infinity ...
in [Math]
|
Prev: math
Next: The proof of mass vector.
From: David Kastrup on 20 Oct 2005 14:14 Tony Orlow <aeo6(a)cornell.edu> writes: > William Hughes said: >> >> You have made exactly this mistake before. Yes every number >> is a set. No, not every set is a number. For example >> {peach, apple, plum, fiddle} is a set but not a number. > > Huh! I coulda swore that was 4! Sure. Your incapacity of grasping the difference between set and cardinality and members is legendary. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 20 Oct 2005 14:23 Randy Poe said: > > Tony Orlow wrote: > > David R Tribble said: > > > Tony Orlow wrote: > > > >> I already showed you the bijection between binary *N and P(*N). > > > >> What didn't you like about it? It is valid. > > > > > > > > > > David R Tribble said: > > > >> No, you showed a mapping between *N and R, which is equivalent > > > >> to a mapping between *N and P(N). That's easy. > > > > > > > > > > Tony Orlow wrote: > > > >> No, it was specifically a bijection between two sets of infinite binary > > > >> strings representing, on the one hand, the whole numbers in *N starting > > > >> from 0, both finite and infinite, in normal binary format, and on the other > > > >> hand, the specification of each subset of whole numbers in *N, where each > > > >> bit which, in the binary number, represents 2^n denotes membership of n in > > > >> the subset. This is a bijection between the whole numbers in *N and P(*N), > > > >> using an intermediate bijection with a common set of infinite binary > > > >> strings. > > > > > > > > > > David R Tribble said: > > > >> But that's an incomplete mapping, because there are not enough infinite > > > >> binary strings in *N to enumerate all of the subsets of *N. Try it, > > > >> if you don't believe me. > > > > > > > > > > Tony Orlow wrote: > > > > Not enough infinite binary strings? How many in *N and in P(*N)? > > > > Are you saying that I cannot construct a bijection between the two on an > > > > element-by-element basis which continues infinitely through the set of > > > > binary strings? > > > > > > That's exactly what I'm saying. > > > > > > card(*N) = c, but card(P(*N)) = 2^c, and c < 2^c. > > At what point does the bijection break down? > > What is "the" bijection? "The" bijection I offered between *N and P(*N), using the intermediate set of infinite binary strings, remember? It's described, above, in this very post. > The proof that card(S) < card(P(S)) > shows that no bijection exists. That proof has been given a > few times in this thread and I see you're trying to get through > it. Whether you believe the proof or not yet, you do realize > that the proposition is "Let f(S) be any mapping from S to > P(S). Then f can't be a bijection." Right? Yes, the proof purports to show no bijection can exist by positing a complete infinite set and deriving a contradiction from the fact that the element mapping to it is not in it. But the sets are infinite, and you do not reach the end of either. For any set, there exists a number mapping to it, and every number maps to a unique set. The fact that it has no end is no different from an y other infinite bijection. So, I ask again, what rule of bijection did I break, and where does this bijection break down, since it seems to hold for all finite cases? Did I break some specific bijection-forming rule? > > If the proof is correct (as it is), then ALL bijections break > down. The proof consists of showing that no matter what > mapping you choose, there's an unmapped element of P(S). Yes, assuming you have enumerated the entire infinite, unending set, with all its elements, one of which must map to the set. The proof ignores the fact that for any element declared as the last, the one corresponding to the last bit in the subset bit string, there is always another that is larger. This is one of the hallmark inconsistencies in this theory. You are trying to prove something about the string that maps to the entire set? It's like the largest finite. Any such string represents a natural larger than the set, which leads to the question of which set that natural is first a member of, since it's not a member yet. Most of the contradictions in the proofs you use are derived in one form or another from the "largest finite" contradiction. I recommend refraining from proofs by contradiction for a while, and seeing what you can prove directly. That