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From: William Hughes on 19 Oct 2005 09:40 albstorz(a)gmx.de wrote: > David R Tribble wrote: > > Albrecht S. Storz wrote: > > >> [...] > > >> Since there is no biggest number and since there is no infinite number, > > >> the size of the set of numbers in form of sets of #s is undefined as > > >> the biggest natural number is undefined. > > >> > > >> But the sequence of the sets of # fullfill the peano axiomes. So this > > >> set must be infinite. > > >> > > >> The cardinality of a set is not able to be infinite and "not defined" > > >> at the same time. > > >> This is the contradiction. > > > > > > > David R Tribble wrote: > > >> I don't see the contradiction. The size of the set is "not defined" > > >> to be the same as any natural number, and the set size is obviously > > >> infinite. This is no contradiction, since no natural number is > > >> infinite. > > >> > > >> The thing that is "not defined" is the largest natural, which obviously > > >> does not exist. But the set size is infinite, and is nicely defined > > >> by an infinite cardinal. > > >> > > >> You seem to be mixing the two concepts of "natural" and "cardinal" > > >> numbers to create a supposed contradiction, but that does not work. > > > > > > > Albrecht S. Storz wrote: > > > You are not able to understand that there is no difference between > > > numerals and sets. > > > > I have no problem seeing the correspondence between natural numbers > > and von Neumann sets. But neither of these are the same as > > cardinalities, which are not numbers, but measures (sizes) of sets. > > Natural numbers are sets. Why be so delicate about this? It's not only > a correspondence between them. It's identity. Only a fool is unable to > see that fact if the numbers are shown in unitary (1-adic) system. > > And now the cardinality. The definition of cardinality bases on sets. > The existence of infinite sets bases on definition. The meaning of the > cardinality of an infinite set is unknown. Actually you should learn the definition of cardinality before making statements about it. This might stop you uttering such idiocy as "The meaning of the cardinality of an infinite set is unknown." > 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. >Cardinality is just a artificial concept with no sens and > meaning. Again, learn the definition of cardinality before criticizing it, > Define the existence of unicorns and be happy. After you learn the definition of cardinality, please explain how it is possible for a set not to have a cardinality. - William Hughes
From: William Hughes on 19 Oct 2005 09:50 albstorz(a)gmx.de wrote: > David Kastrup wrote: > > albstorz(a)gmx.de writes: > > > > > Virgil wrote: > > >> In article <1129684276.251366.121150(a)g47g2000cwa.googlegroups.com>, > > >> albstorz(a)gmx.de wrote: > > >> > > >> > David R Tribble wrote: > > >> > > > >> > > > > >> > > Because I can prove it (and it's a very old proof). A powerset of > > >> > > a nonempty set contains more elements that the set. Can you prove > > >> > > otherwise? > > >> > > > >> > This argument is stupid. Is there any magic in the powerfunction? > > >> > > >> "Proofs" are not stupid until they can be refuted. The proof that for an > > >> arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. > > > > > > > > > Even if you think that the powersets of finite and infinite sets > > > have both a greater cardinality than their starting sets, you would > > > not really think it depends on the same cause in both cases. > > > > > > You must proof it independently for finite and for infinite sets. In > > > this sense the argument is stupid. > > > > Uh, no. The proof depends merely on the fact that some value has to > > be either a member of a set, or not. And if some value is in one set, > > bur not another, then those two sets are different. > > > > That's all. Finiteness or infiniteness does not even play into it. > > The proof just constructs a set which differs by the membership of at > > least one particular value with every target set in the assumedly > > complete mapping of set to powerset. > > > > -- > > David Kastrup, Kriemhildstr. 15, 44793 Bochum > > > > I don't know what you are talking about. The proof for finite sets > needs just a complete induction. This will not hold for infinity I > think. Perhaps you are thinking about proving that n<2^n for all natural numbers n? This can be done by induction, and indeed this proof doen not extend to infinite numbers. However, this is not the proof usually used to show that there is no bijection between a set and its powerset. The standard proof does not use induction, nor does it require finiteness. -William Hughes > > > Regards > AS
