Cantor Confusion
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From: mueckenh on 23 Nov 2006 07:40 Virgil schrieb: > In article <1164196369.064190.253870(a)h48g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1164126395.211430.7520(a)h54g2000cwb.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Tell me which point is not accepted: > > > > 1) Every line which contains an index of the diagonal, contains all > > > > preceding indexes of the diagonal too. > > > > 2) Every index of the diagonal is in a line. > > > > 3) In order to show that there is no line containing all indexes of the > > > > diagonal, there must be found at least one index, which is in the > > > > diagonal but not in any line. > > > > > > This one is flat out false, unless one assumes, a priori, a last line > > > and a last member of the diagonal. > > > > We assume not a last line, but we assume that eery line has finitely > > many indexes. And this is true. > > What you allege in (3) does not follow from this. > It is true that given any line there will be a diagonal element not in > that line. > It is false that given a diagonal element there is no ine which contains > it. > > WM again dyslexes his quantifiers. > > > > > > It is certainly false in ZF or NBG, where such an assumption is also > > > false. > > > > For finite indexes it is correct. > > Infinite things are not constrained to behave in all respects like > finite ones. That is because "infinite" means "not finite". But *the lines are all finite*. That's just why I chose the EIT as example. > > > > > > It is quite enough to show that for every index in the diagonal, > > > except 1, there is some line not containing that index. Then no line > > > can contain every index. > > > > Name two finite indexes which cannot be in one line. > > That is not at all relevant to what I said. It is relevant since every line is finite. So we have to deal with finite indexes only. And every line has finitely many indexes. So we have to deal with finitely many indexes only, as far as the lines are concernded. (That's why you raised that silly argument with the diagonal loger than every line.) So your magic belief is easily destroyed. Regards, WM
From: mueckenh on 23 Nov 2006 07:40 Virgil schrieb: > In article <1164196369.064190.253870(a)h48g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1164126395.211430.7520(a)h54g2000cwb.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Tell me which point is not accepted: > > > > 1) Every line which contains an index of the diagonal, contains all > > > > preceding indexes of the diagonal too. > > > > 2) Every index of the diagonal is in a line. > > > > 3) In order to show that there is no line containing all indexes of the > > > > diagonal, there must be found at least one index, which is in the > > > > diagonal but not in any line. > > > > > > This one is flat out false, unless one assumes, a priori, a last line > > > and a last member of the diagonal. > > > > We assume not a last line, but we assume that eery line has finitely > > many indexes. And this is true. > > What you allege in (3) does not follow from this. > It is true that given any line there will be a diagonal element not in > that line. > It is false that given a diagonal element there is no ine which contains > it. > > WM again dyslexes his quantifiers. > > > > > > It is certainly false in ZF or NBG, where such an assumption is also > > > false. > > > > For finite indexes it is correct. > > Infinite things are not constrained to behave in all respects like > finite ones. That is because "infinite" means "not finite". But *the lines are all finite*. That's just why I chose the EIT as example. > > > > > > It is quite enough to show that for every index in the diagonal, > > > except 1, there is some line not containing that index. Then no line > > > can contain every index. > > > > Name two finite indexes which cannot be in one line. > > That is not at all relevant to what I said. It is relevant since every line is finite. So we have to deal with finite indexes only. And every line has finitely many indexes. So we have to deal with finitely many indexes only, as far as the lines are concernded. (That's why you raised that silly argument with the diagonal loger than every line.) So your magic belief is easily destroyed. Regards, WM
From: mueckenh on 23 Nov 2006 07:40 Virgil schrieb: > In article <1164196369.064190.253870(a)h48g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1164126395.211430.7520(a)h54g2000cwb.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Tell me which point is not accepted: > > > > 1) Every line which contains an index of the diagonal, contains all > > > > preceding indexes of the diagonal too. > > > > 2) Every index of the diagonal is in a line. > > > > 3) In order to show that there is no line containing all indexes of the > > > > diagonal, there must be found at least one index, which is in the > > > > diagonal but not in any line. > > > > > > This one is flat out false, unless one assumes, a priori, a last line > > > and a last member of the diagonal. > > > > We assume not a last line, but we assume that eery line has finitely > > many indexes. And this is true. > > What you allege in (3) does not follow from this. > It is true that given any line there will be a diagonal element not in > that line. > It is false that given a diagonal element there is no ine which contains > it. > > WM again dyslexes his quantifiers. > > > > > > It is certainly false in ZF or NBG, where such an assumption is also > > > false. > > > > For finite indexes it is correct. > > Infinite things are not constrained to behave in all respects like > finite ones. That is because "infinite" means "not finite". But *the lines are all finite*. That's just why I chose the EIT as example. > > > > > > It is quite enough to show that for every index in the diagonal, > > > except 1, there is some line not containing that index. Then no line > > > can contain every index. > > > > Name two finite indexes which cannot be in one line. > > That is not at all relevant to what I said. It is relevant since every line is finite. So we have to deal with finite indexes only. And every line has finitely many indexes. So we have to deal with finitely many indexes only, as far as the lines are concernded. (That's why you raised that silly argument with the diagonal loger than every line.) So your magic belief is easily destroyed. Regards, WM
From: Han de Bruijn on 23 Nov 2006 07:46 mueckenh(a)rz.fh-augsburg.de wrote: [ exposition of WM's binary tree snipped: just look it up ] > If you don't understand this simple and clear exposition, then there is > no hope that you will be able to think any further. Affirmative. It's not a difficult argument altogether. > I really don't know what further information could be required. Han de Bruijn
From: mueckenh on 23 Nov 2006 07:49
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > > If there is no bijection with one line, then there must be an element > > of the diagonal outside of every line. > > No. Not unless the one line > contains every element from every line. This holds for every finite line. Every finite line contains every element from every preceding line. There are only finite lines. > > > (Because what can be done with > > several *finite* lines in their linear order, can be done with one > > line). > There are only finite lines. > What can be done with all lines in their linear order > cannot be done with one line. There are only finite lines. Each has as many elements as to count its position. Everything is finite. Fine. > > No, you cannot use the fact that a bijection with one line > must exist to prove that the set of all lines doesn't exist, Of course I can. > then turn around and use the fact the the set of all lines > doesn't exist to prove that a bijection with one line must exist > Drop simply worshipping the "infinite". Regards, WM |