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After reading this page, I still have some questions about Lindenbaum algebras. Assume that the scope is a propositional language with a denumerable set X of propositonal variables.

  1. In that case, the Lindenbaum algebra of propositonal language is treated, i.e. the Boolean algebra of equivalence classes of formulas modulo the relation "logical equivalence". I wonder if the Lindenbaum algebra is freely generated by the set of propositional variables or by the set of equivalence classes of propositional variables. Is there a difference? At first, I thought it should be more correct to say that the algebra is generated by the same kind of objects as its elements (i.e. by equivalence classes, not variables).

  2. In the case of the Lindenbaum algebra of some set T of formulas, i.e. the Boolean algebra of equivalence classes of formulas modulo the relation "logical equivalence under T", is the algebra freely generated, and by what? By {$[x]: x\in X$}, where $[x]$ is the equivalence class of those logically equivalent to X? Or by {$[x]: x\in X$}, where $[x]$ is the equivalence class of those logically equivalent to x under T? Moreover, how should I demonstrate such situation?

  3. The lindenbaum algebra can be seen as the quotient algebra of the algebra of formulas (aka. term algebra, word algebra) modulo logical equivalence, or logical equivalence under T. Since the algebra of formulas itself is freely generated by its propositional variables, can we deduce that the Lindenbaum algebra is freely generated (as a universal algebra) because it is the quotient of a free algebra?

  4. If we work in propositional language with finitely many propositional variables, the Lindenbaum algebra (of the language, or of T) is finitely generated, and thus finite. What about a language with uncountably many propositional variables? Is there any interesting notion about the Lindenbaum algebra to study in this case (e.g. cardinalities, subalgebras, morphisms)? (In the denumerable case, Lindenbaum algebras are all isomorphic, and so I guess there's nothing much to say.)

I'm sorry if I've asked too many details. I'd appreciate any answer or advice. Thank you.

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1 Answer 1

(Ad. 1.) Lindenbaum algebra is generated by the equivalence classes which can be easily seen from it's construction. From the definition of set of generators of a given structure, the generators are contained in the set of all elements of that structure. (Ad. 2.) For a given subset T of language L Lindenbaum algebra modulo T is generated by a set of equivalence classes $[x]$={$y \in L: Cn(T)\vdash (x\sim y)$}. The result is quotient of Lindenbaum algebra which need not to be free. Look - (Ad. 3.) A quotient of free algebra need not to be free, consider as an example infinite free Boolean algebra and any not free finite Boolean algebra. The latter is quotient of the former. These facts come from correspondence bettwen homomorphisms and subalgebras. (Ad. 4.) Finitly generated algebra need not to be finite - consider Heyting algebra generated by one element http://en.wikipedia.org/wiki/Heyting_algebra#Examples. From algebraic point of view this topic is threated well in available free online book by Burris&Sankappanavar - "A course in universal algebra". EDIT: Particularly interesting aspect of Lindenbaum algebras is their dependence on a given consequence operator.

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Thank you. Perhaps I have to spend more time to work out my problems. –  knight Jul 1 '12 at 18:01

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