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: Tony Orlow on 12 Apr 2007 15:25 MoeBlee wrote: > On Mar 31, 5:18 am, Tony Orlow <t...(a)lightlink.com> wrote: >> Virgil wrote: >>> In article <1175275431.897052.225...(a)y80g2000hsf.googlegroups.com>, >>> "MoeBlee" <jazzm...(a)hotmail.com> wrote: >>>> On Mar 30, 9:39 am, Tony Orlow <t...(a)lightlink.com> wrote: >>>>> They >>>>> introduce the von Neumann ordinals defined solely by set inclusion, >>>> By membership, not inclusion. >>> By both. Every vN natural is simultaneously a member of and subset of >>> all succeeding naturals. >> Yes, you're both right. Each of the vN ordinals includes as a subset >> each previous ordinal, and is a member of the set of all ordinals. > > In the more usual theories, there is no set of all ordinals. > Right. Ordinals are...ordered. Sets aren't. >> In >> this sense, they are defined solely by the "element of" operator, or as >> MoeBlee puts it, "membership". Members are included in the set. Or, >> shall we call it a "club"? :) >> >> Anyway, my point is that the recursive nature of the definition of the >> "set" > > What recursive definition of what set? > Oh c'mon! N. ala Peano? (sigh) What kind of question is that? >> introduces a notion of order which is not present in the mere idea >> of membership. > >> Order is defined by x<y ^ y<z -> x<z. > > Transitivity is one of the properties of most of the orderings we're > talking about. But transitivity is not the only property that defines > such things as 'partial order', 'linear order', 'well order'. > It defines order, in general. >> This is generally >> interpreted as pertaining to real numbers or some subset thereof, but if >> you interpret '<' as "subset of", then the same rule holds. > > Yes, the subset relation on any set is a transitive relation. > Yes. It's related. >> I suppose >> this is one reason why I think a proper subset should ALWAYS be >> considered a lesser set than its proper superset. It's less than the >> superset by the very mechanics of what "less than" means. > > A proper subset is less than a proper superset of it, in the sense > that the proper superset has all members of the proper subset plus at > least one more. Yes. It's true. It is not always the case though that a set is not 1-1 > with some proper subset of itself. In the finite, the two aspects > coincide, but not in the infinite. That's just the way it is in the > more usual set theories. That does not stop you from formulating a > different theory though. > > MoeBlee > Right. There can always be a 1-1 correspondence defined between a set with no end and its proper subset with no end, even if that correspondence is so complicated so as to defy all attempts to define it. Perhaps that is the motivation behind Choice. But, where that violates the "lesserness" of the proper subset, it cannot be considered a comprehensive comparison of "size". Another level of analysis needs to be applied to get "measurable" results. ToeKnee
From: Tony Orlow on 12 Apr 2007 15:27 MoeBlee wrote: > On Mar 31, 5:39 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> In order to support the notion of aleph_0, one has to discard the basic >> notion of subtraction in the infinite case. That seems like an undue >> sacrifice to me, for the sake of nonsense. Sorry. > > For the sake of a formal axiomatization of the theorems of ordinary > mathematics in analysis, algebra, topology, etc. > > But please do let us know when you have such a formal axiomatization > but one that does have cardinal subtraction working in the infinite > case just as it works in the finite case. > > MoeBlee > Sorry, MoeBlee, but when I produce any final product in this area, cardinality will be a footnote, and not central to the theory. As I work on other things, so do I work on this. Take care. ToeKnee
From: Tony Orlow on 12 Apr 2007 15:28 Virgil wrote: > In article <4611182b(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > >> It's true that the set of consecutive naturals starting at 1 with size x >> has largest element x. > > Not unless x is less than or equal to some natural. Prove it, without assuming it. > >> Is N of that form? > > It is if x >= aleph_0. Then aleph_0 e N. Sorry. Tony
From: Tony Orlow on 12 Apr 2007 15:30 Lester Zick wrote: > On Mon, 2 Apr 2007 16:12:46 +0000 (UTC), stephen(a)nomail.com wrote: > >>> 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. > > Is that true? > > ~v~~ Yes, Lester, Stephen is exactly right. I am very happy to see this response. It follows from the assumptions. Axioms have merit, but deserve periodic review. 01oo
From: Tony Orlow on 12 Apr 2007 15:34
Virgil wrote: > In article <461e7764(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Lester Zick wrote: > >>> And what is your definition of "infinite"? >>> >>> ~v~~ >> "greater than any finite" >> > > And is TO's definition of finite "less than infinite"? Hi Virgil. Sweet greetings. Finite is properly defined by bijection. A finite set cannot be bijected with a subset, and a finite real lies between two finite naturals, each the size of a finite set. (no signature) |