An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 25 Jul 2006 19:56 In article <1153861897.581734.123090(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > If *every* digit can be indexed, then I need not put "finite" between > > > "every" and "sequence", because every indexing sequence is finite. A > > > digit which is not a finite sequence remote from the first digit cannot > > > be indexed and cannot be covered. > > > > Your deleting of "finite" (or "indexing") here is precisely what I said, > > the > > negation of the axiom of infinity. Any indexing sequence is finite. > > Right. > > But the sequence of *all* digits of 0.111... is neither an indexing > > sequence, > > nor a finite sequence. And it is precisely *that* sequence that can not be > > covered by a finite natural. Remember: there is a set that contains all > > naturals. There is (clearly) no natural that is the largest in that set. > > So there is also no last digit in the sequence of digits of 0.111..., where > > the digits are indexed by naturals. > > But every natural is finite. Every digit which can be indexed by > naturals is *by definition* in the true list (not the extended list) The "true list" of all indices is the one represented by the unending "0.111...", since every ending list is too short. > and every digit which cannot be indexed is not in the true list. So > your attempt to explain that a number which is not in the true list > nevertheless could be indexed is completely unfounded. Thus "muecken" acknowledges the infiniteness of the set of indices( naturals) and simultaneously declares it finite. > > > A union of finite sequences is a finite sequence. > > > > An arbitrary union of finite sequences is not necessarily a finite > > sequence. > > This is similar to: the union of finite sets is a finite set. But this > > argues that the union of {1}, {1, 2}, {1, 2, 3}, ... is a finite set. > > Consider the elements of those sets as indices to the elements of the > > sequences. This is only true when N is finite, but that is the negation > > of the axiom of infinity. > > True is that the elements of n are finite. Hence there union cannot be > infinite WRONG! The axiom of infinity in both ZF and NBG says otherwise, and until "mueckenh" can come up with better objections to them than he so far has, they are at least as valid as his own contrary assumptions. > because every union of natural numbers is a natural number. Who says? While any finite union is finite, there is nothing that requires unions of finite natuals to be finite unions, and the apparently consistentf ZF and NBG allow it. > And every natural number is finite. You always intermingle size and > number (Anzahl). It is "mueckenh" who continually conflates the size of any individual natural with the size of the population of all naturals. > My logic is very clear: > (every digit of n can be indexed) ==> (n is in the list) > not (n is in he list) ==> not (every digit of n can be indexed) And it is as clearly wrong as it is possible for it to be. "Mueckenh" again conflates the size of individual naturals with the size of the collection of all of them. (Every finite ordinal can be indexed) ==> (The set of indices is not a finite ordinal)
From: Dik T. Winter on 25 Jul 2006 19:57 In article <1153829288.635491.96140(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > I need not count to w. If a column contains one 1 then the *+sum is 1. > This is a definition like that of Cantor's list. Depends. I have no idea (and it is very dissimilar), whatever way I look at it, I find: (*+){k = 1 .. n} Ak = An and I can not substitute oo for n (so as to include all Ak), because there is no Aoo. Now, when we look at the Ak just as strings of symbols we can define it for all Ak: (*+){k in N} Ak = K where for all p in N, K[p] = 1, but there is still no implied definition. > > The last conclusion requires proof (and it is false). What is the case > > is that for every Ai the indexing of digits will terminate at some point, > > and the indexing of digits of 0.111... will never terminate when we start > > at the first digit and go on. Because there will always be a further > > digit. (As there will always be a further An.) > > But each one will be finite. You mean the index of the digit and the An. Yes. > (every digit of n can be indexed) ==> (n is in the list) Where is the proof? There is an easy proof of: (n is in the list) ==> (every digit of n can be indexed) but not of the converse. > > > The last digit of 0.111... has no ordinal number. > > > > Indeed, but the reason is that there is no last digit. w is still a > > potential infinity, you can not count to it (for one thing, it has > > no predecessor). > > Why does the left column of my extended list differ from the right one? > There are the same numbers given, only their representation is > different. You do not show here what you are referring to, but I think I understand what you mean. And I do not think I have ever stated that they did differ. But we are talking about lists, not about extended lists. And in what way is what you state a rebuttal of what I did state? I am simply claiming that the last digit of 0.111... does not exist, and so it also can not have an ordinal number. > > > Therefore the set of > > > its 1's and the set N has no ordinal number. Therefore it is wrong to > > > define omega as the LUB of the naturals *and simultaneously as their > > > cardinality*. > > > > Eh? The ordinal number of the ordered set N (in the natural order) is w, > > the cardinal number of N is aleph-0 (as is the cardinal number of all > > countable sets). > > The least ordinal of a class is also its cardinal number. If you say N > has the cardinal number omega you are not wrong today. Even Cantor did > so at his later times. Perhaps. I do not know whether that is really true for all sets. But what you stated is still wrong. Formally aleph-0 is defined as the equivalence class of sets that biject with the set of naturals. w is defined as the order type of N (and because N is well-ordered), it also becomes the ordinal number of N. You need proof to show that all ordinal numbers of a class (what is a class?) are for sets that are in the same equivalence class with respect to bijection. But perhaps I am wrong, but I think that today the distinction between w and aleph-0 is made more clear than in Cantor's later times. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 25 Jul 2006 20:07 In article <1153863159.167151.152210(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Michael Stemper schrieb: > > > >Is July 11, 2002 contemporary enough? Thomas Jech: Set Theory > > >Stanford.htm, Stanford Encyclopedia of Philosophy: Until then, no one > > >envisioned the possibility that infinities come in different sizes, and > > >moreover, mathematicians had no use for "actual infinity." > > > > Okay, so your own reference shows that philosophers don't think that > > mathematicians use the term "actual infinity". That sounds correct. > > T. Jech is one of the leading set theorist. > He says up to the advent of set theory there was no use for actual > infinity. Does he say that there is no use for actual infinity within current set theory? I doubt it.
