Give me ONE digit position of any real that isn't computable up to that digit position!
in [Logic]
|
Prev: THE PROBLEM WITH THE NATURAL PHILOSOPHY ALLIANCE
Next: No, dumbass, nobody is going to give you "one digit position of any real
From: Transfer Principle on 10 Jun 2010 16:43 On Jun 10, 11:33 am, Leland McInnes <leland.mcin...(a)gmail.com> wrote: > On Jun 9, 7:50 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > I want to see a peaceful discussion in which the OP > > discusses why they work in the alternate theory, while > > the others post why they prefer ZFC to the theory, and > > so on. > Just in case I'll get the ball rolling on a discussion of synthetic > differential geometry/smooth infinitesimal analysis as compared to > classical calculus. The prime difference here is that SDG supposes a > non-punctiform continuum -- you can't pick out individual points from > it -- which squares nicely with the intuitive notion of a continuum as > an indivisible whole that it ultimately incapable of being described > in terms of discrete/distinguishable points. This is, of course, > incompatible with classical calculus and the classical real line which > is nothing but distinguishable points. Now, to have such a notion of a > continuum, you need to make some sacrifices, like, for instance, > forgoing the law of excluded middle. On the upside you can develop a > purely synthetic geometric calculus in which standard results fall out > naturally as simple algebra involving infinitesimals. Indeed, you can > build up a very natural differential geometry in which, for example, > thinking about a vector field on a manifold as infinitesimal flow > across the manifold versus an infinitesimal element of a symmetry > group of the manifold is just a matter of lambda abstraction. > So there we go; some positive comments about a theory that is not > formulatable in ZFC (I suppose it might be, but it would not be > pretty). I doubt you'll see many comments though, because I doubt I've > said much that is especially controversial. For one thing, the vast majority of posters who do mention infinitesimals are criticized for doing so. MR is the most well-known infinitesimalist here on sci.math (though currently, MR is posting against negative numbers, rather in favor of nonzero infinitesimals like his "smallest quantity"). AP is another infinitesimalist. TO mentions such numbers from time to time. RF has his "iota." Would any of these posters be open to the smooth infinitesimals described by McInnes? I can't be sure of this. For one thing, these infinitesimals are said to be nilpotent. It's hard to say whether RF, TO, MR would accept them. AP is unlikely to accept any infinitesimals that don't have digits, unless there's way to define digits for these smooth infinitesimals. Also, I'm not sure whether any of them want to give up Excluded Middle. Still, McInnes has started the type of discussion that I'd like to see. The response I'd like to see is one which defends classical analysis against these smooth infinitesimals -- and I mean something more like "Smooth infinitesimals are bad because they contradict LEM" than "There are no nonzero infinitesimals, and anyone who thinks so is a --" (five-letter insult).
From: MoeBlee on 10 Jun 2010 17:38 On Jun 10, 3:43 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > Also, I'm not sure whether > any of them want to give up Excluded Middle. I give it up every day I take to work in intuitionsistic logic. What really is your damn problem? It seems much more psychological than mathematical to me. MoeBlee
From: MoeBlee on 10 Jun 2010 17:46 On Jun 10, 3:43 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > the vast majority of posters who do > mention infinitesimals are criticized for doing so. I love infinitesimals. I love non-standard analysis. I love the mathematical logic from which non-standard analysis came about. I love IST. I love the whole comparison between non-standard analysis and standard analysis. I love that analysis can be approached in either way and in many more ways. REALLY. And I'm not worried in the least that I'll be criticized for saying such things. And I can't fathom what is your real problem that you've developed your bizarre obssession to show (contrary to fact, contrary even to my POSTINGS) that I'm intolerant of such things. MoeBlee
From: herbzet on 10 Jun 2010 18:33 Transfer Principle wrote: > FredJeffries wrote: > > A couple days ago Herb posted: > > > What would be instructive would be to see at what point in Cantor's > > > proof that |S| < |P(S)| that the proof fails in NF(U). > > Maybe YOU could start the ball rolling towards a counterexample by > > responding to that. > > The set theorist Randall Holmes was a webpage on the > NF(U) theories: > > math.boisestate.edu/~holmes/holmes/nf.html > > On that page, we scroll down and see the following: > > "Cantor's paradox of the largest cardinal: Cardinal > numbers are defined in NF as equivalence classes of > sets of the same size. The form of Cantor's theorem > which can be proven in Russell's type theory asserts > that the cardinality of the set of one-element > subsets of A is less than the cardinality of the > power set of A. Note that the usual form > (|A| < |P(A)|) doesn't even make sense in type > theory. It makes sense in NF, but it isn't true in > all cases: for example, it wouldn't do to have > |V| < |P(V)|, and indeed this is not the case, > though the set 1 of all one-element subsets of V is > smaller than V (the obvious bijection x |-> {x} has > an unstratified definition!)." So we have |1| < |V| = |P(V)|. > So we can see what's going on here. NF(U) is based > on the idea of a Stratified Comprehension and types > on the variables. In particular, in the formula > "xey," y must be of a higher type than x. Thus, a > set A and its power set P(A) usually don't even > have the same type. Thus we *can* write A e P(A), because that formula *is* stratified, at least sometimes. Does that ever fail, perhaps when we take A = V? > But if we have "xey" and "xez," the y and z can be > the same type. In particular, we can let y be P(x) > and z be {x}, so that P(x) and {x} can have the > same type. This is why Holmes often refers to the > singleton subsets above. > > According to Holmes, Cantor's proof does show that > P1(A), the set of singleton subsets of A, does have > a smaller cardinality than P(A). But P1(A) often > doesn't have the same size as A, That is, it is smaller. In those cases, what, I wonder, are the elements x of A that don't have unit sets {x}? > and Holmes gives > an explicit example of such a set -- the set V of > all sets. Such sets are called "non-Cantorian." Still unclear on where Cantor's proof |A| < |P(A)| fails. Does it not in general make sense to assume (for the reductio) that there is a bijection f:A-->P(A)? If we assume there is such a bijection, will there not exist a set D of elements x of A such that ~(x e f(x))? Etc. > George Greene pointed out earlier that the idea of > having sets that aren't the same size as their set > of singleton subsets (which I admit is a very > counterintuitive concept) renders NF(U) not worth > considering except theoretically. > > But then again, there may be some posters who > consider Cantor's theorem to be even more > counterintuitive than card(P1(V)) < card(V), and > such posters should be given the choice of having > a set theory with non-Cantorian sets. > > And if NF(U) is still undesirable for those posters, > then they can look for still other theories with a > maximal (universal) set. All I want is for everyone > to have the opportunity to _choose_ a set theory > that best reflects his own intuition. You're quite the democrat! -- hz
From: herbzet on 10 Jun 2010 18:47
herbzet wrote: > Transfer Principle wrote: > > FredJeffries wrote: > > > > A couple days ago Herb posted: > > > > What would be instructive would be to see at what point in Cantor's > > > > proof that |S| < |P(S)| that the proof fails in NF(U). > > > Maybe YOU could start the ball rolling towards a counterexample by > > > responding to that. > > > > The set theorist Randall Holmes was a webpage on the > > NF(U) theories: > > > > math.boisestate.edu/~holmes/holmes/nf.html > > > > On that page, we scroll down and see the following: > > > > "Cantor's paradox of the largest cardinal: Cardinal > > numbers are defined in NF as equivalence classes of > > sets of the same size. The form of Cantor's theorem > > which can be proven in Russell's type theory asserts > > that the cardinality of the set of one-element > > subsets of A is less than the cardinality of the > > power set of A. Note that the usual form > > (|A| < |P(A)|) doesn't even make sense in type > > theory. It makes sense in NF, but it isn't true in > > all cases: for example, it wouldn't do to have > > |V| < |P(V)|, and indeed this is not the case, > > though the set 1 of all one-element subsets of V is > > smaller than V (the obvious bijection x |-> {x} has > > an unstratified definition!)." > > So we have |1| < |V| = |P(V)|. > > > So we can see what's going on here. NF(U) is based > > on the idea of a Stratified Comprehension and types > > on the variables. In particular, in the formula > > "xey," y must be of a higher type than x. Thus, a > > set A and its power set P(A) usually don't even > > have the same type. > > Thus we *can* write A e P(A), because that formula *is* > stratified, at least sometimes. Does that ever fail, > perhaps when we take A = V? > > > But if we have "xey" and "xez," the y and z can be > > the same type. In particular, we can let y be P(x) > > and z be {x}, so that P(x) and {x} can have the > > same type. This is why Holmes often refers to the > > singleton subsets above. > > > > According to Holmes, Cantor's proof does show that > > P1(A), the set of singleton subsets of A, does have > > a smaller cardinality than P(A). But P1(A) often > > doesn't have the same size as A, > > That is, it is smaller. In those cases, what, I > wonder, are the elements x of A that don't have > unit sets {x}? > > > and Holmes gives > > an explicit example of such a set -- the set V of > > all sets. Such sets are called "non-Cantorian." > > Still unclear on where Cantor's proof |A| < |P(A)| > fails. Does it not in general make sense to assume > (for the reductio) that there is a bijection f:A-->P(A)? Maybe that. In general, f would be a set of ordered sets <x, f(x)> and this would be illegal, since f would contain sets of elements of different types. ??? > If we assume there is such a bijection, will there not > exist a set D of elements x of A such that ~(x e f(x))? > > Etc. [snip] -- hz |