An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: David R Tribble on 21 Aug 2006 18:55 Tony Orlow wrote: David R Tribble wrote: >> Is K a list of natural numbers? Pay attention. K is the sequence of >> digits required to represent the naturals in binary. The set of binary >> strings of length n has size 2^n. If n is aleph_0, then your countably >> infinite string of digits produces uncountably many possible strings. >> We've been through this. You cannot cull bits or strings in any way to >> reduce this uncountably infinite set to produce the countably infinite >> set of naturals you claim exists. In other words, there is no TYPE of >> number which could represent the size of K. > David R Tribble wrote: >> Yes, we've been though all this before and you still won't listen to >> reason. >> >> It takes ceil(log2(k)+1) binary digits to encode each natural k, which >> we'll call D(k). So: >> D(1) = 1, D(2) = 2, D(3) = 2, D(4) = 3, etc. >> >> Now add up all those D(k) for all natural k (all k in N), and call it >> t, the total number of binary digits for all naturals: >> t = sum D(k), for all k in N. > Tony Orlow wrote: > Why would you sum them all, when each sequence of bit positions include > those before it? To prove your claim (that countably infinite string of digits produces uncountably many possible strings) is false. Pay attention. David R Tribble wrote: >> Obviously, since each k is a finite natural, then each D(k) is a finite >> natural as well, i.e., each natural k requires a finite number of >> binary digits to represent it. Adding each finite D(k) to the finite >> sum D(1)+D(2)+...+D(k-1) up to that point yields a countable finite >> total number of digits for each k. > David R Tribble wrote: > Right, in your theory, which yields an uncountable number of binary > strings, ala 2^aleph_0. I'm proving exactly the opposite. Pay attention. David R Tribble wrote: >> There are a countably infinite number of naturals. Since the sum >> of a countable number of countable values is itself countable, t is >> countable. > Tony Orlow wrote: > So, what is ceil(log2(aleph_0)+1) again? Is that countably infinite? It's meaningless. > You have missed the point entirely, as usual. Oh well. I just proved your claim to be false. I think that's on point.
From: David R Tribble on 21 Aug 2006 19:05 Virgil wrote: >> So definitions of operations are valid only for their original domains >> of definition and new definitions are required for different domains of >> definition. > Tony Orlow wrote: >> Bull. The rules for arithmetic manipulation can be applied without >> problems (for anything but transfinite mathology) to variables of >> infinite value. > Virgil wrote: > Not in mathematics, they can't. Indeed. So Tony, what is 1/0 + 2/0? I suppose you will tell us that it is obviously 3/0. Which is obviously not the same as 4/0. Virgil wrote: >> Anyone wishing to extend an operation only defined on one domain to a >> larger domain must give a clear definition of what that extended >> definition is to mean. TO has not done this. > Tony Orlow wrote: >> The domain is the real line. > Virgil wrote: > Which specifically does not include any points at infinity, and at best > only defines arithmetic on the finite numerical values associated with > geometrical points, and on nothing else. And that's only because we can equate real numbers to points on a geometric "line" (which actually involves other concepts such as "distance" and metric spaces). Show us where Big'un exists as a point on the real line. Is it somewhere near 1/0? Or maybe near 3/0?
From: Dik T. Winter on 21 Aug 2006 18:47 In article <1156189552.184903.323170(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1155999560.728582.29520(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > But I now understand that the series: > > > > sum{n = 1 .. oo} (-1)^n/n > > > > does not converge according to your logic. > > > > > > For every natural number n we have n/n = 1. But the notation lim n --> > > > oo is not clear. > > > > You do not know the convention that taking powers goes before division? > > I overlooked the "(-1)^". You never overlook something? Oh, yes, sometimes I do. > > But I now understand that the series: > > sum{n = 1 .. oo} ((-1)^n)/n > > does not converge according to your logic. > > It does to any practicable approximation (not by my logic but in > reality). Mathematics is not about reality... > > And why is the notation lim n --> oo not clear. You have defined it just > > a few days ago: > > > > > The same is meant as in sequences and series like: > > > > > lim [n --> oo] (1/n) = 0 > > > > > n becoming arbitrarily large, running through all natural numbers, but > > > > > always remaining a finite number < aleph_0. > > That is correct as long as n is a natural number. You said (see above) that the notation was not clear, while it was clearly a natural number. > But often some expression like 0.111... is ivolved where not all digit > positions can be indexed by natural numbers and n is understood to > approach something I do not know. Back to that again. I define it as K: K[p] is 1 for all p in N. Where is the limit? On the other hand in the notation 0.111... for 1/9, there is a limit involved: lim{n -> oo} sum{k = 1 .. n} 1/10^k what is not clear about that limit? > > > > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings > > > > > where the remainings are |{n+1,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 > > > > > > > > That makes no sense. As I read it, we have: > > > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings = 1 + 1 = 2. .... > The remainings yield 1. That is correct. How do you *define* remainings? And I see now that you switched from your original, which read: > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 (because > omega is a fixed, well defined and well deteremined quantum, according > to Cantor) What are the remainings in this case? > > > > > > But I will allow an error here. But I was thinking about: > > > > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > > > > which might be written as: > > > > lim{n -> oo} |{1, 2, 3, ..., n}|/|{2, 4, 6, ..., 2n}| > > > > > > = 