Cardinality once more
in [Logic]
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From: zuhair on 2 Jan 2010 17:56 On Jan 2, 4:44 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jan 2, 8:30 am, zuhair <zaljo...(a)gmail.com> wrote: > > > Suppose we add Cardinality symbolized by "| |" as a primitive to the > > language of ZF minus Reg., so"| |" would be a primitive one place > > function symbol. Let's > > name this theory ZF minus Reg.(=,e,| |) to differentiate it from > > ZF minus Reg. > > > And suppose we add the following axiom to the list of axioms of > > ZF minus Reg.(=,e,| |). > > > Axiom of Cardinality: > > > For all x,y : |x|=|y| <-> Exist f ( f:x-->y , f is bijective ) > > > So now we have the theory ZF minus Reg.+Axiom of Cardinality(=,e,| |).. > > Okay, so far. You can find this (Tarskian, if I recall) approach (even > whether or not we have regularity) in certain textbooks as a temporary > way of dealing with cardinality without using choice (or even > regularity): > > card(x) = card(y) <-> x equinumerous with y > > It may be adopted as a temporary axiom (let's call it "Tarski's > axiom") until we add choice to prove a corollary of the numeration > theorem (consider all formulas closed as appropriate): > > E!y(y is an ordinal & y equinumerous with x & Az((z ordinal & z > equinumerous with x) -> ~zey)) > > (i.e, there is a unique y that is the least ordinal equinumerous with > x) > > or it may be added as a temporary axiom until we add regularity (if we > didnt' already have regularity) to prove: > > E!yAt(tey <-> (t equinumerous with x & Az(z equinumerous with x -> ~ > rank(z) e rank(t)))) > > (i.e. there exists the set of all sets that are equinumerous with x > and are of lowest rank among sets equinumerous with x) > > I gather that you pretty much already know that, but I'm including it > for context, along with mentioning, that with choice and regularity > combined, the two approaches above pick out the same set to serve as > card(x). > > > So for example I want to come with a "definition of Cardinality" in ZF > > minus Reg. > > suppose that was, for a specific formula Phi(x): > > > Card(x) <-> Phi(x) > > That is not a definition of a function symbol. It is not even a well > formed formula. On the left of the biconditional you have a TERM but > the left of a biconditional must be a FORMULA. > > One form for definining a function symbol is: > > card(x) = y <-> Phi(x y) > > where we have the thereom already proven: > > E!y Phi(x y) > > In case you don't have that theorem already proven, then (in a theory > with '0' as a (primitive or defined) 0-place function symbol) you can > use "the Fregean method": > > card(x) = y <-> ((AxE!y Phi(x y) & Phi(x y)) or (~AxE!y Phi(x y) & > y=0)) > > So let's assume you've adopted such an approach. > > > Then I will add the axiom: > > > Axiom of Cardinality: For all x Exist y ( y=Card(x) ). > > If 'card' (primitive or defined) is now in the language (primitive or > extended by definitions), then the above formula is a theorem of logic > alone anyway; you don't need it as an additional axiom. This has been > mentioned many many times in discusssions on sci.logic that I think > you probably have read. > > > So we'll have ZF minus Reg. + Axiom of Cardinality. > > > Now what I want is that > > > ZF minus Reg. + Axiom of Cardinality Equi-interpretable with > > ZF minus Reg. + Axiom of Cardinality (e,=,| | ). > > > Can that be done? > > If we correct your formulation, then, if I correctly understand your > intent, your question becomes: > > Is there a formula Phi (in the language of set theory extended to > include 'card') such that: > > ZF(minus regularity)+"card(x) = y <-> ((AxE!y Phi(x y) & Phi(x y)) or > (~AxE!y Phi(x y) & y=0))" > = > ZF(minus reguarity)+Tarski's axiom ? > > MoeBlee Welcome Moe, It has been long time since I've read anything for you. This matter with Cardinality has been going on for around one month here. Let me present my idea in cross manner, Cardinality is a one place function on sets such that for every set x , every set y we have Card(x)=Card(y) <-> Exist f (f:x-->y, f is bijective) Lets call this: the *basic requirement for Cardinality*. Now Cardinality can be *defined* or can be a *primitive* concept. With defined cardinality here I am referring specifically to a definition by a formula in first order logic with identity and epsilon membership as the sole primitives i.e. in FOL(=,e). This as you said takes the form Card(x)=y <-> Phi (x,y) So lets say that Phi(x y) is a *cardinality defining formula* if card (x) defined after this formula will satisfy the basic requirement for Cardinality. remember Phi(x y) is a formula in FOL(e,=) Now my question was the following: Is their a cardinality defining formula Phi(x y) such that ZF(minus Reg.) + "For all x Exist y ( y=card(x) ) were card(x) <-> Phi (x y)" is Equi-interpretable with ZF(minus Reg.) + Tarski's axiom. Of course the first theory is in FOL(e,=) While the second is in FOL(e,=,| |). Is that possible? Zuhair
