An uncountable countable set
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From: David R Tribble on 9 Oct 2006 14:32 Virgil wrote: >> The property of not being an infinite natural holds for the first >> natural, and holds for the successor of each non-infinite natural, so >> that it must hold for ALL naturals. > Tony Orlow wrote: > It holds for all finite naturals, .... because all naturals are finite ... > but if there are an infinite number of > naturals generating using increment, .... because the increment operation is "applied" an infinite number of times, "generating" an infinitude of finite naturals ... > then there are naturals which are > the result of infinite increments, which must have infinite value. Where's your proof? What is an "infinite increment" (or "infinite successor")?
From: David R Tribble on 9 Oct 2006 14:33 Tony Orlow wrote: > No, the inductive proof of an equality applies to all n, finite or > infinite. But "is finite" is an inequality, equivalent to "<oo". > lim(n->oo: n)=oo, not <oo. You can only increment a finite value a > finite number of times before you get infinite values out of it. How many times?
From: Ross A. Finlayson on 9 Oct 2006 15:00 David Marcus wrote: > Ross A. Finlayson wrote: > > David Marcus wrote: > > > I thought you said there was a contradiction in ZF. In the context of > > > ZF, the Burali-Forti argument shows that there is no set of all > > > ordinals, but does not lead to a contradiction. So, do you still say > > > there is a contradiction in ZF? If so, what is it? > > > > {For any x: x is a set} = emptyset <=/=> {For any x: x is a set} = U > > (V, L) > > > > That says, for any x, that's the empty set, and, for any x, that's the > > universal set, it seems sufficient to show the universe non-empty. > > > > There is no set of ordinals nor cardinals in ZF. Yet, because there's > > the axiom of infinity, those infinite ordinals/cardinals as there ever > > would be are claimed to exist, basically where they're all hereditarily > > finite, those ordinals of the cumulative hierarchy. > > Sorry, but I don't follow. Are you saying this is a contradiction within > ZF? By "within" I mean that ZF proves this contradiction. > > -- > David Marcus Hi David, You seem like you know what you're talking about, which is good. Basically, yes, I say the existence of the (universal) quantifier, where the word universal is in parentheses because there's the mutually implicit existential quantifier, that the existence of the universal quantifier in ZF lead to illustration of a contradiction derivable from ZF. I have some other arguments along those lines as well, of similar tack, where I advocate axiom-free natural deduction, as a return of sorts to a more "naive" set theory with post-Cantorian acknowledgement. That is to say, there are thousands of pages more of my opinion about these matters readily available. Ross
From: Tony Orlow on 9 Oct 2006 15:36 Ross A. Finlayson wrote: > David Marcus wrote: >> Ross A. Finlayson wrote: >>> David Marcus wrote: >>>> I thought you said there was a contradiction in ZF. In the context of >>>> ZF, the Burali-Forti argument shows that there is no set of all >>>> ordinals, but does not lead to a contradiction. So, do you still say >>>> there is a contradiction in ZF? If so, what is it? >>> {For any x: x is a set} = emptyset <=/=> {For any x: x is a set} = U >>> (V, L) >>> >>> That says, for any x, that's the empty set, and, for any x, that's the >>> universal set, it seems sufficient to show the universe non-empty. >>> >>> There is no set of ordinals nor cardinals in ZF. Yet, because there's >>> the axiom of infinity, those infinite ordinals/cardinals as there ever >>> would be are claimed to exist, basically where they're all hereditarily >>> finite, those ordinals of the cumulative hierarchy. >> Sorry, but I don't follow. Are you saying this is a contradiction within >> ZF? By "within" I mean that ZF proves this contradiction. >> >> -- >> David Marcus > > Hi David, > > You seem like you know what you're talking about, which is good. > > Basically, yes, I say the existence of the (universal) quantifier, > where the word universal is in parentheses because there's the mutually > implicit existential quantifier, that the existence of the universal > quantifier in ZF lead to illustration of a contradiction derivable from > ZF. Hi Ross - By "mutually implicit", do you refer to the fact that "for all p, x is true" is equivalent to "there does not exist p, such that x is false"? Like "and" and "or", we can get away with only one of them, in conjunction with "not". Is that what you mean? > > I have some other arguments along those lines as well, of similar tack, > where I advocate axiom-free natural deduction, as a return of sorts to > a more "naive" set theory with post-Cantorian acknowledgement. > > That is to say, there are thousands of pages more of my opinion about > these matters readily available. > > Ross > I'll take three copies. :)
From: Tony Orlow on 9 Oct 2006 15:37
David R Tribble wrote: > Tony Orlow wrote: >> No, the inductive proof of an equality applies to all n, finite or >> infinite. But "is finite" is an inequality, equivalent to "<oo". >> lim(n->oo: n)=oo, not <oo. You can only increment a finite value a >> finite number of times before you get infinite values out of it. > > How many times? > Less than any infinite number of times. |