1 by mathematics and by set theory. > > > > > > > or as: > > > > lim{n -> oo} |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| > > > > > > = 2 by mathematics but 1 by set theory. > > > > As set theory does not know about limits, > > Now we know about limits in set theory. > > > I fail to see that, but > > even if set theory had knowledge about limits, I would state: > > No, also 2 by set theory. What you fail to see is that > > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > > is defined by neither mathematics, nor by set theory. Both do give > > *no* definition of aleph-0/aleph-0. The cardinal numbers do not > > form a field, only a ring (and not even a division ring). So > > arbitrary division is not defined, definition is not even possible. > > Division was possible and was practised in fact long before rings and > fields were known. Yes? On cardinal numbers? > Do you agree that aleph_0 = aleph_0 ? Yup. > Do you agree that division by n is a process which leads to a partition > of a number (of units) into n equal shares (and possibly a remaining)? No. Or perhaps, but in that case in a way you would not agree with: aleph-0 = 2 * 0 + aleph-0 = 2 * 1 + aleph-0 = 2 * 2 + aleph-0... Which one should I chose? Or aleph-0 = aleph-0 * 1 = aleph-0 * 2 = aleph-0 * 3... Again, which one should I chose, and why? Note again that in the cardinal numbers there is no definition of either subtraction or division. If you want them you have to define them properly. > Exactly that is done by a bijection. Each share here is one unit. For a > process like > |{1, 2, 3, ..., 2n}| / |{2, 4, 6, ..., 2n}| > we have at first a bijection between the sets > {1, 2, 3, ..., n} and {2, 4, 6, ..., 2n} and then we consider the > remaining which yields another complete bijection. Therefore the result > is 1 + 1 = 2. For *finite* cardinals. Let's see: |{1, 2, 3, ...}| / |{2, 4, 6, ...}| I can do (among many others): (1) I start with a bijection between {1, 2, 3, ...} and {2, 4, 6, ...}; so the quotient is 1 without remainder. (2) I start with a bijection between {1, 3, 5, ...} and {2, 4, 6, ...}; so the quotient is 1 with a remaining that also yields a bijection, so the total quotient is 2 without remainder. (3) I start with a bijection between {2, 3, 4, ...} and {2, 4, 6, ...}; so the quotient is 1, with remainder 1. Depending on how I chose my bijections I can get every finite quotient with every finite (non-negative) remainder. So, which one should I chose? And, what is: |{2, 3, 5, 7, 11, 13, ...}| / |{1, 4, 6, 8, 9, 10, 12, ...}| ? > One must have a very restricted mind to believe that without fields and > rings division was impossible. With fields it gets a proper definition. With rings you have to provide a definition before you can use it. I do not say it is impossible, I do say that you have to provide a definition. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 Aug 2006 18:58 In article <1156190138.941958.282830(a)m73g2000cwd.googlegroups.com> "David R Tribble" <david(a)tribble.com> writes: > Tony Orlow wrote: > > But Monsieur, what about the injection from P(N) into N, via the bit > > strings which denote set membership, each of which also corresponds to a > > binary natural? Tsk, tsk. Mustn't forget that one! Remember, the only > > set which doesn't map is the entire set, and that maps to the largest > > natural, that is, ...1111 with all bits in finite positions. > > ... as well as all the infinite subsets of N. You keep forgetting > about those, don't you? No, there is a subtlety. He states that ...111 is a natural (as would all infinite strings be). What I am wondering is what bit strings he uses when that natural ...1111 is in a subset. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 Aug 2006 19:49
Some clarification. In article <J4CnBH.DqI(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <J472w4.Lt1(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > > In article <1155885821.815144.187270(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > To be more correct. The first proof is not relying on the > > > > representation of reals, the second (diagonal) proof is not > > > > about reals at all. .... [ About Cantor's proofs:] I was not even completely correct in that statement. After completely reading his first proof and his second proof (the diagonal proof), I am quite sure that both his proofs were targeted at the following theorem: There are sets with a cardinality larger than that of the natural numbers. The first part of his first proof shows that a complete ordered field has cardinality larger than the natural numbers. In his proof he did not rely an any properties of the reals other than that they form a complete ordered field (he uses reals to exemplify). (The second part shows that the cardinality of the set of subsets of a set is strictly larger than that of the original set. This proof comes close to the proof provided by Hessenberg, but is, in my opinion, a bit less strict.) What he states in the quote provided by Mueckenheim is that that proof can be easily used to show that the cardinality of any set of reals in an interval is larger than the cardinality of the natural numbers. His second proof is about sets of sequences, and his diagonal proof shows that the cardinality of the set of infinite sequences with two possible elements is strictly larger than the cardinality of the natural numbers. Zermelo has indicated how that can be converted to a proof that the reals have cardinality larger than that of the natural numbers. But Cantor had *no* intention to prove that at that point at all, and did nowhere write that it could be applied for that purpose. Cantor's one and only purpose was a proof of the theorem: There are sets with cardinality larger than that of the natural numbers. Has the book by Zermelo on Cantor's work ever been translated into English? If not, the non-German speakers are certainly missing something. As with the untranslated book by O. Perron on continued fractions. (BTW, with Zermelo's book also the non-French speaking are also missing something; parts of it are in French.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |