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From: cbrown on 29 Dec 2006 21:15 Dik T. Winter wrote: > In article <1167393772.865867.58640(a)73g2000cwn.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > A path is an (ordered) set. If there is no element distinguishing it > > from every other set, then it is not different from every other set. > > Each path can be distinguished from each other path by some edge. There > is *no* edge that distinguishes a particular path from all other paths. > Back to that again. Quantifier dyslexia. > > Similar to: each natural number written in decimal can be distinguished > from each other natural number by some (actually in most cases many) > decimal digit. For each natural number there is *no* decimal digit that > distinguishes it from all other decimal digits. Even more trivially: let A = {a,b}, B = {b,c} and C = {c,a}. For each set, there is no element that distingushes it from the other sets; yet for each set can be distinguished from the others. Cheers - Chas
From: mueckenh on 30 Dec 2006 10:23 cbrown(a)cbrownsystems.com schrieb: > Have you given up on your "rational relation" proof that |N| = |R|? No, why should I? It is correct. (A series with a first but no last term can be reversed to have a last but no first term - without oosing its value.) But it is easier to see, and should be visible even for such as you, that the tree built from the union of all finite trees (with representations of rational paths) is the same as the complete infinite tree. Regards, WM
From: mueckenh on 30 Dec 2006 10:28 cbrown(a)cbrownsystems.com schrieb: > Dik T. Winter wrote: > > In article <1167393772.865867.58640(a)73g2000cwn.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > A path is an (ordered) set. If there is no element distinguishing it > > > from every other set, then it is not different from every other set. > > > > Each path can be distinguished from each other path by some edge. There > > is *no* edge that distinguishes a particular path from all other paths. > > Back to that again. Quantifier dyslexia. > > > > Similar to: each natural number written in decimal can be distinguished > > from each other natural number by some (actually in most cases many) > > decimal digit. For each natural number there is *no* decimal digit that > > distinguishes it from all other decimal digits. > > Even more trivially: let A = {a,b}, B = {b,c} and C = {c,a}. For each > set, there is no element that distingushes it from the other sets; yet > for each set can be distinguished from the others. If you want to talk to the topic presently discussed, then you should recognize the elementary constraints: Paths are linearly ordered sets. Your example misses the clue. Regards, WM
From: Bob Kolker on 30 Dec 2006 10:33 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > >> Have you given up on your "rational relation" proof that |N| = |R|? > > > No, why should I? It is correct. (A series with a first but no last > term can be reversed to have a last but no first term - without oosing > its value.) No it isn't. The Reals have a larger cardinality than the integers. Bob Kolker
From: mueckenh on 30 Dec 2006 10:39
Dik T. Winter schrieb: > In article <1167392633.967611.252860(a)n51g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > Apparently your limits work only for real numbers in lists but no for > > > > real numbers in trees. > > > > > > Oh, they do. If the trees are properly defined. > > > > Imagine a tree which contains only paths of rational numbers in > > infinite representation, i.e., ending with a period of 000... or > > 111.... > > > > The first rational tree contains only the paths 0.000... and 0.111... > > > > 0. > > / \ > > 0 1 > > / \ > > 0 1 > > ............. > > > > The second tree contains the paths 0.000..., 0.0111..., 0.1000..., and > > 0.111... > > > > and so on until all rational numbers are represented. > > At what point is 1/3 represented? In the binary tree it is not represented, but it could be represented by using a ternary tree. But this is the same with decimal numbers. There is no representation of 1/3, unless it were in the infinite. But if you insist on an existing decimal representation of 1/3 in the infinite, then also the union of all finite binary trees contains such an existing binary representation of 1/3. > > > Take the union of all these rational trees. It contains all paths > > representing rational numbers. > > No. It contains all paths representing rational numbers for which the > denominator is a power of two. Every decimal number gives a power of 10. > > > What distinguishes this union of rational trees from the complete tree? > > That some rational numbers are missing. If the union contains all > rational numbers, there must be one of the rational trees that contains > (for instance) 1/3. There is none. My question is only the following: Do you think that the union of all rational binary trees is the complete infinite binary tree or not? > > > There is no node and no edge of the complete tree which is not > > in the union of all rational trees. As a path is an ordered set of > > edegs (and nodes), two paths are not different unless they differ by an > > edge (or node). Hence, there is no path in the complete tree which is > > not in the union of all rational trees. > > Where is 1/3? At what point does it get into the union of rational > trees? > My question is only the following: Do you think that the union of all rational binary trees is the complete infinite binary tree or not? > > That is the reason why the cardinal number of the paths in both trees > > cannot be different. > > That is the reason why the cardinal number of the paths in both trees > *is* different. > With a different number and configuration of nodes (or edges) or with exactly the same? > > But if we subtract the rational paths from the complete tree, then > > nothing remains but the empty set which obviously is the set of all > > paths representig irrational numbers. > > > > This shows that irrational numbers are nothing (but a chimera). > > And apparently a lot of rational numbers are nothing? A lot of binary representations, yes. But every rational number has some representation - in contrast to irrational numbers. Regards, WM |