# Tagged Questions

**5**

votes

**4**answers

233 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

**1**answer

349 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

**0**answers

276 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

**1**answer

304 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

**2**answers

569 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

**1**answer

572 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

**4**answers

615 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

**3**answers

479 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

**2**answers

308 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 ...