According to the answer by François G. Dorais, we know that a logic $\mathfrak L$ is compact iff its Stone space of the Lindenbaum–Tarski algebra of the empty theory (w.r.t the deductive system) is compact. It sounds great, but a bit confused me: The Lindenbaum–Tarski algebras depend on concrete deduction systems, whereas the compactness of a logic seems only respect to the semantics. So how to understand it? And is the compactness of logics really independent from deduction systems over them?

Update According to Lin's notice, there are two kinds of compactness, semantic and syntactic. The syntactic one is equivalent to the compactness of the Stone space of the Lindenbaum–Tarski algebra of empty theory w.r.t the deductive system. For the semantic one, I have an idea.

Let $\mathfrak F(\mathcal L) :=(\wp(\mathcal L),\mathbf 0,\mathbf 1,\land,\lor)$, s.t.

1)$\mathbf 0= \mathcal L$;

1')$\mathbf 1= Cn(\emptyset)$;

2)$\Phi \land \Psi=Cn(\Phi \cup \Psi)$;

2')$\Phi \lor \Psi=Cn(\Phi) \cap Cn(\Psi)$.

In which $\Phi$ and $\Psi$ are arbitrary formula sets; and $Cn(\Phi)$ denote the logical consequence set of $\Phi$, that is, $Cn(\Phi)=\{\varphi \in \mathcal L|\Phi \vDash \varphi\}$.

It is not very hard to prove that $\Dashv \vDash$(let's use $\Leftrightarrow$ instead) is a congruence of $\mathfrak F(\mathcal L)$(note:$\Phi \Leftrightarrow \Psi$ iff $Cn(\Phi)=Cn(\Psi)$), and $\mathfrak F(\mathcal L)/\Leftrightarrow$ is a complete lattice. It looks like Lindenbaum–Tarski algebra, except it is in semantics and involves formula sets rather than sentence sets.

Now if we can define a $\lnot$ to make it to become a Boolean Algrbra, then it seems its Stone space is compact iff the logic is semantically compact. Is it true?


  • 1
    $\begingroup$ There are two ways of formulating a compactness theorem: syntactic and semantic. The semantic version is of course independent of the deductive system. $\endgroup$ – Zhen Lin Mar 27 '13 at 10:19
  • $\begingroup$ Do you really want arbitrary formulas instead of sentences? When talking about $\vDash$ do your models come with interpretations of variables? $\endgroup$ – Miha Habič Mar 27 '13 at 12:04
  • $\begingroup$ @MihaHabič Yes. $\endgroup$ – Popopo Mar 27 '13 at 14:45
  • 1
    $\begingroup$ I think your top element shouldn't be $\emptyset$, at least if you want equalities in your definitions of the operations. Consider that $Cn(\emptyset)$ is not empty. $\endgroup$ – Miha Habič Mar 27 '13 at 18:19
  • $\begingroup$ Let me reconsider it. $\endgroup$ – Popopo Mar 27 '13 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.