From: Tony Orlow on 19 Oct 2005 10:04 stephen(a)nomail.com said: > Tony Orlow <aeo6(a)cornell.edu> wrote: > > stephen(a)nomail.com said: > >> Tony Orlow <aeo6(a)cornell.edu> wrote: > >> > David R Tribble said: > >> >> > >> >> But you have not provided a mapping between any set and its powerset, > >> >> infinite or otherwise. > >> > Have too. > >> > >> No you have not Tony. > > > Have, too! > > >> The proof that there does not exist > >> a bijection between a set and its power set is quite short. > > Then it shouldn't take too much looking to see where it goes wrong.... > >> > >> 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 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. However, given the definition of bijections, it is easy to map any well-ordered infinite set to its power set through the binary representation, such that each element's position starting at element 0, represented as a binary natural, also represents a subset of the naturals, where the bit in the 2^n position denotes membership of the nth element. You may have a discrepancy in the values of the naturals and the values in the subsets, but you have a complete bijection nonetheless. Discrepancies in value don't bother you elsewhere. Why is this bijection so different? > > > If y is a member of w, then all you can say is that y does not map to any > > subset which contains itself. Y does not map to w. Neither does x. > > What is x? If y is a member of w, then y is in f(y), and > by the definition of w, w is not a member of w. Yes, there is no element y in any given set S which maps to S. That element would be element 2^|S|-1, outside the scope of S. > > >> > >> If y is not an element of w, then y is not in f(y), which > >> means it is an element of w. > > If y is not a member of w, then y maps to a subset which contains y as a > > member, and y IS in f(y). Is this possible? We'll see what you think..... > > >> > >> These are both contradictions. So y cannot be an element > >> of w, and it cannot not be an element of w. So y > >> cannot exist. > > Neither of those possibilities causes a contradiction. You are getting confused > > with your double negatives. If w is the set of all elements which do not map to > > subsets containing themselves, then being a member of w means simply that y > > does not map to a subset containing y, which is perfectly possible. Not being a > > member of w means that an element maps to a subset which DOES contain itself. > > Where is the contradiction? > > The contradiction is that if y is in w, then it is not in w, > and if y is not in w, then it is in w. That is a pretty > obvious contradiction. Well, y is in w, as are all the elements. y does not map to w. For any given set, there is no element in the set which maps this way to the set itself. And yet, when you have infinite sets such as this, can't I just map element 2^S-1 to subset S? > > You apparently failed to grasp the very important fact that w=f(y). > So once again, is y in w? Not if w=f(y). If w=f(y), then y is not in S, and therefore not in w. > > If y is in w, then y is in f(y), because w=f(y). > If y is in f(y), then y is not in w, because w is defined > as containing the elements x in S such that x is not in f(x). > w cannot contain y, because y is in f(y). > > If y is not in w, then y is not in f(y), because w=f(y). > If y is not in f(y), then y is in w, because w is defined > as containing the elements x in S such that x is not in f(x). > w must contain y, because y is not in f(y). Got it. y is not in S, and therefore not in w. > > >> > >> So there is at least one element in P(S) which is > >> not in the image of f, so f is not an onto function, > >> and it is not a bijection. > > Sorry, not so. > > Yes. The fact that you cannot understand a simple > proof does not make the proof invalid. I got a little confused, but you still haven't proven anything like the impossibility of a bijection with the power set. In other cases bijections are performed without regard to such discrepancies. > > >> > >> What is wrong with this proof in your opinion? > > You confused yourself with double negatives. If I misinterpreted any of what > > you said, please clarify. > > You apparently misintrepted all of it. not entirely. > > Do you understand that we are assuming that w=f(y)? No, I forgot that in figuring out what we were talking about specifically. I don;t see anything in it proving bijection impossible. If w=f(y) then y is not in S, assuming some limit to S. But, S goes on forever and ever, and we can always borrow for our bijection, having sets containing elements with at most the log2 of the value of the element that maps to it. As long as there's a 1-1 correspondence, what's the problem? > Do you understand that y is in w if and only if y is not in f(y)? Yes. > Do you understand that > y is in w if and only if y is not in w > is a contradiction? If y is not in S, then y is not in w. > > Stephen > -- Smiles, Tony
From: stephen on 19 Oct 2005 10:42 Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: >> Tony Orlow <aeo6(a)cornell.edu> wrote: >> > stephen(a)nomail.com said: >> >> Tony Orlow <aeo6(a)cornell.edu> wrote: >> >> > David R Tribble said: >> >> >> >> >> >> But you have not provided a mapping between any set and its powerset, >> >> >> infinite or otherwise. >> >> > Have too. >> >> >> >> No you have not Tony. >> >> > Have, too! >> >> >> The proof that there does not exist >> >> a bijection between a set and its power set is quite short. >> > Then it shouldn't take too much looking to see where it goes wrong.... >> >> >> >> 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 > 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. No, that is not clear at all. It is entirely possible that an element maps to a set that contains itself. However no element maps to w. You still do not get it. Here is a simple mapping from N to P(N). 