Cantor Confusion
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From: Han de Bruijn on 13 Oct 2006 05:14 cbrown(a)cbrownsystems.com wrote: > Is there a way, in the language of set theory, to state "although there > is no largest set which is finite, there exists no set which is > infinite"? Of course! That's exactly what set theory /is for/. Huh? Sorry, I don't get it. Han de Bruijn
From: Han de Bruijn on 13 Oct 2006 05:35 Virgil wrote: > In article <b8869$452f4a39$82a1e228$32738(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >> >>How about: I'm not interested in "the end". I don't know where the end >>is. And I don't care as well. As long as the end is somewhere where it >>causes a uncertainity which is acceptable for my purpose. > > But what if "the end" isn't anywhere because there isn't one? > > As soon as you posit an end, you run into problems. You would be much > better off saying that all such questions about an end to the naturals > are unanswerable, and stick to what you can explicitly construct. Both approaches run into problems. Either you accept infinities, either you accept a little bit of Physics: uncertainity and inexactness. Guess you know what my choice is. Guess I know what your choice is. Han de Bruijn
From: William Hughes on 13 Oct 2006 06:01 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > > > > > > 0.1 > > > > > > > > > > 0.11 > > > > > > > > > > 0.111 > > > > > > > > > > ... > > > > > > 3) Every set of unary numbers which covers 0.111... to a finite > > > > > position N can be replaced by a single unary number. > > > > > > > > > > > > > > This holds for every finite position N. If 0.111... has only finite > > > > > positions, then (3) holds for every position. As 0.111... does not > > > > > consist of mre than every position, it holds for the whole number > > > > > 0.111.... > > > > > > > > > > > > > So we have > > > > > > > > for every digit position N, there exists a set of > > > > unary numbers which covers 0.111... to position > > > > N > > > > > > > > and > > > > > > > > for every digit position N, if there exists a set of unary > > > > numbers which covers 0.111... to position N, > > > > there exists a single unary number, M, > > > > such that M covers 0.111... to position N > > > > > > > > this implies > > > > > > > > for every digit position N, > > > > there exists a single unary number, M, > > > > such that M covers 0.111... to position N > > > > > > > > > > > > this does not imply > > > > > > > > there exists a single unary number M such that for every digit > > > > position N, M covers 0.111... to position N > > > > > > Why shouldn't it? > > > > Because > > > > for every digit position N, > > there exists a single unary number, M, > > such that M covers 0.111... to position N > > > > can be true even if M depends on N. The statement > > > > there exists a single unary number M such that for every digit > > position N, M covers 0.111... to position N > > > > cannot be true if M depends on N. Therefore the first does > > not imply the second. > > I did not claim that the first statement implied the second. I did not > claim that one letter depends on the other. I claim that the second > statement is true for every finite N. And that is obvious. Take just M > = N. > Hence every finite position of 0.111... will be covered by one single > number M of the list. This statement does not refer in any respect to > the number of digit positions. All finite digit positions are covered > by one finite number M. If 0.111... has only finite digit positions, > then the whole number is covered. > > > >If every digit position of 0.111... is a finite > > > position then exactly this is implied. > > > > No, the fact that every digit position of 0.111... is finite does not > > mean that M does not depend on N, and that is what we need. > > Why should we need that? Why should the number of finite positions be > of any influence, if we know that the theorem is true for any finite > position? > > You think we cannot name the M, so it cannot exist? You know we cannot > construct a well-order of the reals. But nevertheless it does exist, > according to Zermelo's proof. And here, there *is* such an M, for all > finite positions, according to a very simple proof, independent of > whether or not we can name it. No. We do not have one M. We have a lot of different M(N)'s, one for each digit position. We can show that if there is a last N, call it K, then all the M(N) can be replaced by a single M(K). But there is a last N iff the number of digit positions in 0.111.... is finite. - William Hughes - William Hughes > > The problem is that you believe in an actually infinite series of > finite numbers. This belief leads to intermingling digit position and > number of digits just according to the needs and, after all, it leads > to such self-contradictive statements as we have seen here with "the > vase". > > > What we need is not that "every digit position of 0.111... is finite" > > but "there are a finite number of digit positions in 0.111..." > > These two statements are not the same. > > Whether they are the same or not is a question which I will not discuss > in this context. But if they are not the same, then the second one (or > its negation) is at least completely irrelevant with respect to the > problem in question. This depends only on a particular digit position > and not at all on the observation that there may be further positions. > > Regards, WM
From: Dik T. Winter on 13 Oct 2006 07:24 In article <1160675848.377420.163220(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1160646886.830639.308620(a)c28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > If every digit position is well defined, then 0.111... is covered "up > > > to every position" by the list numbers, which are simply the natural > > > indizes. I claim that covering "up to every" implies covering "every". > > > > Yes, you claim. Without proof. > > 1) "Covering up to n" > means > 2)"covering n" > and "covering the predecessors of n". > Therefore we need not prove (2) if (1) is true. Yes. But you claim: (3) "covering 0.111...", not covering n. > > You state it is true for each finite > > sequence, so it is also true for the infinite sequence. > > It is true *for every finite position*. I do not at all care how many > such positions there are. The obvious covering of (2) by (1) does not > depend on frequency. But were is the "covering up to 0.111..."? -- 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 13 Oct 2006 07:31
In article <1160675932.010700.124010(a)m7g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1160647755.398538.36170(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > It is not > > > contradictory to say that in a finite set of numbers there need not be > > > a largest. > > > > It contradicts the definition of "finite set". But I know that you are > > not interested in definitions. > > We know that a set of numbers consisting altogether of 100 bits cannot > contain more than 100 numbers. Therefore the set is finite. The largest > number of such a set cannot be determined, as far as I know. That set is indeterminate. Just use Ascii notation. The string "Graham's number" fits in 100 bits. > Could you determine it? Or would you prefer to define that such ideas > do not belong to mathematics? Then I would not be interested in that > definition. It is an indeterminate set. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |