyway, nothing in mathematics would change, but surely the interest of the resulting propositions would diminish, should the word "size" disappear. People think (and IMHO in error) that the definition captures correctly the concept "size".
From: Dave Rusin on 20 Jul 2005 16:01 Bored today, I peek in again at this year's Tiresome Poster of The Year winner and find, to my surprise, that there has been a tiny bit of progress. Something like actual definitions have been offered. Imagine! In article <MPG.1d4863d52071fde5989f51(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >> Tony has any number* of "proofs" that an infinite set of natural >> numbers must include "infinite naturals", but these are generally >> circular. The one from "information theory" says that since there can >> only be a finite number of strings of finite length (even if the length >> has no limit), then to get an infinite set of numbers, you must include >> some that are infinitely long. The bit after "since" is a restatement >> of what he purports to prove, but he ignores people pointing this out. > >Ahem! That is another misrepresentation. The bit after "since" is a statement >about symbolic systems, and is a fact outside of the natural numbers. True! (The last bit, I mean) -- maybe even truer than you think. >Given a set of symbols of size S, (To be precise, it's the SET which has size S, not the symbols themselves. Here "set ... of size S" should mean that S is a cardinal; I don't know whether Master Orlow is going to use a _finite_ set of symbols or not, but in fact what he's about to say is TRUE even for non-finite sets S.) >one can construct a set of all strings of length L, True! (But this time "set ... of length L" does NOT mean the _set_ has "length" L; this time it's the strings IN the set that have length L. Isn't "natural" language a bear? Everyone says things ambiguously this way, but only T.O. is actually tripped up by the ambiguity.) We'd better be clear what "strings" are. It's not enough to have a bunch of symbols; the symbols have to come in a certain order, e.g., "1A$" is not the same string as "A1$". So unless you want a very generous notion of what "strings" are, you'll have to interpret "length L" as meaning that L is an _ordinal_, or at the very least that L is a linearly ordered set (not necessarily well-ordered, I suppose). For strings of finite length, this is unnecessarily fussy since the finite ordinals and finite cardinals are identical, and we can just say "...of length 3" without much fuss and bother. On the other hand, there is a very clear difference between what one might call "strings of length 1+omega" and "strings of length omega+1" -- one of them has a zeroth term, then a 1st, 2nd, 3rd, etc; the other instead has a 1st term, then a 2nd, 3rd, etc. AND a last term. Failure to notice the distinction has led our hero to spout all kinds of gobbledygook. But again, yes, for any ordered set L one can indeed construct the set of all L-strings of elements of S. (It's isomorphic to, or by some definitions equal to, the set of functions L --> S .) >and the set of strings has size S^L. True again! Even when S and L are not finite (assuming "... has size ..." means "... has cardinality ..."). Indeed, this is the usual _definition_ of what cardinal exponentiation is. As a bonus, we get to wink at the subtleties of the last section, since if L and M are two ordinals of the same cardinality, then S^L = S^M. >This is a fact, True! >which [mal-formed sentence alert: we've got a subject to a sentence for which no main verb follows. I think what was intended was that the verb "proves" was to follow "...(S is finite)," below. Or something like that.] >when combined with >the fact that digital strings are strings on a finite alphabet (S is finite), True! -- sort of. I'm not sure who uses the term "digital strings". Do they have in mind strings of a _particular_ finite length? Of arbitrary but finite lengths? Or what? Hmm, we'll let this slide for a moment. >S^L can only be infinite if L is infinite. True again! Indeed, ( S^L is finite ) iff ( ( S is finite AND L is finite ) or ( L is empty ) or ( S is empty) ) . >Therefore, an infinite set of >digital numbers MUST contain numbers with infinite numbers of digits. Aha! Tony wins! Yes indeed, it is true -- IF "set of digital numbers" means "set of strings in S^L, where S = {0, 1, ..., 9}" (I'm guessing here), then indeed, such a set cannot be infinite if L is finite. That's absolutely correct! If we specify _A_ finite set L and consider the set of digit strings of length L, then the set of all of those is finite. Sure enough! BUT --- what has that got to do with the set of natural numbers? No one has ever said that each natural number is a digit string of some _single_ finite length L. (Indeed, no one but hacks ever says that natural numbers are digit strings in the first place, but never mind that now.) It is true that EACH natural number n (individually) can be represented by a digit string of SOME finite length L(n). But (duh!) there is no single finite ordinal L which is at least as large as every one of these, in other words, bigger numbers need more digits. Duh. If you want to consider all the natural numbers at once as being members of a single set S^L so that you can apply the previous line of reasoning, you'll have to use an infinite L . You're welcome to do so, and you may choose for example to use the ordinal omega , so that today's date would be ...0002005 - ...0007 - ...00020 That's pretty cumbersome but not wrong. It's damning for your line of reasoning because now the "L" in your arguments is not a finite set, so you can't conclude N is finite. It's also a potently misleading notation because S^L will include other things besides natural numbers -- things like ...01010101 which are most emphatically not natural numbers. And yes, you can use other, larger ordinals too, embedding the natural numbers into S^(omega+1), for example --- a set which includes not only the previous non-numbers like ...010101 (which I would now assume is shorthand for 0 ...010101) but also things like 1...000000 . This last bit of freedom allows you to make an even bigger fool of yourself by continually changing your choice of ordered set L, thus changing the set of things about which you make your wild claims. (For example, even using L = omega+1 does not allow an interpretation of what "101...0101" means. In which S^L does this string lie, grandmaster?) Summary: To his credit, Tony has actually said some correct things about sets of the form S^L . But he mistakenly assumes that the set of ("finite") natural numbers is a subset of S^L for some finite L, and based on some of his other screed seems to believe that the natural numbers coincides with S^L for some infinite-but-never-quite-specified-and- indeed-of-time-shifting-value ordered set L. So his correct statements about S^L have no bearing on questions about the set of natural numbers. HTH! HAND! dave
