The Definition of Points
in [Math]
|
Prev: Guide to presenting Lemma, Theorems and Definitions
Next: Density of the set of all zeroes of a function with givenproperties
From: Mike Kelly on 13 Apr 2007 07:32 On 12 Apr, 20:17, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 2 Apr, 17:12, step...(a)nomail.com wrote: > >> In sci.math Mike Kelly <mikekell...(a)googlemail.com> wrote: > > >>> On 2 Apr, 15:49, Tony Orlow <t...(a)lightlink.com> wrote: > >>>> Mike Kelly wrote: > >>>>> That isn't a fact. It's true that the size of a set of naturals of the > >>>>> form {1,2,3,...,n} is n. But N isn't a set of that form. Is it? > >>>> It's true that the set of consecutive naturals starting at 1 with size x has largest element x. > >>> No. This is not true if the set is not finite (if it does not have a > >>> largest element). > >>> It is true that the set of consecutive naturals starting at 1 with > >>> largest element x has cardinality x. > >>> It is not true that the set of consecutive naturals starting at 1 with > >>> cardinality x has largest element x. A set of consecutive naturals > >>> starting at 1 need not have a largest element at all. > >> To be fair to Tony, he said "size", not "cardinality". If Tony wishes to define > >> "size" such that set of consecutive naturals starting at 1 with size x has a > >> largest element x, he can, but an immediate consequence of that definition > >> is that N does not have a size. > > >> Stephen > > > Well, yes. But Tony wants to use this line of reasoning to then say > > "and, therefore, if N has size aleph_0 then aleph_0 is the largest > > element, which is clearly bunk". This is where his claims that > > "aleph_0 is a phantom" come from. But, obviously, this line of > > reasoning doesn't apply to all notions of "size". > > mike, what I am saying is that, assuming inductive proof works for all > equalities, this equality holds: |N|=max(N). I guess inductive proof doesn't work for all equalities then, because it's obvious to my inerrant intuition that |N| != max(N) because max(N) doesn't exist. Thankyou for a clear demonstration of why anyone with even vaguely sensible intuitions would reject the idea that "inductive proof works for all equalities". > If this is the case, then > aleph_0 cannot be infinite, while also being a member of N, and > therefore is self-contradictory, and cannot exist any more than a > largest finite. If |N|=max(N) is the case then "|N|" does not denote cardinality, so what you are saying has nothing to do with aleph_0. -- mike.
From: Bob Kolker on 13 Apr 2007 10:06 Mike Kelly wrote: > > a) A consecutive set of naturals starting with 1 with size X can not > have any maximum other than X. > b) A consecutive set of naturals starting with 1 with size X has > maximum X. Whoa! If a -consectutive-(!!!) set of naturals with 1 as its least element has cardinality X (an integer), then its last element must be X. A simple induction argument will show this to be the case. Can you show a counter example? Bob Kolker
From: Bob Kolker on 13 Apr 2007 10:08 Bob Kolker wrote: > Mike Kelly wrote: > >> >> a) A consecutive set of naturals starting with 1 with size X can not >> have any maximum other than X. >> b) A consecutive set of naturals starting with 1 with size X has >> maximum X. > > > Whoa! If a -consectutive-(!!!) set of naturals with 1 as its least > element has cardinality X (an integer), then its last element must be X. > > A simple induction argument will show this to be the case. > > Can you show a counter example? I should have said, for I assumed it, that X is finte. Sorry about that. Bob Kolker
From: Mike Kelly on 13 Apr 2007 10:16 On 13 Apr, 15:06, Bob Kolker <nowh...(a)nowhere.com> wrote: > Mike Kelly wrote: > > > a) A consecutive set of naturals starting with 1 with size X can not > > have any maximum other than X. > > b) A consecutive set of naturals starting with 1 with size X has > > maximum X. > > Whoa! If a -consectutive-(!!!) set of naturals with 1 as its least > element has cardinality X (an integer), then its last element must be X. > > A simple induction argument will show this to be the case. > > Can you show a counter example? > > Bob Kolker N -- mike.
From: Mike Kelly on 13 Apr 2007 10:16
On 13 Apr, 15:08, Bob Kolker <nowh...(a)nowhere.com> wrote: > Bob Kolker wrote: > > Mike Kelly wrote: > > >> a) A consecutive set of naturals starting with 1 with size X can not > >> have any maximum other than X. > >> b) A consecutive set of naturals starting with 1 with size X has > >> maximum X. > > > Whoa! If a -consectutive-(!!!) set of naturals with 1 as its least > > element has cardinality X (an integer), then its last element must be X. > > > A simple induction argument will show this to be the case. > > > Can you show a counter example? > > I should have said, for I assumed it, that X is finte. Sorry about that. > > Bob Kolker Ah. Fair enough. -- mike. |