should help tremendously. After all, once you derive a contradiction, the question remains as to its exact source. > > > It certainly works for all finite cases, does it not? > > Absolutely not. Let S = {1,2,3}. > Then P(S) has 8 elements: > {}, > {1}, {2}, {3}, > {1,2}, {2,3}, {1,3}, > {1,2,3} > > There is no bijection between the 3 elements of S and the > 8 elements of P(S). You snipped what i was referring to as working. I believe it was that no element is a member of the set to which it maps. This is certainly the case here and for all finite cases. > > > What makes you think it falls apart at some point? > > It falls apart for EVERY set. No, in the infinite case, the bijection exists for every natural and for every subset. It is a perfect 1-1 correspondence. Please name a natural or a subset which is left out of the mapping. > > > For which element is there not a corresponding subset? > > It breaks apart in the other direction: there is (at least > one) subset for which there is no corresponding element. Name it please. > > First you have to define your mapping rule. Given any mapping > rule, a subset can be identified for which there is no > corresponding element. I gave the mapping rule: natural <-> binary string <-> subset, ordered naturally from 0. > > - Randy > > -- Smiles, Tony
From: Tony Orlow on 20 Oct 2005 14:28 imaginatorium(a)despammed.com said: > > Tony Orlow wrote: > > stephen(a)nomail.com said: > > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > stephen(a)nomail.com said: > > <showing that there cannot be a bijection from a set to its power > set...> > > > > >> Let f be a function from S to P(S). > > > > Our proposed mapping bijection.... > > > >> > > > >> Define the set w as follows: > > > >> > > > >> w= { x : x in S and x is not in f(x) } > > > > So, w is the set of all elements which are not members of the subsets which > > > > they map to through f(x).... > > > >> > > > >> Clearly w is a subset of S, and so w is an element of P(S). > > > > Clearly...but properly? > > > > > > >> We now show that w is not in the image of f. That is, > > > >> there does not exist a y such that f(y)=w. > > > > So, there can be no y such that it maps to the subset of all elements which do > > > > not map to subsets containing themselves? We'll see..... > > > >> > > > >> Suppose such a y exists. If such a y exists, it must > > > >> either be an element of w, or not. > > > > One or the other. I'll accept the excluded middle.... > > > > > > >> > > > >> If y is an element of w, then y is in f(y), which means > > > >> it is not an element of w. > > > > If y is in w, this means y is a member of the set of elements which map to > > > > subsets which do not contain themselves. This means that y is in S but not in f > > > > (y). So, indeed, it IS a member of w. There is no reason why both x and y > > > > cannot map to subsets which do not contain themselves. > > > > > > If y is in w, then y is in f(y), because w=f(y). Remember, > > > we are assuming there exists a y such that f(y)=w. However > > > w is defined such that y can only be in w, if y is not in f(y). > > > Okay, I hit this one at the end of the day, and got confused halfway through it > > Yes, that sounds fairly typical. You are about to show that eleven > million (whatever, I've forgotten the numbers already, and I made them > up anyway) mathematicians have been getting it wrong for 100 years with > an 8-line proof. But you hit it at the end of the day and got confused. > > > and forgot that you're assuming y is mapped to w. Sorry about that. It is clear > > that no element in the set maps to a subset that contains itself, as I > > illustrated below. If f(y)=w, then y can't be in w, but then that means y IS in > > f(y), which means it's in w. Got it. > > > > That certainly causes a bit of a contradiction, based on the largest-finite > > kind of argument, since you are noting that whatever subset you choose, it > > never contains the natural that maps to it, and the question remains what > > number maps to the set containing just that natural. You are assuming some > > completed w, where some identifiable y is the number that maps to it. But that > > number has to be larger than any of the elements in w. So, you draw a > > contradiction from the assumption that y is in N and also maps to N. Clearly, > > the power set is LARGER than the set. > > Well, that was a jumble, wasn't it? What is a "completed" w? Normal set > theory talks about sets, and a set contains its members. It contains > all of its members, all the time, never contains anything else, never > becomes tired, "unidentifiable", or "tenuous", just sits there > containing all the elements it contains. (Notice that this immediately > means that normal set theory can't accommodate things like sets that > contain different collections of all of something, one of the direct > consequences of the "infinite numbers are just the same as finite > numbers only bigger" crank line.) > > Notice also how you are again unable to consider abstract sets, and > keep mumbling about "numbers". Why on earth should the y be "larger" > than any element in w? We've proved that y cannot exist at all, but > there is no a priori reason any notion of "largeness" is involved. If > we were considering the set of all finite simple groups or the set of > all Platonic solids, there would be no "larger", but the proof applies > exactly the same. Did I not give what appears to be a valid bijection between the NUMBERS in *N and the SETS in P(*N)? If the bijection is not valid, please state what mistake I made in constructing it. When I say the element y is "larger", perhaps what I should say is its position is greater than the size of S, that is, it is in the set, but not in the segment of the set considered as S. When you ask for an element that maps to the entire set, that assumes you have a bitstring that that element should be equal to. How many 1's are in that bit string? Where does it end? > > > However, given the definition of bijections, it is easy to map any well-ordered > > infinite set to its power set .... > > Blah blah blah. This one line after a proof that no such bijection > exists. Well, it's no wonder you can't do mathematics. It's no wonder that mathematics is in the confused state it's in on this matter, from what I see. But hey! So's everything is this primitive world. > > Brian Chandler > http://imaginatorium.org > > -- Smiles, Tony
From: Tony Orlow on 20 Oct 2005 14:31 David R Tribble said: > Albrecht Storz wrote: > > So, if there is an infinite set there is an infinite number. > > Do you mean that an infinite set (or natural numbers) must contain an > infinite number as a member (which is false)? Or do you mean that > the size of an infinite set is represented by an infinite number > (which is partially true)? > > He means the latter. He is equating set sizes with numbers in general. -- Smiles, Tony
From: William Hughes on 20 Oct 2005 14:45
Tony Orlow wrote: > William Hughes said: > > > > albstorz(a)gmx.de wrote: > > > William Hughes wrote: > > > > > > > > > > > > Coincidently natural numbers and cardinalities are undistinguishable = > > in > > > > > finity. > > > > > > > > They are very similar, but they are not quite "undistinguishable". > > > > A natural number is a set, a cardinality is an equivalence class. > > > > > > You make me hopefull. Some experts make "=E4=E4=E4h", "h=F6mm" and > > > "=FC=FC=FCh" if I said "A natural number is a set." One sees a > > > correspondence between natural numbers and von Neumann sets after all. > > > You are free to say "A natural number is a set." without "=E4=E4=E4h-", > > > "h=F6mm-" and "=FC=FC=FCh-" comments. Be lucky. You are right. > > > And natural numbers don't behave in any other way than sets. So, if > > > there is an infinite set there is an infinite number. If there is no > > > infinite number there is no infinite set. And vic versa. > > > > > > > > You have made exactly this mistake before. Yes every number > > is a set. No, not every set is a number. For example > > {peach, apple, plum, fiddle} is a set but not a number. > Huh! I coulda swore that was 4! No, four is is the set {0,1,2,3} [or {1,2,3,4} for Fortran programmers] Yes the cardinality of {peach, apple, plum, fiddle} is 4, but a cardinality is not a set, it is an equivalence class. > > > Just because you have a set does not mean you have a number. > So, not every set has a size? You should make up you mind on this one. However, even if you wish to assign a "size" to every set, there are some sets for which this "size" is not a number. > > > So yes, there is an infinite set. But this does not mean > > that this set is a number. Indeed, no infinite set is a number. > It's not aleph_0? What is that thing anyway? aleph_0 is a cardinal. Cardinals and numbers are different. -William Hughes |