Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 13 Dec 2006 10:59 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Unbounded is potentially infinite but it is not necessarily actually > > infinite. > > > > Please give an example of a set you consider to be > actually infinite. > Impossile. There is no actual infinty as is shown by our discusson abou the IET. So I had to lie if being forced to name an actually infinite set. Regards, WM
From: mueckenh on 13 Dec 2006 11:04 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > A_1 = {1} > > > > > A_2 = {1,2} > > > > > A_3 = {1,2,3} > > > > > > > > > > B = {1,2,3} > > > > > > > > > > then B is contained in the last A_i. If there is no last A_I, then > > > > > there is > > > > > no A_i that contains B > > > > > > > > That has nothing to do with "last". > > > > > > If A_i contains B, then A_i contains any A_j. > > > Therefore A_i is "last". > > > > > No comment? If B contains any A_i then B is the last. Regards, WM
From: mueckenh on 13 Dec 2006 11:07 Franziska Neugebauer schrieb: > >> Mind and brain and representation of (abstract) entities therein is > >> still off topic in sci.math. > > > > This decision is > > 1) wrong > > 2) not yours. > > Which _mathematical_ institution is doing research in the field of > "representation of (abstract) entities in mind or brain"? New aspects are new because they have not been treated yet. Regards, WM
From: William Hughes on 13 Dec 2006 11:36 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > A_1 = {1} > > > > > > A_2 = {1,2} > > > > > > A_3 = {1,2,3} > > > > > > > > > > > > B = {1,2,3} > > > > > > > > > > > > then B is contained in the last A_i. If there is no last A_I, then > > > > > > there is > > > > > > no A_i that contains B > > > > > > > > > > That has nothing to do with "last". > > > > > > > > If A_i contains B, then A_i contains any A_j. > > > > Therefore A_i is "last". > > > > > > > > No comment? > > If B contains any A_i then B is the last. B cannot be the last A_i, as B is not one of the A's (B corresponds to the diagnonal, the A_i correspond to the lines) - William Hughes > > Regards, WM
From: cbrown on 13 Dec 2006 13:14
mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1165923504.410525.226600(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > So you can not show a bijection (in your opinion), but nevertheless you > > > > state that you have given a surjection. Do you not think you are > > > > contradicting yourself a bit? > > > > > > I give a surjection on all existing paths. > > > > But not from the edges. But from parts of edges. > > > > > > > No? Even if there are enough parts to gather more than a whole edge to > > > > > be mapped on every path? > > > > > > > > No. Each part of an edge may map to a single path, that does *not* give > > > > a map from each edge to a single path. > > > > > > Since infinity = infinity became an allowed equation, 1/2 + 1/2 = 1 can > > > no longer be true? > > > > So if the first half of ab edge maps to, say, 1/2 and the other half to, > > say, 3/4. To what single path maps the edge itself? > > Why isn't it sufficient to collect the shares of two edegs for every > path? That shows that there are not more paths than edges. Can you > explain your objection to factions? > > I previously asked you to "show" how you perform this mapping, given that we agree to the following: (1) Edges are countable. (2) Paths can be bijected with the reals. (3) B(e) is the (uncountable) set of all paths to which edge e belongs. (4) Shares (of edges in the original diagram) are denoted (p,e) for some path p and some edge e where e belongs to p. (5) S(e) is the (uncountable) set of all shares of edge e. (6) C(p) is the (countable) set of all shares which belong to path p. (7) 1 + 1/2 + 1/4 + ... +1/2^n + ... = 2 Your claim (which you seem unable to completely articulate) seems to be that there are two "full edges" whose "shares" are exactly the shares of edges composing a particular path p. But a path p is composed of a countable number of shares; so a "full edge" can have at most a countable number of shares. Meanwhile, an edge in the original diagram consists of an uncountable number of shares. So full edges can only be said to be "the same as" edges in the original diagram if you /assume/ that countable is the same as uncountable. But that is what you are trying to prove - you can't simply assume it in a proof of it. Cheers - Chas |