Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Han de Bruijn on 20 Oct 2006 07:04 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? Han de Bruijn
From: Franziska Neugebauer on 20 Oct 2006 07:48 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? F. N. -- xyz
From: Han de Bruijn on 20 Oct 2006 07:57 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. Two members are equal if they have the same members ... Right? Han de Bruijn
From: Dave Seaman on 20 Oct 2006 08:08 On Fri, 20 Oct 2006 01:24:35 GMT, Dik T. Winter wrote: > > Am I right that "sets of first cardinality" are the finite sets, "sets of > > second cardinality" are the denumerable sets, and so on? I have not > > found an explanation of that terminology, although "numbers of the second > > class" are clearly explained. > No. The first cardinality is aleph-0. The second cardinality is aleph-1. > Finite sets just have there size as cardinality without an ordinal attached > to it. So when he states > 'sets of the first cardinality can only be "counted" with ordinals of the > second class' > he is nearly correct. Hou would have been correct if he had omitted the > worde 'only'. So then the set of ordinals of the first cardinality is identical to the second number class. > > > > I read that to mean that the correspondence goes like this: > > > > > > 0 -> w > > > > 1 -> w+1 > > > > 2 -> w+2 > > > > > I think not. > ... > > What German word would you use to describe the contents of the above > > table? Notice that the table does not have an end. > Abz?hlen. But now I see what you mean, and I also see how you come to > your interpretation. First: > w?hrend die Mengen erster M?chtigkeit nur durch (mit Hilfe von) Zahlen > der zweiten Zahlenklasse abgez?hlt werden k?nnen > might indeed mean that you are counting sets rather than elements of sets. > That statement is ambiguous. But the successor: > die Abz?hlung bei Mengen zweiter M?chtigkeit > leaves a lot of doubt. If there was 'von' rather than 'bei', I would agree > with your interpretation. But I think it is only a native German speaker > who could explain precisely what was written. Hmm. Doesn't the table illustrate one set of numbers 'bei' the other? -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Franziska Neugebauer on 20 Oct 2006 08:13
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. OK. > 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 |