An uncountable countable set
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From: Virgil on 18 Jul 2006 14:48 In article <1153223462.224954.305750(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > 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. > > Those who choose to close their eyes enjoy always the mercy of not > being forced to see the sheer misery of mathematics. If it makes you so miserable why do you keep doing it? You remind one of the small boy who kept hitting himself on the head with a hammer. Because it felt so good when he stopped.
From: Virgil on 18 Jul 2006 14:51 In article <1153224254.540565.150600(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > 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. > > Unknown word. > > > > > 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? > > To contribute a 1 to each column. > > > > 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? > > I told you recently: In a list like mine a *+ sum is defined. This * > sum is equal to the largest number of the list. Not unless there IS a largest number in the list. An in your list of terminating strings there is no such thing.
From: Virgil on 18 Jul 2006 14:59 In article <1153224623.508718.110530(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > 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. > > The axiom is not responsible for the well-order of *all* elements of > the set, because well-order is a stepwise process, you see? What I see is that well -ordering is a property of a set as a whole and not a property of tis members separately from the set. A well ordering of a set requires the existence of an order relation on the set ->as a whole<- with the special property that the induced order on every non-empty subset identifies a first element of that non-empty subset. Any total ordering of a finite set is automatically a well ordering, but not so for infinite sets.
From: Virgil on 18 Jul 2006 15:04 In article <1153225087.356970.296750(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1153145345.281803.129520(a)35g2000cwc.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Franziska Neugebauer schrieb: > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > ... > > > > > 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. > > > 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? > > > > Yes, right. And now the remainder of the proof, please? > > If the *+ sum of the list is a unary representation of some 1's, then, > by the construction of the unary numbers of the list, this sum must be > a unary numer of the list. This is so whether you pretend it would not > be so or not. For every finite list of finite strings of WM's, the so called "*+ sum" is the last member, so that "*+" is equivalent to "last of". But what if there is no "last of"? WM still says there still is one even when there isn't one.
From: Dik T. Winter on 18 Jul 2006 20:04
In article <1153242093.641744.4400(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Sorry, we are now talking about 0.999..., please remain with the argument. > > You said that the definition is silly. Why is the definition above silly? You again refrain from answering questions. The definition of 0.999... is as follows: 0.999... = lim{n -> oo} sum{k = 1 .. n} 10^(-k) You said that definition is silly. What is silly about it? > > And that there are undefined digits in irrational numbers does not bother > > me in the least. I have no idea why you have a problem with that. > > In this case not the digits are undefined but the indices are > undefined. See the discussion of the *+ sum of my list of unary > numbers. what indices in the limit above are undefined? > > But to go on: > > > > > And therefore we have *always* undefined digits in any > > > irrational number and in the diagonal of Cantor's list. There are > > > *always* most of its digits unknown, i.e., always there are more digits > > > unknown than are known. No matter how small an epsilon you select. > > > > For Cantor's diagonal no epsilon is needed, neither is their for irrational > > numbers. The only thing we need to know is that the digits are all in the > > range [0, 9], > > The only thing we need to know for the binary tree is that no path can > split without two additional edges. Why don't you use this kind of > abstract mathematics in that case too? Because, and I state it again, there is a difference between sequential processes and simultaneous processes. Four your count of edges and paths you need limits, but those limits do not exist. > > > Hence you cannot prove that the digits of the diagonal are all > > > different from those of the line numbers. > > > > Why not? For the proof no exhaustion is needed. Just like (in another > > thread) the proposition > > For the binary tree no exhaustion is requird either. It suffices to see > that > > | > o > /\ > > is the ever repeating element which cannot be outwitted. Again, you need something like a limit here, because adding nodes is a sequential process. > > For all p there is an n such that An[p] = K[p] > > does not need exhaustion. In the case of the list and the diagonal the > > similar statement is > > For all p there is an n such that An[p] != D[p] > > proving that D is different from all An. BTW, in the first case also the > > proposition > > For all p there is an n such that An[p] != K[p] > > proving that K is also different from all An. > > In the first and second case take n = p, in the third case take n = p + 1. > > This is *not* a proof by exhaustion. It is simply stating a fact, with > > an easy proof. > > > > > You can prove it for each one > > > but you cannot prove it for all. > > > > If something is true for each one, it is true for all. > > It is true for each natural, that it cannot index all 1's of 0.111... . Also, it is true for each natural that it cannot index all 1's of 0.1111. That means that for all integers it is impossible to index all 1's of 0.1111. On the other hand, with the *set* if integers it is possible to index all 1's of 0.1111, and also of 0.111... . That is, when with "to index" we mean "point to an individual element" as in K[p]. The reason is that each natural can index one and only one digit. On the other hand, you on occasion use index to mean something completely different. For that I could use the word "cover" where A covers B if in all positions where B has a 1, A also has a 1. Now in the finite case we find that the digits of 0.1111 are covered by all naturals greater than or equal to 4. But the digits of 0.111... are not covered with any natural. But nobody has stated that that is possible, it is only you who *claims* that that should be possible, without giving any reasonable proof of it. > > Let's reason > > with the excluded middle (because that would complicate matters). Assume > > it is not true for all, then there must be some for which it is not > > true (say z). On the other hand, it is true for each one, so also > > true for z. A contradiction, hence the assumption is false. And in > > the cases above you can proof it for each one in one sweep, because > > you prove it for an arbitrary element. > > It is true for each natural, that it cannot index all 1's of 0.111... . Again, you mean cover here. I would like to ask you to refrain from such confusing use of words. And that is indeed true. But that does *not* mean that For all p there is an n such that An[p] = K[p] is false. That is true, take n = p. > > If stepwise exhaustion is impossible, your reordering could be completed. > > But you can with your *+ operation give the proof in one sweep, but you > > failed to answer to my question: "what happens at infinity?". > > 1) There is o infinity. Ever Transposition has a natural number. Sorry, that was loosely speaking. What happens if you add *all* of them together? You still fail to tell what happens in that case. (It is quite easy to tell it, and you need a limit, but you are unwilling, or perhaps unable to tell it. When I ask you what the meaning of *+ ... is you do not give a reasonable answer.) Answer: for each digit position, when there is an Ap with a one in that digit position, the identical digit position in K is also a 1. And that answers the question. The result is: for all p K[p] = 1 which we notate as K = 0.111..., because: for all p there is an n such that An[p] = 1. > 2) The same happens as with Cantor's diagonal. There is absolutely no > difference. And indeed, if I write it like I did above, there is absolutely no difference. K is properly defined, and it is easily shown that it is 0.111..., and in the same way the diagonal is properly defined. > > You can exhaust an infinite set if you take out all elements at once. > > You may think that would be possible, but it is not. You always need > the belief expressen by the three "..." I do not believe them, unless there is a proper definition or common usage (that can be given a proper definition). > > That is why the function f(n) = n + 1 |