From: MoeBlee on 2 Jan 2010 18:50 On Jan 2, 3:44 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > Is there a formula Phi (in the language of set theory extended to > include 'card') such that: > > ZF(minus regularity)+"card(x) = y <-> ((AxE!y Phi(x y) & Phi(x y)) or > (~AxE!y Phi(x y) & y=0))" > = > ZF(minus reguarity)+Tarski's axiom ? Correction: I didn't mean that Phi may have the symbol 'card' in it. I should have said: Is there a formula Phi (in the language of set theory) such that: ZF(minus regularity)+"card(x) = y <-> ((AxE!y Phi(x y) & Phi(x y)) or (~AxE!y Phi(x y) & y=0))" = ZF(minus reguarity)+Tarski's axiom ? MoeBlee
From: MoeBlee on 2 Jan 2010 19:10 On Jan 2, 4:56 pm, zuhair <zaljo...(a)gmail.com> wrote: > Cardinality is a one place function on sets such that for every set > x , every set y > we have > > Card(x)=Card(y) <-> Exist f (f:x-->y, f is bijective) > > Lets call this: the *basic requirement for Cardinality*. Yes, as I said, that is a well-known idea, going back at least to Tarski. > Now Cardinality can be *defined* or can be a *primitive* concept. > > With defined cardinality here I am referring specifically to a > definition by a formula in first order logic with identity and epsilon > membership as the sole primitives i.e. > in FOL(=,e). > > This as you said takes the form > > Card(x)=y <-> Phi (x,y) > > So lets say that Phi(x y) is a *cardinality defining formula* if card > (x) defined after this formula will satisfy the basic requirement for > Cardinality. Yes, I said I'm with you there, still a basic set theoretical path. > remember Phi(x y) is a formula in FOL(e,=) Right. I shouldn't have said 'card' can be in Phi. So I just corrected my post in this way: Is there a formula Phi (in the language of set theory) such that: ZF(minus regularity)+"card(x) = y <-> ((AxE!y Phi(x y) & Phi(x y)) or (~AxE!y Phi(x y) & y=0))" = ZF(minus reguarity)+Tarski's axiom ? > Now my question was the following: > > Is their a cardinality defining formula Phi(x y) such that > > ZF(minus Reg.) + "For all x Exist y ( y=card(x) ) were card(x) <-> Phi > (x y)" > > is Equi-interpretable with > > ZF(minus Reg.) + Tarski's axiom. Your formulation is not correct. (1) You ignored what I reminded you of: card(x) <-> Phi(x y) is not well formed. (2) If I'm not mistaken (I'd welcome someone more expert commenting), equi-interpretability would be of concern if the two theories have different languages. But your theories do have the same language. Whether you consider 'card' as primitive in one language but "defined" (yours is not in a correct definitional form) and thus added to another language, they end up both being the same language. So, if I'm not mistaken, to ask a substantive equi-interpretability question, you need one theory to have 'card' in its language' and the other theory not to mention 'card' at all (if that theory is adding 'card' as DEFINED, then any axiom would not NEED to use the symbol 'card'). Now, please, take it a step at a time. We know what one theory is. ZF(-regularity)+Tarski's axiom, with 'card' as primitive. But you must reformulate your definition of 'card' for the other theory. Then you can also add any axioms you want with 'card' in it (though, of course, with 'card' DEFINED, those axioms can also be written withOUT 'card'). Thanks for the welcome back. I hope all is well with you these days. (I actually need not to post so I can concentrate on other things right now, but I couldn't resist this time as I saw a basic mistake you need to correct before you can advance with your question. But now that I've posted, I'm probably headed back to the same old time-eating posting routine!) MoeBlee
From: MoeBlee on 2 Jan 2010 19:20 On Jan 2, 6:10 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > card(x) <-> Phi(x y) > > is not well formed. Maybe you typoed? Maybe you meant card(x) = y <-> Phi(x y). So your other theory is?: ZF(-regularity)+"AxEy(y = card(x) & Axy(card(x) = y <-> Phi(x y))" (Some of those quantifiers might be redundant, but I put them in for clarity.) But, as I mentioned, the part about Ey y = card(x) doesn't add anything substantive. So you might as well just make it: Axy(card(x) = y <-> Phi(x y)) Thus your theory: ZF(-regularity)+"Axy(card(x) = y <-> Phi(x y))" MoeBlee
From: zuhair on 2 Jan 2010 20:35
On Jan 2, 7:20 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jan 2, 6:10 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > card(x) <-> Phi(x y) > > > is not well formed. > > Maybe you typoed? Maybe you meant > > card(x) = y <-> Phi(x y). Yes, of course. Yea, I myself didn't notice this typo really. Thanks for the correction. > > So your other theory is?: > > ZF(-regularity)+"AxEy(y = card(x) & Axy(card(x) = y <-> Phi(x y))" > > (Some of those quantifiers might be redundant, but I put them in for > clarity.) > > But, as I mentioned, the part about Ey y = card(x) doesn't add > anything substantive. So you might as well just make it: > > Axy(card(x) = y <-> Phi(x y)) > > Thus your theory: > > ZF(-regularity)+"Axy(card(x) = y <-> Phi(x y))" were Phi(x y) is a cardinality defining formula of course. Yes, that is the theory. > > MoeBlee |