An uncountable countable set
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From: David R Tribble on 13 Sep 2006 16:27 Mike Kelly wrote: >> There is no bijection between the naturals and the set of all >> binary strings. > David R Tribble wrote: >> Finite binary strings (only). Tony thinks that the set of all possible >> finite bitstrings is the powerset of the naturals (the finite bit >> positions or indices), so there are more (finite) bitstrings than >> bit positions. >> >> Yet he admits that there is a bijection between the two. He >> concludes that this demonstrates a flaw or inconsistency in set >> theory. >> >> In a similar vein, he thinks he has a bijection between the naturals >> (as binary strings) and all the subsets of N. Likewise, he thinks that >> this proves that set theory is flawed. >> >> Obviously he's confused between powerset cardinalities (|P(S)| = 2^|S|) >> and binary bitstrings (bit n = 2^n), probably because of the similarity >> in notation. > Tony Orlow wrote: > You don't have it quite right. While there are countably infinite bit > strings in the power set of the naturals which don't correspond to any > standard finite natural, I contend that these strings cannot represent > infinite values, given that all bit positions are finite, and that they > qualify as finite naturals, ... They can't be finite naturals if they have an infinite number of nonzero digits. That would mean that s=1+2+4+8+... is a finite quantity and does not diverge, which is obviously false. Also, if this were true, then ...111 would be the largest binary natural, but, as you have said yourself, there is no such thing. > ... since each has a successor and precedessor, > generated the same way as for standard finite naturals. If they have an infinite number of digits they cannot be "generated" the same way that finite naturals are. Specifically, incrementing such a value cannot be the same operation as incrementing a natural. > So, when these strings are included as finite naturals, indeed, > that set of strings bijects with the power set of the naturals. That's half right. The set of infinite bitstrings bijects with the set of infinite subsets of N. But infinite bitstrings cannot represent finite naturals. They can represent binary reals in [0,1), though.
From: Virgil on 13 Sep 2006 16:36 In article <1158134126.183169.73090(a)e63g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > If it does not hold you should be able to give an index position > > > > such that it is false. > > > > > > It is not in the list, although we cannot give an index position which > > > is responsible for that fact. All possible indexes are in the list. > > > > But that list is capable of indexing as many more objects as it already > > has. > > And it has many more objects as it already has. Nonsense. No set has more members than it has. > > > > Because it is not in the list which, by definition, contains all > > > numbers which can be indexed and which can index. > > > > Not so. Even if one says it contains all the numbers which are indices > > that list still can index as many again as already indexed. > > Yes, because they are all in the list. So that "Mueckenh" is claiming a set of indices which has more members than it has? Nonsense!!! > But 0.111... is not there. It can be put there. > > > > > > > > > Now we may ask: Is it possible that a list item indexes or > > > > > covers other digits than those which are indexed or covered > > > > > by list items? The answer is: no. > > > > If one insists that a number may only be used to index itself, perhaps, > > but that is not necessary, and when noy necessary the answer becomes yes. > > The axiom of infinty applies to my list. All natural numbers which do > exist (and can be indexed) are there. 0.111... is not there. That is no part of the axiom of infinity. That is a false claim by "Mueckenh" which is not supported by anything. > > > > > > > > Because you assert that all digits of 0.111... can be indexed, which is > > > wrong. Indexing and covering "by all list numbers" is equivalent. Both > > > is true or both is false. > > > > > > It is false that indexing and 'covering "by all list numbers" ' are > > equivalent. Indexing merely means tagging each object with some index > > value, and may be done in many ways for any but the smallest of sets. > > If you deny that indexing and 'covering "by all list numbers" ' are > equivalent, then give me a number of the list (no other naturals do > exist) which can be indexed by all list numbers but which cannot be > covered by all list numbers (or the reverse). Give me your definition of 'covering "by all list numbers" ' first, as unless it means the same as indexing, I have no idea what it does mean. > > > > > > It has been proven by showing that 0.111... does not belong to the set > > > of numbes which can be indexed completely. > > > > Except that an equally valid proof has shown the opposite. > > No proof has shown this. 1 -> 0.111... 2 -> 0.1 3 -> 0.11 4 -> 0.111 .... proves it. > > Regards, WM
From: Virgil on 13 Sep 2006 16:41 In article <1158134221.197566.298180(a)d34g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1157836470.071804.254270(a)i3g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > A set of n elements is *a number*. It need no necessarily be a latin or > > > arabic symbol. > > > > In that case "Mueckenh" has problems because this requires that some > > numbers, as ordinals, are members of themselves which ordinals cannot be. > > The set III is the number 3. If that is impossible according to your > mathematics, then your mathematics is deplorable. Then all mathematic is deplorable, at least in "Mueckenh" 's eyes, except "Mueckenh" 's own. In all mathematics except "Mueckenh"'s, there is a necessary distinction between a number and a numeral. A numeral can represent a number but cannot be a number any more that the name "Mueckenh" can be the person which it names. "Mueckenh" needs a good dose of Korzybski.