From: Dik T. Winter on 25 Jul 2006 20:23 In article <1153826572.582842.304680(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > In set theory "infinity" means "actual infinity". Potential infinity > > > can neither be counted nor be surpassed. > > > > Read Cantor where he wrote that w is not actual infinity but potential > > infinity, because you could not count to it. About the surpassing I am > > not so sure. > > I would have read it if he had written such a nonsense. But he did not. > > Cantor, Werke, p. 195, footnote: Das Zeichen oo, welches ich in Nr. 2 > dieses Aufsatzes [hier S. 147] gebraucht habe ersetze ich von nun an > durch w, weil das Zeichen oo schon vielfach zur Bezeichnung von > unbestimmten [d. h. potentiellen] Unendlichkeiten verwandt wird. > > For people like you he said: > Cantor, Werke, p. 374: "Trotz wesentlicher Verschiedenheit der Begriffe > des potentialen und aktualen Unendlichen, indem ersteres eine > vernderliche endliche, ber alle Grenzen hinaus wachsende Gre, > letztere ein in sich festes, konstantes, jedoch jenseits aller > endlichen Gren liegendes Quantum bedeutet, tritt doch leider nur zu > oft der Fall ein, da das eine mit dem andern verwechselt wird." I will look up what I found. I think I will come back to it, but it needs time. (I was quite struck by his note.) > > > > > The proof is given by the list: > > > > > > > > > > 1 0.1 > > > > > 2 0.11 > > > > > 3 0.111 > > > > > ... ... > > > > > w 0.111... > > > > > > > > That is neither a proof, nor a list. > > > > > > The list is given in the upper part. Call the whole an extended list. > > > It shows that w is not the number of the naturals, because w appears in > > > the sequence of ordinal numbers only once. > > > > That is only a statement, *not* a proof. > > Why is the ordinality of the left column of the extended list another > than the ordinality of the right column? I do not state so. The ordinality of both "lists" is w+1. Why do you think they are different? In general, the ordinal number of any initial segment of ordinal numbers is the smallest ordinal number that is larger than all ordinal numbers in that segment. (This can be done because the ordinal numbers form a well-ordered set.) And note that 0 is also an ordinal number (of the empty set). So the ordinal number of {} is 0, of {0, 1, 2} is 3, etc. Finally, the ordinal number of {0, 1, 2, 3, 4, ..., w} is w+1. And so is the ordinal number of {1, 2, 3, 4, 5, ..., w}, because there is an order preserving bijection between the two. You apparently did not appreciate that 0 is an ordinal number. (And that is independent of Bourbaki.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 26 Jul 2006 12:57
In article <1153897104.622376.322420(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1153862905.806404.152570(a)i3g2000cwc.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > > I think that my "list" has proved that (among other > > > > > proofs). > > > > > > > > Your list proofs nothing. > > > > > > (every digit of n can be indexed) ==> (n is in the list) > > > > I ask for a proof, not for a restatement. Pray prove the above using > > common > > mathematical and logic methods. It is false. > > It is the *definition* of the list! Every digit which can be indexed > can be indexed by a natural number. Every natural number is in the > list. If "every digit which can be indexed can be indexed by a natural number" then it is the digits which are "in" the list, not the natural numbers. The things which are 'in' a list are those things which are 'indexed by' natural numbers. > I think it is in vain to discuss with you if you do not accept the > basics of logic. Curious, that is also our view of discussing things with "mueckenh". > > > So the triangle gets infinitely long but not infinitely broad. This > > > means the Diagonal must bend. > > > > Eh? Where do you conclude *that* from? The triangle gets infinitely long > > and infinitely broad. > > Wrong. Infinitely broad would mean infinite numbers. Why don't you see > such things? Because they are not true. Sensible people often have great difficulty seeing such things. If "mueckenh" claim that the "triangle" does not get infinitely broad, then he is claiming some finite maximum broadness. Then he should be able to give a finite upper bound on that broadness. Well, "mueckenh", can you? > Infinitely broad would mean infinite numbers. Then infinitely long would also. > So what? In an infinite tree there is no terminaton. But according to > the axiom of infinity "all levels of the tree exist". > > > So while > > in all cases the number of edges is less > > it is larger by a factor of 2. Only in finite cases. For infinite binary trees: There is a natural bijection between edges and finite binary strings. There is a natural bijection between paths and endless binary strings. > Correct! I agree with you. I did not say that they had different > cardinality but that they have equal cardinality. Then "mueckenh" is claiming a surjection from a set to its power set. |