Cantor Confusion
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From: Franziska Neugebauer on 20 Oct 2006 08:16 Han de Bruijn wrote: > Franziska Neugebauer wrote: >> Han de Bruijn wrote: >>>Franziska Neugebauer wrote: >>>>Han de Bruijn wrote: >> >> [...] >> >>>>>Good! Now given two such members x and y. What does x = y mean? >>>> >>>>x = y :<-> Az(z e x <-> z e y) >>> >>>Of course. Because all members are sets. But I think this is an >>>infinite recursion with the equality definition. Does it end >>>somewhere? >> >> Where do you spot recursion? > > Two sets are equal if they have the same members. The "two sets" are one set if they have the same members. Let A und B refer to the "two sets". If A = B then the "two sets" are one set. > Two members are equal if they have the same members ... Right? Two variables x and y refer to the same member (set) if x = y. Once again: Where do you spot recursion? F. N. -- xyz
From: Dik T. Winter on 20 Oct 2006 08:17 In article <J7EuKz.E74(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <eh89qg$jdf$1(a)mailhub227.itcs.purdue.edu> Dave Seaman <dseaman(a)no.such.host> writes: .... > Yes, I found it also very strange. That is why I did remark on it. The > only thing I can imagine is that he first started to use the common meaning > and later switched over to a fairly non-standard meaning. .... > Abz?hlen. But now I see what you mean, and I also see how you come to > your interpretation. I think that quoting some context of my original quote will make it a lot clearer. I will do it in an English translation only (and where I use 'count', Cantor used 'abz?hlen', * means italics in original) : A very extensive *class* of point-sets are those that are of what I called the *first* cardinality [later called aleph-0] and that I called *countable* sets in sections 1 to 4. The latter expression *can* in a more restrictive sense also be used to distinguish these sets from those of a *higher* cardinality, but in section 5 I have *shown* and *put upfront* that, strictly spoken, one also could also *always say* about sets of the *second*, *third* or *higher* cardinality that they are *countable*; the *difference* is only that, while sets of the *first* cardinality can be counted only *through* (*with the aid of*) numbers of the *second* number-class, the counting of sets of the *second* cardinality *only through* numbers of the *third* number-class, with sets of the *third* cardinality *only* through numbers of the *fourth* number-class, etc. I think that from the first part it is clear that in the second part Cantor is speaking about counting individual sets. And so my opinion remains that the text is in error. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dave Seaman on 20 Oct 2006 08:39 On Fri, 20 Oct 2006 12:17:48 GMT, Dik T. Winter wrote: > In article <J7EuKz.E74(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > > In article <eh89qg$jdf$1(a)mailhub227.itcs.purdue.edu> Dave Seaman <dseaman(a)no.such.host> writes: > ... > > Yes, I found it also very strange. That is why I did remark on it. The > > only thing I can imagine is that he first started to use the common meaning > > and later switched over to a fairly non-standard meaning. > ... > > Abz?hlen. But now I see what you mean, and I also see how you come to > > your interpretation. > I think that quoting some context of my original quote will make it a lot > clearer. I will do it in an English translation only (and where I use > 'count', Cantor used 'abz?hlen', * means italics in original) : > A very extensive *class* of point-sets are those that are of what I > called the *first* cardinality [later called aleph-0] and that I called > *countable* sets in sections 1 to 4. The latter expression *can* in > a more restrictive sense also be used to distinguish these sets from > those of a *higher* cardinality, but in section 5 I have *shown* and > *put upfront* that, strictly spoken, one also could also *always say* > about sets of the *second*, *third* or *higher* cardinality that they > are *countable*; the *difference* is only that, while sets of the > *first* cardinality can be counted only *through* (*with the aid of*) > numbers of the *second* number-class, the counting of sets of the > *second* cardinality *only through* numbers of the *third* number-class, > with sets of the *third* cardinality *only* through numbers of the > *fourth* number-class, etc. > I think that from the first part it is clear that in the second part > Cantor is speaking about counting individual sets. And so my opinion > remains that the text is in error. So it appears. Thanks for your comments. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Virgil on 20 Oct 2006 14:15 In article <b008d$453887ef$82a1e228$31075(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Virgil wrote: > > > In article <6cf73$45387e07$82a1e228$27759(a)news1.tudelft.nl>, > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>Ha! Mathematicians can't even define their most frequently used symbol, > >>which is the equality ' = '. And that is a prerequisite for their "is". > > > > The equal sign, "=", has many meanings, which differ depending on > > context, so that there cannot be one Procrustian meaning that fits all > > contexts. > > > > For sets A and B, A = B means that for > > all x, x is a member of A if and only if x is a member of B. > > Good! Now given two such members x and y. What does x = y mean? In ZF it means "for all z, z is a member of x if and only if z is a member of y". And so on, ad infinitum.
From: Virgil on 20 Oct 2006 14:26
In article <540b7$4538ad26$82a1e228$12664(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Franziska Neugebauer wrote: > > > Han de Bruijn wrote: > > > >>Virgil wrote: > >> > >>>For sets A and B, A = B means that for > >>>all x, x is a member of A if and only if x is a member of B. > >> > >>Good! Now given two such members x and y. What does x = y mean? > > > > x = y :<-> Az(z e x <-> z e y) > > Of course. Because all members are sets. But I think this is an infinite > recursion with the equality definition. Does it end somewhere? It is equivalent to x = y :<-> Az(not(z e x) <-> not(z e y)) so what is allegedly infinitely recursive about it? |