5
votes
4answers
184 views

Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?

This question comes after the comments in the recent related question Sigma-complete Lindenbaum algebras?, but in its current form is sufficiently different in my opinion, and so I decided to follow ...
1
vote
1answer
271 views

Compactness and completeness in Gödel logic

The standard proof of the completeness theorem in first-order Gödel logic is based on a first-order countable language. I want to know that is there any proof of the completeness theorem in ...
2
votes
0answers
246 views

Constructing the Stone Space of a Distributive Lattice

Does anyone have a good reference for the method of giving a topology to a distributive lattice as outlined in M.H. Stone's "Topological representation of distributive lattices and Brouwerian logics"? ...
1
vote
1answer
280 views

Complete De Morgan algebra

Recall that an algebra $(A,\sim)$ is a De Morgan algebra if $A$ is a bounded distributive lattice and $\sim$ is a unary operation which satisfies: ${\sim} (x\vee y)={\sim} x\wedge {\sim} y$ and ...
4
votes
2answers
561 views

additive functions on a lattice

Given a lattice $L$. Can one classify all functions $f:L\rightarrow \mathbb{R}$, that satisfy $f(a \wedge b)+f(a\vee b) = f(a)+f(b)$. Some examples are the the set of all finite subsets of a given ...
4
votes
1answer
564 views

Can the Knaster-Tarski theorem be proved using the Schroeder-Bernstein theorem?

The reverse can be done easily and the proof is well known I am wondering if the exact same argument can be used to prove reverse as well.
6
votes
4answers
602 views

`Topos' with alternate subobject lattice?

We know that for any topos E, and for any object A in E, the subobjects of A, Sub(A), form a Heyting lattice. Does anybody know of any sort of modification of the definition of a topos that makes ...
2
votes
3answers
462 views

Countable atomless boolean algebra covered by a larger boolean algebra

Suppose Q is an atomless countable boolean algebra, and B is an arbitrary atomless boolean algebra. Q is unique modulo isomorphisms. There is a subalgebra in B that is isomorphic to Q. There is ...
3
votes
2answers
305 views

Semilattices in atomless boolean algebras

Let S be a bounded semilattice without maximal elements. Can we always construct an atomless boolean algebra B, containing S as a subsemilattice, such that S is cofinal in B-{1}? That is, for every ...