From: Tony Orlow on 20 Jul 2005 16:22 Daryl McCullough said: > Tony Orlow writes: > > > >Dik T. Winter said: > > >> Back on your horse again. Tell me about the binary numbers (extended to the > >> left with 0's) where the leftmost 1 is in a finite position. Are all those > >> numbers finite? Are there only finitely many of them? > > > >yes and yes > > What definition of "finite" are you using? > > Once again, if your claims had any merit whatsoever, then you > would be able to rephrase them in a way that does not rely on > unorthodox meanings of terms. Rephrase your claim without using > the word "finite" or "infinite". Is that possible? > > -- > Daryl McCullough > Ithaca, NY > > Are you talking to me or Dik? he's the one who used the terms. Do you need me to define "yes"? No one else seems to have a definition for "infinite" as a stand-alone word. Finite means with an end, and infinite means without end, as I have said. What don't you understand? If the leftmost possible significant digit is at position n, then the largest number possible in base b is b^n. If b and n are known, then b^n is known. If b and n are finite, then b^n is finite. This is just using the standard definition of finite, independent of cardinality. It's not rocket science. -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 16:26
malbrain(a)yahoo.com said: > > > Tony Orlow (aeo6) wrote: > > David Kastrup said: > > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > > > > > David Kastrup said: > > > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > >> > > > >> > David Kastrup said: > > > >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > >> >> > > > >> >> > Alec McKenzie said: > > > >> >> >> "Stephen J. Herschkorn" <sjherschko(a)netscape.net> wrote: > > > >> >> >> > > > >> >> >> > Can anti-Cantorians identify correctly a flaw in the proof > > > >> >> >> > that there exists no enumeration of the subsets of the > > > >> >> >> > natural numbers? > > > >> >> >> > > > >> >> >> In my view the answer to that question a definite "No, they > > > >> >> >> can't". > > > >> >> >> > > > >> >> >> However, the fact that no flaw has yet been correctly > > > >> >> >> identified does not lead to a certainty that such a flaw > > > >> >> >> cannot exist. Yet that is just what pro-Cantorians appear to > > > >> >> >> be asserting, with no justification that I can see. > > > >> >> >> > > > >> >> > Even though every subset of the natural numbers can be > > > >> >> > represented by a binary number where the first bit denotes > > > >> >> > membership of the first element, the second bit denotes > > > >> >> > membership of the second element, etc? > > > >> >> > > > >> >> Well, what number will then represent the set of numbers dividable by > > > >> >> three? > > > >> > 100100...100100100100 > > > >> > > > >> > Of course, you will argue that this infinite value is not a > > > >> > natural number, since all naturals are finite, but that is > > > >> > clearly incorrect, as it is impossible to have an infinite set of > > > >> > values all differing by a constant finite amount from their > > > >> > neighbors, and not have an overall infinite difference between > > > >> > some pair of them, indicating that at least one of them is > > > >> > infinite. > > > >> > > > >> You have not shown such a thing, and of course it would be > > > >> inconsistent with the Peano axioms defining the naturals. > > > > > > > > That is simply not true. > > > > > > Sulking won't help. > > Don't be a jerk. > > > > > > > There is nothing in Peano's axioms that states explicitly that all > > > > natural numbers are finite. > > > > > > It is an immediate consequence. > > Read on. > > > > > > > The fifth axiom, defining inductive proof, is used to prove this > > > > theorem, but it is a misapplication of the method. > > > > > > An axiom is not a "misapplication". > > No, the proof is a misapplication of the axiom. > > Yes, you're right, the axiom is a statement of agreement. We agree via > the axiom that the definition of ALL is EACH-AND-EVERY. Each and every > natural number is finite. That's our concept of the natural numbers. > We agree that we need only deal with a finite number of steps to prove > something about the infinite set of natural numbers. > > (...) > > > Now, I have heard the argument that inductive proof does not prove things for > > an infinite number of steps, but only a finite number, but if this is the case, > > then it does not prove anything for an entire infinite set of natural numbers. > > That's not what our axiom says. We get to prove things about the > infinite set of natural numbers in a finite number of steps. Applying > the axiom of induction is a single step in a leap of faith that we > agree to before hand. > > > Either you agree that there are an infinite number of steps involved, or that > > the set of naturals is finite, or that inductive proof does not prove a > > property true for all natural numbers as Peano stated. > > That's not what our axiom says. It says that induction covers all the > natural numbers in a single step, a single leap-of-faith. That's really not the case. It is a recursive proof where the property is proven true for each element depending on its truth for the preceding element. f(n)->f(n+1), for n=1 to oo. Otherwise, how do you think it proves things for each and every n in N? > > karl m > > -- Smiles, Tony |