Give me ONE digit position of any real that isn't computable up to that digit position!
in [Math]
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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
From: Marshall on 10 Jun 2010 20:03 On Jun 10, 1:25 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > All I want is for everyone > to have the opportunity to _choose_ a set theory > that best reflects his own intuition. I want a pony. Marshall
From: FredJeffries on 10 Jun 2010 23:27 On Jun 10, 3:33 pm, herbzet <herb...(a)gmail.com> wrote: > Transfer Principle wrote: > > 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}? I haven't had time to study this, but I would guess that if V is the set of all sets then there is no unit set {V} ?
From: George Greene on 11 Jun 2010 17:56 On Jun 4, 5:55 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > A computer can calculate ANY digit sequence up to INFINITE length. No, it can't. > Sci.math will make a minor correction there, a computer can only calculate > all digit sequences up to ALL finite lengths. Sci.logic will make this correction too. The fact that you think it is "minor" means you should give up and go home. > This is like saying a computer can only calculate all digit sequences up to INFINITE finite lengths. No, it ISN'T like that. It isn't like that because "this" actually makes sense and is grammatical in English, whereas yours, "INFINITE finite lengths", is completely nonsensical. There is no such thing as an "INFINITE finite" anything, lengths or otherwise. INFINITE and "finite" ARE CONTRADICTORY or mutually exclusive. Everybody who can actually speak English knows that.
From: herbzet on 11 Jun 2010 23:04
FredJeffries wrote: > herbzet wrote: > > Transfer Principle wrote: > > > > 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}? > > I haven't had time to study this, but I would guess that if V is the > set of all sets then there is no unit set {V} ? I don't know, Fred, I don't know much about NF either, but that seems like a good guess. I think that the set universe V in NF is symmetrical between "large" and "small" sets -- that is, the complement of any set A exists, V - A. This is different from ZFC in that the complement of a set in ZFC doesn't exist -- or at least, it can't be proved to exist, I'm unclear on that point. It is usually said that the complement is "too large" to be a set. My guess is that the sets in NF that aren't equinumerous with the set of their respective singleton subsets are all among the "large" sets, but that's just a wild guess on my part. While we're here, I might as well mention that talking about ZF(C) set theory and other set theories is a more-or-less permanent fixture of the scene here in sci.logic, but I must confess I've never been clear about why that is. If one is especially interested in a consistent foundation for mathematics, then I guess that would explain all the interest, but it's never been made clear to me why it would be of particular interest to the study of logic in general. Consequently, I've never put much effort into mastering all the nit-picking details of any set theory. I get the impression that perhaps set theory is roughly equivalent to the study of higher-order logics. Maybe that's the reason people talk about it so much here. -- hz |