An uncountable countable set
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From: Han de Bruijn on 29 Sep 2006 03:50 Virgil wrote: > In article <b1fd9$451b85a3$82a1e228$11085(a)news1.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>Virgil wrote: >> >>>In article <d12a9$451b74ad$82a1e228$6053(a)news1.tudelft.nl>, >>> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >>> >>>>Randy Poe wrote, about the Balls in a Vase problem: >>>> >>>>>It definitely empties, since every ball you put in is >>>>>later taken out. >>>> >>>>And _that_ individual calls himself a physicist? >>> >>>Does Han claim that there is any ball put in that is not taken out? >> >>Nonsense question. Noon doesn't exist in this problem. > > What has noon to do with my question? > > According to my recollection of the problem, there was a specific time > (prior to noon) at which any ball which had been put into the vase was > taken out. > > Does Han dispute my recollection? Maybe we're not talking about the same version of the SuperTask. So: where can I find this recollection of yours? Han de Bruijn
From: Han de Bruijn on 29 Sep 2006 03:53 Virgil wrote: > In article <1159437062.473100.294820(a)k70g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >>Virgil schrieb: >> >>>Several sets may all have the common property of being pairwise >>>bijectable, but if any of their members are distinguishable from those >>>of another set then the sets are equally distinguishable. >> >>Each one of the sets expresses, represents, and *is* the same >>(cardinal) number. > > Then one apple and one orange are the same because they have the same > cardinality. In _that_ respect, with respect to counting: definitely, yes! Han de Bruijn
From: stephen on 29 Sep 2006 03:51 Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: > stephen(a)nomail.com wrote: >> In TO-matics, it is also possible to end up with >> an empty vase by simply adding balls. According to TO-matics >> >> ..1111111111 = 1 + 1 + 1 + 1 + ... >> >> and >> ..1111111111 + 1 = 0 >> >> So if you just keep on adding balls one at a time, >> at some point, the number of balls becomes zero. >> You have to add just the right number of balls. It is not >> clear what that number is, but it is clear that it >> exists in TO-matics. > Commonly known with digital computers as "overflow" ? > If TO-matics is an idealization of overflow, then it _is_ consistent > anyway. Sad for you :-( > Han de Bruijn Why sad for me? Tony is the one who seems to deny that you can ever end up with 0 balls, yet at the same time seems to think overflow is possible. How is that consistent? Stephen
From: Han de Bruijn on 29 Sep 2006 03:57 Virgil wrote: > In article <1159437157.109258.97400(a)b28g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >>Virgil schrieb: >> >>>In article <1159186907.615747.304410(a)h48g2000cwc.googlegroups.com>, >>> mueckenh(a)rz.fh-augsburg.de wrote: >>> >>>>1/3 is a number, properly defined, for instance, by the pair of numbers >>>>1,3 or 2,6 or 3,9 etc. But 0.333... is not properly defined because you >>>>cannot index all positions, you cannot distinguish the positions of >>>>this number from those with finite sequences (and you cannot >>>>distinguish them from other infinte sequences which could exist, if one >>>>could exist). >>> >>>Def: 0.333... = lim_{n -> oo} Sum_{k = 1..n} 1/3^n >> >>Definitions (even correct definitions unlike this one) don't guarantee >>existence (I used above "to be properly defined" but I meant "to >>exist"). Example: The set of all sets is defined but is not existing. > > Def, corected: 0.333... = lim_{n -> oo} Sum_{k = 1..n} 3/10^k > > And it exists because there is a real number (actually a ratonal number) > L = 1/3 such that for every epsilon greater than 0, there is a largest > n such that | L - sum_{k+1..n}| >= epsilon. Shouldn't that be "such that | L - sum_{k = 1..n}| <= epsilon" ? And why that "largest n" instead of just "n" ? Han de Bruijn
From: Han de Bruijn on 29 Sep 2006 04:04
Virgil wrote: > In article <76b59$451ba0bd$82a1e228$18077(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>mueckenh(a)rz.fh-augsburg.de wrote: >> >>>Virgil schrieb: >> >>>>>>You stated that you needed counting to determine the successor. That is >>>>>>false. The successor is defined without any reference to counting. >>>>> >>>>>The successor function *is* counting (+1). >>>> >>>>Not to those who can't count. Successorship does not require numbers, it >>>>only requires "next". >>> >>>How far would those who cannot count be able to find "the next"? >> >>And how do you distinguish "the next" from something previous? > > By pointing at them separately. > >>This is >>not a joke. Many young children don't find it trivial that you shouldn't >>count a thing twice. > > But they are much less prone to mistaking who has more marbles, or > whatever, which argues that injection, surjection and bijection are more > basic than counting. Have two bags with say a hundred marbles in it and _make_ the bijection. I wish you good luck. And, BTW, I would like to have a computer program which does the job, properly. Video circuit attached. Han de Bruijn |