details re Z-R |- PA consistent
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From: Transfer Principle on 3 Aug 2010 23:50 On Aug 3, 12:54 pm, FredJeffries <fredjeffr...(a)gmail.com> wrote: > On Aug 2, 8:56 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > As I wrote earlier, RF considers the universal set of part > > theory to be dual to the empty set of set theory. And so > > we take the Empty Set Axiom of ZFC: > > Ex (Ay (~yex)) > > And so we replace each occurrenceof "~yex" with the dual > > formula "xsy" to obtain the following: > > Universal Set Axiom: > > Ex (Ay (~xsy)) > I don't understand. If we replace "~yex" with "xsy" don't we get > Ex (Ay (xsy)) ? Obviously, I must have made an error somewhere. In correcting my error, let me re-emphasize why I'm doing this in the first place. RF tells us that part theory is a sort of mirror image to set theory, which to me means that "x is a structure of y" in part theory corresponds to "y is an element of x" in set theory. Like "is an element of," "is a structure of" is to be primitive. But this doesn't simply mean that we take each axiom of ZF and blindly replace each appearance of "yex" with "xsy." If we did, the resulting theory would be trivially equivalent to ZF as every model of ZF would be a model of the new theory, with "is a structure of" mapped to the inverse of "e." This isn't going to give us TO-infinitesimals, which is RF's desiderata. So instead, the approach we take is, we replace each "yex" in a ZF axiom with "xsy," and then compare the axiom to RF's desiderata to see whether they match. If not, then we modify it until it does match or reflect the desiderata. So here are my corrections: Dual Axioms (Correction): To form the dual of a ZF axiom, replace "yex" with "xsy." ZF Axiom of Extensionality: Axy (x=y <-> Az (zex <-> zey)) RF Dual Axiom of Extensionality: Axy (x=y <-> Az (xsz <-> ysz)) There's no problem here since this dual axiom doesn't contradict any of RF's desiderata. So we proceed: ZF Axiom of Empty Set: Ex (Ay (~yex)) RF Dual Axiom of Empty Set (Correction): Ex (Ay (~xsy)) Since we already have Extensionality, we know that there's a _unique_ set x such that Ay (~xsy). But now we check with RF's desiderata -- RF wants the dual of the empty set to be a _universal_ set. But it's not obvious why this set need necessarily be a universal set. I explain this more fully in the previous post, but the trick is to add an extra axiom -- and this axiom is motivated by the fact that when we eventually define infinitesimals, "s" restricted to the infinitesimals corresponds to "<=", and we expect "<=" to satisfy Trichotomy. So we add an axiom: RF Trichotomy Axiom: Axy (xsy v ysx) From these three axioms, we can prove that there exists a set V such that Ay (ysV) -- and so V is a universal set, which matches RF's desiderata.
From: FredJeffries on 4 Aug 2010 15:32 On Aug 3, 8:23 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > But what if someone wants to discuss, say, positive infinitesimals, > which don't exist in classical analysis? As soon as they mention > infinitesimals, the others in the thread will either say that they > don't exist, or ask for a definition -- and they mean a _formal_ > definition, one that's eliminable to _primitives_. > > And so, as much as such posters want simply to do interesting > mathematics regarding infinitesimals without worrying about > formal theories, their opponents force them to consider formal > theories -- ones that prove that their infinitesimals even exist. Bunk. Tristam Needham in his wonderful book "Visual Complex Analysis" throws infinitesimals all over the place without ever giving a formal definition. I have yet to see a review criticizing him for this. http://books.google.com/books?id=ogz5FjmiqlQC
