An uncountable countable set
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From: Virgil on 17 Jul 2006 17:20 In article <1153145345.281803.129520(a)35g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > >> You may recall that every sequence member has trailing zeros. > > > > > > Indeed. I recall. Therefore the linearity of the list numbers enforces > > > a column with zeros. > > > > Nice try, but non sequitur. There is no such column. Otherwise show one > > or prove its existence. > > The proof requires logic. Therefore I am afraid you will not accept it. WM's version of logic is quite different from the sort used in ZF of NBG or any other part of mathematics. > Consider the columns spanned by the digit positions of 0.111... Either > there is a column with only zeros, or there is at least one 1 in each > column, or ? If you mean to list 0.0, 0.1, 0.11,0,111,... so that the 0's line up vertically, then every "column" has a least one 1. And each column to the right of the '.' column has the same "number" of 1's as all the others.
From: Virgil on 17 Jul 2006 17:26 In article <1153147907.669663.129980(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Set theory lives by contradiction. Some exhaustions > of nfinite sets are accepted other are not. But except from tradition > there are no other reasons. Therefore, set theory is folklore. > > Regards, WM The difference is that our "traditions and folklore", which we chose to call axioms and definitions, are logically consistent, as far as anyone can tell, whereas WM's is full of self-contradictons.
From: Virgil on 17 Jul 2006 17:29 In article <1153148551.942037.97110(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > This is the deep dilemma of > set theory: There is no actually infinite set of finite numbers. But the existence of this "dilemma" can only be established by assuming it. So for those who do not chose to assume it, it does not exist.
From: Franziska Neugebauer on 17 Jul 2006 17:44 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> > It reads: Either there is a column with only zeros, or there is at >> > least one 1 in each column spanned by the digit positions of >> > 0.111... , isn't it? >> > But you will "argue": Nice try but there is neither nor. >> >> What you call a proof is an unproven proposition. preliminaries: -------------- aleph_0 def= | omega | 0.111... =def (a_i) having a_i = 1 A i e omega a_ij means the well known matrix of figures > IF aleph_0 does exist, THEN 0.111... covers aleph_0 columns. This is as meaningful as If i exists then sqrt(-1) is i. > IF 0.111... covers aleph_0 columns, THEN aleph_0 columns do exist. This is as meaningful as If sqrt(-1) is i then i exists. > IF aleph_0 columns do exist THEN we can consider their contents. This is as meaningful as If i exists then we can consider its value. > IF we can consider the contents of each column, THEN we can ask how > many 1's are therein. Lotta questions. > IF we can ask how many 1's are in each one, THEN the answer can be > "zero 1's" or "not zero 1's". We can. > IF the answer is in each case is "not zero 1's", THEN in each column > at least one 1 must be present. This is the case, since every a_jj = 1 j e N by definition. > However, there is no natural numbers with this property, Could you precicely _define_ which /property/ you are talking about? For every column j e N a_mj has the 1 in position m(j) = j, since a_jj = 1 A j e N. Where exactly lies your problem? > because 0,111... has more 1's than each natural number. You are riding a dead horse. F. N. -- xyz
From: Virgil on 17 Jul 2006 17:57
In article <1153168957.805313.57460(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > Now you use an entirely new term: "can be exhausted". I think you mean > > that you can take out elements one by one and doing this at some stage > > the infinite set becomes empty. Howver, I think, that if that can be > > done, that there is a last element you can take out. And, according > > to the axiom of infinity, that is not possible, so infinite sets can > > not be exhausted in this sense. > > But in another sense? In the sense of having a set of all of them, as per the axiom of infinity, an axiom does it. |