From: Clive Tooth Subject: Re: Help!! Need proof for the existence of the set {a, b} Date: Sun, 20 Aug 2000 18:33:31 +0100 Newsgroups: sci.math Summary: [missing] Stephen Montgomery-Smith wrote: > Remind me what the replacement axiom is. Does it say > that if I have a formula F(x,y) such that I can prove > that given x there is at most one y satisfying the > formula, and if I have a set A, then there is a set > B = {y:F(x,y) and x in A} ? I believe that 'ZF minus Replacement' is called Zermelo set theory. Here are some definitions of Replacement: ------------------------------------------------------ Paul R Halmos "Naive Set Theory" ... the replacement axiom is not mentioned. ------------------------------------------------------ Krzysztof Ciesielski "Set Theory for the Working Mathematician", page 17: Replacement scheme axiom (Fraenkel 1922; Skolem 1922) For every formula phi(s,t,U,w) with free variables s, t, U, and w, set A and parameter p if phi(s,t,A,p) defines a function F on A by F(x)=y <-> phi(x,y,A,p), then there exists a set Y containing the range F[A] of the function F, where F[A]={F(x): x in A}. ------------------------------------------------------ Judith Roitman "Introduction to Modern Set Theory", page 43: Replacement Axiom Schema. Ranges of definable functionals exist: If phi is a formula so that Ax,y,z if phi(x,y) and phi(x,z), then y=z, then AwEs(s={y:Ex in w phi(x,y)}). ------------------------------------------------------ Peter J Cameron "Sets, Logic and Categories", page 114: 8. (Replacement Axiom) Let psi be a first-order formula with two free variables x and y which 'defines a partial function': that is, for all x, there is at most one y which satisfies the formula. Let a be any set. Then there is a set b consisting of all those y such that psi(x,y) holds for some x in a: that is, b={f(x):x in a}, where f is the 'function defined by psi'. ------------------------------------------------------ Yiannis N Moschovakis "Notes on Set Theory", page 170: 11.1. (VIII) Replacement Axiom. For each set A and each unary definite operation H, the image H[A] =_df {H(x)|x in A} of A by H is a set. ------------------------------------------------------ -- Clive Tooth http://www.pisquaredoversix.force9.co.uk/ End of document ============================================================================== From: Clive Tooth Subject: Re: Help!! Need proof for the existence of the set {a, b} Date: Sun, 20 Aug 2000 17:39:09 +0100 Newsgroups: sci.math Stephen Montgomery-Smith wrote: > Remind me what the replacement axiom is. Does it say > that if I have a formula F(x,y) such that I can prove > that given x there is at most one y satisfying the > formula, and if I have a set A, then there is a set > B = {y:F(x,y) and x in A} ? Yes. > If so, does the following work? > Let A = powerset of powerset of empty set. > > Let F(x,y) be > (x = empty set and y = a) or y = b Well, if x=empty we have _two_ y's: a and b. Better might be: Let F(x,y) be (x = empty set and y = a) or ( x not= empty set and y = b) > > Then I see that B = {a,b}. > > Well set theory is not my expertize - I hope I am > not being naive here. > > aliguevara wrote: > > > > TO: anyone who knows the answer to my question > > > > FROM: Alex Guevara > > EMAIL: aguevara@helios.acomp.usf.edu > > > > Does anyone out there know how to prove "the pair set axiom" from "the power > > set axiom" and "the axiom of replacement" in set theory? -- Clive Tooth http://www.pisquaredoversix.force9.co.uk/ End of document ============================================================================== From: hrubin@odds.stat.purdue.edu (Herman Rubin) Subject: Re: Help!! Need proof for the existence of the set {a, b} Date: 21 Aug 2000 07:56:36 -0500 Newsgroups: sci.math In article <39A00327.6E40955A@math.missouri.edu>, Stephen Montgomery-Smith wrote: >Remind me what the replacement axiom is. Does it say >that if I have a formula F(x,y) such that I can prove >that given x there is at most one y satisfying the >formula, and if I have a set A, then there is a set >B = {y:F(x,y) and x in A} ? >If so, does the following work? >Let A = powerset of powerset of empty set. >Let F(x,y) be >(x = empty set and y = a) or y = b >Then I see that B = {a,b}. You would have to change the definition of F(x,y) to be (x = empty set and y = a) or (x != empty set and y = b); as stated, F(x,y) is true if x = empty set and y is either a or b. >Well set theory is not my expertize - I hope I am >not being naive here. There is a somewhat larger problem here. Unless some other axioms are used, how does one know that there is any set whatever? It is needed that A has at least two elements, one of which is the empty set. In most models, this can be done by using the axiom of infinity. >aliguevara wrote: >> TO: anyone who knows the answer to my question >> FROM: Alex Guevara >> EMAIL: aguevara@helios.acomp.usf.edu >> Does anyone out there know how to prove "the pair set axiom" from "the power >> set axiom" and "the axiom of replacement" in set theory? -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399 hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558