From: Virgil on 13 Sep 2006 16:44 In article <1158134368.616455.314180(a)d34g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Therefore it would be short sighted to believe that mental ideas were > not physics. It is even more short sighted to believe that mental ideas are physics.
From: David R Tribble on 13 Sep 2006 16:45
Tony Orlow wrote: >> Yes, I am including infinite values on the number line, since it's >> "infinitely long". > David R Tribble wrote: >> Yet another thing you have to define or prove. Where do these >> "infinite values" appear on your real number line? > Tony Orlow wrote: > Further from 0 than any finite number. Then how are they "on" the same "line"? David R Tribble wrote: >> Are these infinite values connected (in the mathematical sense) to >> the finite values on the line? If not, how is it a "line"? > Tony Orlow wrote: > Yes, they are on the line, with all finite values between them and 0. That's not what "connected" means. If they are not connected, they are simply two separate sets, which means it is more correct to visualize them as two unconnected (but "ordered") "lines". Tony Orlow wrote: >> You are mistakenly applying >> the standard cardinalistic fact that the number or reals in [0,1] is the >> same as the number of all reals. That is false in my system. > David R Tribble wrote: >> Since it's rather easily proved true in standard theory, you need >> to provide that missing proof of yours showing that it's false in >> your system. For instance, you have to demonstrate that there >> are more reals in (0,2] than in (0,1] while showing that both >> intervals are dense. Good luck with that. (Hint: assume that >> there are only a finite number of reals in any interval.) > Tony Orlow wrote: >> Given the axiomatic statement that there are Big'un reals in every unit >> interval, that 2*Big'un>Big'un, and that (0,1] U (1,2] is (0,2], that's >> done. > David R Tribble wrote: >> Without proving the second part about denseness. So we can >> conclude that you're assuming that there are only a finite number >> of reals in any interval (as I suspected). > Tony Orlow wrote: > No, Big'un is an infinite unit. Given denseness derived from the axiom > of internal infinity, ExeR EzeR x<z -> EyeR x<y<z, there are an infinite > number of reals in the interval. You still have to put the two together, proving that the number of reals in [0,2] is more than the number of reals in [0,1] AND at the same time that both intervals are dense in the reals. You have done no such thing so far. Just stating two axioms is not the same as proving that they are consistent with each other. Your axiom of denseness, by itself, can be used to show that any two real intervals have the same number of points. Your definition of Bigun contradicts this, so you have to show how both can be true at the same time. David R Tribble wrote: >> Unless you have a definition or axiom about denseness that >> is consistent with your theorem above? (The standard accepted >> definition of "dense" is not, of course.) > Tony Orlow wrote: > I don't understand why you assume Big'un is finite, when it's an > infinite unit. Because your statement about Bigun can't be consistent with your axiom of denseness if it's not finite. But please, show us your proof. |