1 -> {1} 2 -> {1,2} 3 -> {1,2,3} 4 -> {1,2,3,4} ... Every element is contained in the subset that it is mapped to. For this mapping, w={}, and no element is mapped to {}. > 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, No. Try again. >> > If y is a member of w, then all you can say is that y does not map to any >> > subset which contains itself. Y does not map to w. Neither does x. >> >> What is x? If y is a member of w, then y is in f(y), and >> by the definition of w, w is not a member of w. > Yes, there is no element y in any given set S which maps to S. That element > would be element 2^|S|-1, outside the scope of S. >> >> >> >> >> If y is not an element of w, then y is not in f(y), which >> >> means it is an element of w. >> > If y is not a member of w, then y maps to a subset which contains y as a >> > member, and y IS in f(y). Is this possible? We'll see what you think..... >> >> >> >> >> These are both contradictions. So y cannot be an element >> >> of w, and it cannot not be an element of w. So y >> >> cannot exist. >> > Neither of those possibilities causes a contradiction. You are getting confused >> > with your double negatives. If w is the set of all elements which do not map to >> > subsets containing themselves, then being a member of w means simply that y >> > does not map to a subset containing y, which is perfectly possible. Not being a >> > member of w means that an element maps to a subset which DOES contain itself. >> > Where is the contradiction? >> >> The contradiction is that if y is in w, then it is not in w, >> and if y is not in w, then it is in w. That is a pretty >> obvious contradiction. > Well, y is in w, as are all the elements. y does not map to w. For any given > set, there is no element in the set which maps this way to the set itself. And > yet, when you have infinite sets such as this, can't I just map element 2^S-1 > to subset S? What is element "2^S-1"? S is a set. Element "2^S-1" means nothing to me. >> >> You apparently failed to grasp the very important fact that w=f(y). >> So once again, is y in w? > Not if w=f(y). If w=f(y), then y is not in S, and therefore not in w. If y is not in S, then it is not part of the bijection. f is function from S to P(S). It makes no sense to plug in a value that is not in S into f. >> >> If y is in w, then y is in f(y), because w=f(y). >> If y is in f(y), then y is not in w, because w is defined >> as containing the elements x in S such that x is not in f(x). >> w cannot contain y, because y is in f(y). >> >> If y is not in w, then y is not in f(y), because w=f(y). >> If y is not in f(y), then y is in w, because w is defined >> as containing the elements x in S such that x is not in f(x). >> w must contain y, because y is not in f(y). > Got it. y is not in S, and therefore not in w. No. You still do not get it. >> >> >> >> >> So there is at least one element in P(S) which is >> >> not in the image of f, so f is not an onto function, >> >> and it is not a bijection. >> > Sorry, not so. >> >> Yes. The fact that you cannot understand a simple >> proof does not make the proof invalid. > I got a little confused, but you still haven't proven anything like the > impossibility of a bijection with the power set. In other cases bijections are > performed without regard to such discrepancies. There is no bijection between a set and its powerset. It does not matter what the set is, it does not matter if it is finite, or infinite. >> >> >> >> >> What is wrong with this proof in your opinion? >> > You confused yourself with double negatives. If I misinterpreted any of what >> > you said, please clarify. >> >> You apparently misintrepted all of it. > not entirely. >> >> Do you understand that we are assuming that w=f(y)? > No, I forgot that in figuring out what we were talking about specifically. I > don;t see anything in it proving bijection impossible. If w=f(y) then y is not > in S, assuming some limit to S. But, S goes on forever and ever, and we can > always borrow for our bijection, having sets containing elements with at most > the log2 of the value of the element that maps to it. As long as there's a 1-1 > correspondence, what's the problem? If nothing in S maps to w, and w is in P(S), then f is not a bijection. It is that simple. Nothing in S maps to w. If you think otherwise, tell us what element of S is mapped to w. You still do not understand the proof at all. It is a relatively simple proof. Here are some examples that might help you out. Although you will likely snip them as "tedious nonsense." First, look at the finite case. Let S={a,b,c}. Lets look at f(a) = { a } f(b) = { a, c } f(c) = { a, b, c } w = { x : x not in f(x)}, so w={b}. {b} is not in the image of f. We can try again. f(a) = { a } f(b) = { b } f(c) = { a, b, c } Now w={}, which is not in the range of f. How about f(a) = { } f(b) = { a } f(c) = { a, b } Now w={a,b,c}. We can keep playing this game, but w will never be in the image of f. Of course in the finite case it is obvious that there cannot be a bijection from S to P(S) because P(S) has 2^|S| elements, and when |S| is finite it is obvious that 2^|S| > |S|. However the above proof makes no mention or use of the set being finite, and it applies equally well to finite and infinite sets. Lets look at the natural numbers and the "bijection" albstorz proposed: 1 -> N 2 -> {} 3 -> {1} 4 -> N\{1} 5 -> {2} 6 -> N\{2} 7 -> {1,2} 8 -> N\{1,2} 9 -> {3} 10 -> N\{3} 11 -> {1,3} 12 -> N\{1,3} 13 -> {4} 14 -> N\{4} 15 -> {1,4} 16 -> N\{1,4} 17 -> {2,3} 18 -> N\{2,3} 19 -> {5} 20 -> N\{5} .... In this case w={ 2,3,5,7,9,11,13,15,17,19 .... } Where does this show up in teh above list? If you claim it shows up in position y, then is y in w or not? Stephen
From: albstorz on 19 Oct 2005 11:14
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 "äääh", "hömm" and "üüüh" 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 "äääh-", "hömm-" and "üüüh-" 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. Regards AS |