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- From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
- Subject: Re: Boolean Algebras
- Message-ID: <1992Nov14.025504.16516@CSD-NewsHost.Stanford.EDU>
- Sender: news@CSD-NewsHost.Stanford.EDU
- Organization: Computer Science Department, Stanford University.
- References: <42536@gremlin.nrtc.northrop.com> <Bxo1Hw.6qA@dcs.ed.ac.uk>
- Date: Sat, 14 Nov 1992 02:55:04 GMT
- Lines: 24
-
- In article <Bxo1Hw.6qA@dcs.ed.ac.uk> mxh@dcs.ed.ac.uk (Martin Hofmann) writes:
- >In article <42536@gremlin.nrtc.northrop.com>, jbarnett@nrtc.northrop.com (Jeff Barnett) writes:
- >>
- >> Is there a free boolean algebra such that the cardinality of
- >> the algebra (not the cardinality of its generators) is that
- >> of the continuum?
- >>
- >> Jeff Barnett
- >
- >I guess the powerset of the natural numbers ordered by inclusion provides an example.
-
- For any infinite X, the power set of X, and for that matter the set of
- all finite and cofinite subsets of X, are Boolean algebras with atoms.
- Hence they cannot be free Boolean algebras.
-
- (In any equationally defined class, the free algebra F(X) on a set X
- consists of the equivalence classes of those formulas whose variables
- are drawn from X. Claim: for Boolean algebras, if X is infinite F(X)
- has no atoms. Proof: Let the equivalence class [p] be an atom where p
- is some (necessarily finite) Boolean formula, and let Q be a Boolean
- variable not appearing in p. Then 0 < p&Q < p, contradicting [p]'s
- atomicity.)
- --
- Vaughan Pratt A fallacy is worth a thousand steps.
-