From: hrubin@odds.stat.purdue.edu (Herman Rubin) Subject: Re: Least upper bound axiom implies field? Date: 15 Aug 1999 15:34:02 -0500 Newsgroups: sci.math Keywords: ordered groups with LUB property In article <36aa776e.518234730@nntp.lni.net>, Paul Miller wrote: >On Fri, 22 Jan 1999 16:21:59 GMT, ullrich@math.okstate.edu wrote: <>In article <36a7e79d.350320840@nntp.lni.net>, <> pmiller@*DIESPAMMERS*.lni.net (Paul Miller) wrote: <>> This came up in my analysis class today: Assume K is a ring and that a <>relation <>> < is defined on K so that if x,y,z are in K, and x> possesses the property that every subset S which is bounded above (upper bound <>> being defined in terms of the relation <) has a least upper bound in K, does <>it <>> follow that K is a field? <> Is Z a field? >I thought of this at work just a few minutes after I posted it. Unfortunately >I don't have internet access there, so I couldn't cancel the original posting. >:-( >However, I think one might be able to guarantee K is a field if we suppose the >existence of a subset M which is dense in K. Of course we need to make the >technical correction that x < y implies x and y are distinct (so that we may not >take the relation = as the given relation), and that the subset S given above >must be nonempty. The restriction kx > x if k > 0 and kx < x if x < 0 seems >necessary as well. >Theorem: if a is not 0, then exactly one of a and -a is less than 0. >Proof: By assumption, since a is not 0, either a > 0 or a < 0. If K is an ordered group with the least upper bound property, it is either isomorphic to the integers or the reals. Multiplication does not add any problems. -- 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 ============================================================================== From: "G. A. Edgar" Subject: Re: Least upper bound axiom implies field? Date: Tue, 24 Aug 1999 14:39:33 -0400 Newsgroups: sci.math > > Maybe you'd like to define what you mean by "ordered group" -- > if it is a group with an order relation such that x < y > iff ux < uy then something seems to be wrong (unless you > assume also that the group is commutative). There are examples > of noncommutative groups with the above one-sided translation > property for an ordering. > x < y implies axb < ayb Theorem of Iwasawa: Any complete lattice-ordered group is Abelian. [Probably from the late 1940's.] -- Gerald A. Edgar edgar@math.ohio-state.edu Department of Mathematics telephone: 614-292-0395 (Office) The Ohio State University 614-292-4975 (Math. Dept.) Columbus, OH 43210 614-292-1479 (Dept. Fax) ============================================================================== From: hrubin@odds.stat.purdue.edu (Herman Rubin) Subject: Re: Least upper bound axiom implies field? Date: 24 Aug 1999 21:01:02 -0500 Newsgroups: sci.math In article <37C2D27F.48B82712@virginia.edu>, Charles H. Giffen wrote: >Herman Rubin wrote: >[snip] >> If K is an ordered group with the least upper bound property, >> it is either isomorphic to the integers or the reals. Multiplication >> does not add any problems. >Maybe you'd like to define what you mean by "ordered group" -- >if it is a group with an order relation such that x < y >iff ux < uy then something seems to be wrong (unless you >assume also that the group is commutative). There are examples >of noncommutative groups with the above one-sided translation >property for an ordering. The least upper bound property is extremely powerful. Let us use + as the group operation, and let x be any positive element. Then the multiples of x are isomorphic to the integers. Now let y be any positive element of the group not a multiple of the element x. As usual, y cannot exceed all the multiples of x, nor x all the multiples of y. So the multiples of x and y either have common values, or interlace each other. If they have common values, it is easy to show that they are both multiples of some element z. If this only continues for a finite number of steps, the group is isomorphic to the integers. Now if it continues for an infinite number of steps, this would give a group isomorphic to a dense subset of the rationals. But a least upper bound completion would give a subgroup which is isomorphic to the reals, and then using the least upper bound property would keep out anything else. If they interlace, a similar argument can be used to show that one must get the real numbers. -- 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