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- Newsgroups: sci.math
- Path: sparky!uunet!noc.near.net!nic.umass.edu!umassd.edu!ipgate.umassd.edu!martin
- From: martin@lyra.cis.umassd.edu (Gary Martin)
- Subject: Re: Order -> Algebraic Structure
- In-Reply-To: rjk@world.std.com's message of Sat, 10 Oct 1992 01:48:43 GMT
- Message-ID: <MARTIN.92Oct10094647@lyra.cis.umassd.edu>
- Sender: usenet@umassd.edu (USENET News System)
- Organization: University of Massachusetts Dartmouth
- References: <BvvuD8.I26@world.std.com>
- Date: Sat, 10 Oct 1992 14:46:47 GMT
- Lines: 29
-
- In article <BvvuD8.I26@world.std.com> rjk@world.std.com (robert j kolker) writes:
-
- Let H be a denumerable completely ordered set, where the ordering is
- dense, and there are no maximum or minumum elements. It is well known that
- any two sets having these properties are order isomorphic.
-
- The set of rationals Q is a fortiori this set (up to an isomorphism).
- Clearly the seemingly innocent densely ordered set inherits its algebraic
- properties via this isomorphism or does it?
-
- Is there someway of showing independent of this coincidental isomorphism ,
- that denumerable,densely ordered -> the algebraic structure of Q, i.e. Q
- is a denumerable field or characteristic 0.
-
- I don't believe so - the field structure on the underlying ordered set is
- _extremely_ non-canonical. The automorphism group of the ordered set is
- enormous (it embeds the full direct product of countably many copies of
- itself!), while the rational field is rigid. In other words, the field
- structure is not invariant under the action of the group of order
- automorphisms, so it isn't definable from the order. I don't know if
- that's what you mean, though.
-
-
-
-
-
- --
- Gary A. Martin, Assistant Professor of Mathematics, UMass Dartmouth
- Martin@cis.umassd.edu
-