Cantor Confusion
in [Math]
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From: WM on 11 May 2007 04:07 On 10 Mai, 22:25, Virgil <vir...(a)comcast.net> wrote: > > 1) for every path p', there is a node where p' leaves p > > and > > 2) there is no node where every path p' has left p > > then you must be able to name a node K(n) such that for nodes K(m < > > n) > > at least two paths p' and p'' are required to accompany p. > > This is true of EVERY path and for each of its nodes in an infinite tree. > > For every node K(n) of P, there is one path p' which accompanies it to > the next node and then branches off, and another, p'', which does not > branch off until the node after that, and so on, ad infinitum. And the Moon is round and January is the first month in the year. So what? I did expect that. And what you said is true. But it is not an argument. Ad infinitum there is a path p' with p, because ad infinitum there is no node where p is single. Therefore, ad infinitum, there is a path which shares with p the node in question and all earlier nodes too. Regards, WM
From: WM on 11 May 2007 04:21 On 10 Mai, 22:25, Virgil <vir...(a)comcast.net> wrote: > > 1) for every path p', there is a node where p' leaves p > > and > > 2) there is no node where every path p' has left p > > then you must be able to name a node K(n) such that for nodes K(m < > > n) > > at least two paths p' and p'' are required to accompany p. > > This is true of EVERY path and for each of its nodes in an infinite tree. > > For every node K(n) of P, there is one path p' which accompanies it to > the next node and then branches off, and another, p'', which does not > branch off until the node after that, and so on, ad infinitum. And the Moon is round and January is the first month in the year. So what? I did expect that. And what you said is true. But it is not an argument. Ad infinitum there is a path p' with p, because ad infinitum there is no node where p is single. Therefore, ad infinitum, there is a path which shares with p the node in question and all earlier nodes too. Regards, WM
From: WM on 11 May 2007 04:37 On 10 Mai, 21:30, Virgil <vir...(a)comcast.net> wrote: > In article <1178796129.155631.279...(a)u30g2000hsc.googlegroups.com>, > > > > Yes. But why only see the one side of the medal? For every path p > > > > there is another path *existing in the tree* which is not different > > > > from p at any node, because p is not single at any node. > > What the above says is that for every path p there is a path p' in the > same tree which is the same as p at every node but somehow not as a > whole the same as p. > > But that is false in ZF or NBG trees. > > Whatever have I said that WM claims would be wrong The line preceding your question is wrong. Also in ZF etc. there is no node which is occupied by path p alone. Hence a path p' is with path p at the node in question and at every earlier node. There are not more nodes than we can ask for. Hint: There need not be a last one in order to ask for all, as we know from Cantor's diagonal argument. Regards, WM
From: Carsten Schultz on 11 May 2007 04:44 WM schrieb: > On 10 Mai, 22:10, Virgil <vir...(a)comcast.net> wrote: >> In article <1178817355.571399.114...(a)y80g2000hsf.googlegroups.com>, > >>> A binary relation is, therefore, determined by giving all ordered >>> pairs of objects in that relation; it dos not matter by what property >>> the set of these ordered pairs is described. >>> !!!!!!!!!!!! But it obviously does matter *that* it is >>> described. !!!!!!!!!!!!!!! >> That depends on what one wants to do with the set. > > H&J could have written: "it does not matter that the set of these > ordered pairs is described by a property" in your opinion? Actually, they might have wanted to say that two relations are the same if their ordered pairs are the same, even if the properties may have seemed quite different in the beginning. But all of this does not matter. How do they /define/ a relation? You seem to have omitted the definition. >> In dealing with sets >> of ordered pairs in general, one need not describe it as anything but a >> set of ordered pairs. > > Did I specify the rule, process, mapping, in short: the formula? When > dealing with the general notion of function, i.e., when not dealing > with a special function you need not specify the formula. But without > a formula or formal description determining by what property the set > of ordered pairs is described, there is no set of ordered pairs, i.e., > *there is no function*. To prove that a function with certain properties exists you need not specify a formula. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: WM on 11 May 2007 04:48
On 10 Mai, 21:36, Virgil <vir...(a)comcast.net> wrote: > In article <1178806498.433405.226...(a)o5g2000hsb.googlegroups.com>, > > > > The axioms of ZF tell how to specify sets in ZF. > > > Yes, how to specify, but they do not specify sets except few. > > And having those few, with intersections, differences, etc., one can > build many without a single "rule" f the sort WM implies are necessary. Just the description of how you apply these axioms to "build" a set, precisely _that is the rule_ or formula. Show me the forest. There are too may trees. I can't see it. > > A set can be specified by specifying its elements. This can be done > > very well, in particular for finite sets, by a list. Yes, the rule or > > formula can be a list listing the elements of a set. > > > > > Do you know what a relation is? Do you know that there is also the > > > > domain and the range required? > > > > There are a domain and range to a relation, but they are consequences of > > > its definition, not antecedents to it. > > > Answer only by yes or no > > The questions were irrelevant to the issue of whether a function is, > first of all, a set of ordered pairs. The function is a formula (see above) together with two sets. Can't you see that these sets are determined by the sets of the x's and the y's in the definition by H&J (and any other meaningful definition of a functionr relation too)? 2.1 Definition A set R is a binary relation if all elements of R are ordered pairs, i.e., if for any z ï R there exist x and y such that z = (x, y). Regards, WM |