An uncountable countable set
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From: MoeBlee on 18 Oct 2006 13:32 Tony Orlow wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> MoeBlee wrote: > >>> P.S. > >>> > >>> I lost the context, but somewhere you (Orlow) posted: > >>> > >>> "ZF and NBG don't handle sequences or their sums, but only unordered > >>> sets" > >>> > >>> Z set theory defines and proves theorems about ordered tuples, finite > >>> and infinite sequences, and infinite summations and infinite products > >>> and many other things like that. > >>> > >>> MoeBlee > >>> > >> Huh! But I thought sets were unordered. > > > > Without the axiom of choice, we prove the existence of certain > > orderings on certain sets. With the axiom of choice, we prove that > > every set has a well ordering. > > > >> If the theory of infinite series > >> is derived from set theory, > > > > I don't say that historically it is derived from set theory. But > > infinite series, such as one finds in ordinary calculus and real > > analysis are definable in set theory and the theorems about them can be > > proven from the axioms of set theory. > > Theorems like, an infinite series with alternating positive and negative > terms can be validly rearranged to have a sum as large or small as one > likes? :) Sorry, I don't buy it. That sounds to me like ambiguity stemming from informal notation. If a sequence is properly defined, then it either converges to a unique value or it does not converge. > > A set theoretic explication of an infinitary thought experiment (which > > is not an explication that occurs IN set theory) differs from your own > > explication of that thought experiment. That is not a contradiction > > between set theory and any theorem regarding infinite series. > > Depends which assumptions you assume regarding such series. No, there is no set theoretic contradiction that has been shown. > If you want to ask which axioms I don't like that lead to my disagreeing > with your conclusion, What conclusion? > then state it as derived from the axioms, and I'll > tell you. State what as derived from the axioms? > It's the Zeno machine that's the problem. You don't have an > axiom for that, really, do you? Where did I ever mention a Zeno machine? MoeBlee
From: MoeBlee on 18 Oct 2006 13:46 Tony Orlow wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> MoeBlee wrote: > >>> Tony Orlow wrote: > >>>> No, set theory confuses the issue with its concentration on omega. > >>> Oh boy, here we go again with "Set theory confuses...". Please just way > >>> which axioms of set theory you reject and which ones you use instead. > >>> > >>> MoeBlee > >>> > >> Uh, what axioms of set theory are specifically involved in your "proof"? > >> I don't remember a deduction from those axioms. Perhaps you could > >> refresh my memory. > > > > My proof of what? > > > > MoeBlee > > > > Regarding the vase-ball problem. I gave no proof regarding the balls in a vase problem. MoeBlee
From: MoeBlee on 18 Oct 2006 14:02 Tony Orlow wrote: > Also, upon which axioms is the definition of cardinality based? The usual definition is: card(x) = the least ordinal equinumerous with x The definition ultimately reverts to the 1-place predicate symbol 'e' (and the 1-place predicate symbol '=', if equality is taken as primitive). For the definition to "work out" ('work out' is informal here) in Z set theory, we usually suppose the axioims of Z set theory plus the axiom of schema of replacement (thus we're in ZF) and the axiom of choice (thus we're in ZFC). However, there is a way to avoid the axiom of choice by using the axiom of regularity instead with a somewhat different definition from just 'least ordinal equinumerous with'. Also, we could adopt a "midpoint" between the axiom schema of replacement and the axiom of choice by adopting the numeration theorem (AxEy y is an ordinal equinumerous with x) instead, which would be a method stronger than adopting the axiom of choice, but weaker than adopting both the axiom of choice and the axiom schema of replacement. As to the more basic axioms of Z, for the definition to "work out", I'm pretty sure we need extensionality, schema of separation (or schema of replacement if we go that way), union, and pairing (pairing is not needed if we have the schema of replacement). I'm not 100% sure, but my strong guess is that we don't need the power set axiom for this purpose. And we don't need the axiom of infinity. Why don't you just a set theory textbook? MoeBlee
From: MoeBlee on 18 Oct 2006 14:46 Han de Bruijn wrote: > But why are the finite ordinals not equivalent with the naturals (I mean > in mainstream mathematics)? The set of finite ordinals IS the set of natural numbers. x is a natural number <-> x is a finite ordinal. Why don't you just read a textbook? MoeBlee
From: Lester Zick on 18 Oct 2006 15:44
On 18 Oct 2006 11:46:47 -0700, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: >Han de Bruijn wrote: >> But why are the finite ordinals not equivalent with the naturals (I mean >> in mainstream mathematics)? > >The set of finite ordinals IS the set of natural numbers. Natural numbers are ordinals? >x is a natural number <-> x is a finite ordinal. > >Why don't you just read a textbook? I've yet to see textbooks call naturals ordinals. They're cardinals not ordinals. Totally different concepts. If textbooks do call naturals ordinals then that's a pretty good reason to ignore textbook definitions. Just because 1 is first, 2 second, . . .etc. is no reason to say 1, 2, . . . etc. are ordinals. First, second, . . . etc. are ordinals. 1, 2, . . . etc. are cardinals. ~v~~ |