From: FredJeffries on 4 Aug 2010 16:30 On Aug 3, 8:50 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > So here are my corrections: > > Dual Axioms (Correction): > To form the dual of a ZF axiom, replace "yex" with "xsy." > > ZF Axiom of Extensionality: > Axy (x=y <-> Az (zex <-> zey)) > > RF Dual Axiom of Extensionality: > Axy (x=y <-> Az (xsz <-> ysz)) > > There's no problem here since this dual axiom doesn't > contradict any of RF's desiderata. So we proceed: > > ZF Axiom of Empty Set: > Ex (Ay (~yex)) > > RF Dual Axiom of Empty Set (Correction): > Ex (Ay (~xsy)) > > Since we already have Extensionality, we know that there's > a _unique_ set x such that Ay (~xsy). But now we check > with RF's desiderata -- RF wants the dual of the empty set > to be a _universal_ set. But it's not obvious why this set > need necessarily be a universal set. > > I explain this more fully in the previous post, but the trick is > to add an extra axiom -- and this axiom is motivated by the > fact that when we eventually define infinitesimals, "s" > restricted to the infinitesimals corresponds to "<=", and we > expect "<=" to satisfy Trichotomy. So we add an axiom: > > RF Trichotomy Axiom: > Axy (xsy v ysx) > Trichotomy seems like overkill. Couldn't you use something (can't remember if it's the Meet or Join) like Axy (Ez (xsz & ysz))
From: MoeBlee on 4 Aug 2010 16:54 On Aug 3, 10:23 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > But what if someone wants to discuss, say, positive infinitesimals, > which don't exist in classical analysis? As soon as they mention > infinitesimals, the others in the thread will either say that they > don't exist, or ask for a definition -- and they mean a _formal_ > definition, one that's eliminable to _primitives_. Calculus with infinitesimals can be formulated in ZFC (even in certain weaker theories). > In another subthread of > this thread, there is a discussion of infinitesimals. The other > posters want to see rigorous formal axioms before they'll even > consider the infinitesimals (as evidenced by their reluctance to > accept RF's and TO's infinitesimals without them). The problem with RF and TO is not really infinitesimals, but a problem with both of them being hopeless cranks (though each in a different way from the other). > But of course, > it'll take a long time to get these axioms correct and sufficiently > rigorous before one can even attempt to derive infinitesimals > from them. Whatever problems they might have, we only have to look to ZFC for axioms that provide infinitesimals. MoeBlee
From: Transfer Principle on 4 Aug 2010 23:49
On Aug 4, 1:30 pm, FredJeffries <fredjeffr...(a)gmail.com> wrote: > On Aug 3, 8:50 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > RF Trichotomy Axiom: > > Axy (xsy v ysx) > Trichotomy seems like overkill. Couldn't you use something (can't > remember if it's the Meet or Join) like > Axy (Ez (xsz & ysz)) I believe that it's "Join." Let's see whether Jeffries's suggestion works. Theorem: Ay (yeV) Proof: Suppose not. Then Ey (~yeV). Using Jeffries's "Join" axiom, we find that Ez (Vsz & ysz) -- z is the "join" of V and y. But then we have Vsz, thereby contradicting the Empty Set Dual axiom. So therefore, Ay (yeV). QED OK, so this "Join" axiom works. The reason that I tried to use Trichotomy earlier is that we'll eventually need it when we reach infinitesimals -- but as I wrote earlier, the four order properties that I mentioned earlier need only hold on R, not on all of V. If all we need to make V universal is this "Join" axiom, then we'll use it instead of Universal Trichotomy. At this point, one might ask, what exactly is the "Join" of V and another set? The only sensible answer is V -- which means that we would need VsV. Thus, we need to amend the Empty Set Dual so that it allows for VsV: RF Universal Set Axiom (Corrected): Ex (Ay (xsy -> x=y)) Incidentally, the "Join" axiom also tells us that a set x satisfying this Universal Set Axiom must be unique. For what if there were two such sets, U and V? Then let z be the Join of U and V. From the argument above, the Join of V and any other set is V, so z=V. But similarly, the Join of U and any other set is U, so z=U. Therefore U=V. QED What about the other axiom Jeffries mentioned, "Meet"? Notice that the dual of the Pairing Axiom (the weaker version of Pairing that I mentioned earlier) is a sort of "Meet" axiom: RF Pairing Dual Axiom: Aab (Ex (xsa & xsb)) But I'm beginning to have second thoughts about this. This tells us that any two objects must "intersect" -- so that no two sets a and b can truly be disjoint. But this could be where infinitesimals come in -- perhaps the intersection of two apparently "disjoint" sets is actually an infinitesimal, or something? This is something that I must think about some more -- we might have to reject this axiom. |