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Let $\mathfrak L$ be a $\mathscr{FOL}$ with completeness and soundness. My question is how many maximal consistent sets on it?

I know that every maximal consistent set can be dealt as an ultrafilter on Henkin model. However, the quantity of ultrafilters on a poset, even on a boolean algebra, cannot be determined by the cardinality of it. Cardinality of the set of ultrafilters on an infinite Boolean algebra

So an extra question is that is there exists an arithmetic to calculate the quantity of ultrafilters on arbitary poset?

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1 Answer 1

Consider the following infinite binary tree. The nodes of the tree are theories (sets of formulas) in the language of first order arithmetic. The root of the tree is the set of Peano Axioms ($PA$). The children of each node $T$ are $T \cup \{GL_T\}$ and $T \cup \{\lnot GL_T\}$, where $GL_T$ is the Goedel sentence of the theory $T$. By Goedel's Imcompleteness Theorem each node is a consistent set of formulas.

Each branch in this tree determines a theory - the union of all nodes, which is consistent by compactness theorem. So the theory corresponding to each branch can be extended to a maximal consistent set. Furthermore it is easy to see that these maximal consistent sets are all different. Now since there are $2^{\aleph_0}$-many branches in this tree, there are at least $2^{\aleph_0}$ maximal consistent sets in the language of arithmetic.

Clearly there are at most $2^{\aleph_0}$ maximal consistent sets, so $2^{\aleph_0}$ is the cardinality of the set of maximal consistent sets in the first-order language of arithmetic.

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Your answer is enlightening, although it is not the final. If you find the final answer, please update. Thank you very much. –  Popopo Jul 4 '12 at 6:49

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