The ultrafilters tag has no wiki summary.

**120**

votes

**0**answers

10k views

### Ultrafilters and automorphisms of the complex field

It is well-known that it is consistent with $ZF$ that the only automorphisms of the complex field $\mathbb{C}$ are the identity map and complex conjugation. For example, we have that ...

**10**

votes

**0**answers

476 views

### Existence (or non) of “definable” ultrafilters

This is a question which I suspect has an absurdly easy answer, but I'm not seeing it.
Let $\langle\cdot,\cdot\rangle:\omega^2\rightarrow\omega$ be your favorite pairing map (for me, this is the ...

**8**

votes

**0**answers

103 views

### Is it possible that all ultrafilters are determined by the meet-semilattice of sub-ultrapowers?

Suppose that $\mathcal{Z}$ is a filter on a set $X$. Let $\Pi(X)$ denote the lattice of all partitions of the set $X$. Then $(\Pi(X),\wedge)$ is a meet-semilattice where
$P\wedge Q=\{R\cap S|R\in ...

**8**

votes

**0**answers

512 views

### A basic question on Stone-Cech compactification of $\mathbb{Z}$

Can the identity isomorphism on the additive group $\mathbb{Z}$ be extended to a non-identity semigroup isomorphism on $\beta\mathbb{Z}$, and still preserves $\beta\mathbb{Z}\setminus\mathbb{Z}$? ...

**7**

votes

**0**answers

114 views

### Extending a Ramsey filter

We take filter to mean filter on $\omega$ containing all cofinite sets. We say a filter $F$ is Ramsey if ZERO does not have a winning strategy in the following infinite game between the two players ...

**5**

votes

**0**answers

207 views

### Double ultrapower of the hyperfinite $II_1$-factor

Let $\omega$ be a free ultrafilter on the natural numbers and $R$ be the hyperfinite $II_1$-factor (the definition of $R$ is recalled in the comments).
Question: Does there exist another free ...

**5**

votes

**0**answers

314 views

### Boolean Prime Ideal Theorem and non-principal ultrafilters

Somewhat related to my other question Existence of non-principal ultrafilters on sets,
is it known whether it is consistent with ZF that every infinite set has a free (non-principal) ultrafilter, but ...

**4**

votes

**0**answers

87 views

### How much do idempotent ultrafilters generate in terms of semigroups?

It is known that the set of ultrafilters on, say, the natural numbers $\mathbb{N}$, can naturally be endowed with the structure of a compact topological left semigroup (which fails to be anything ...

**3**

votes

**0**answers

104 views

### Are irrational multiples of central sets again central?

Let me begin by giving the relevant definitions. A set $A \subset \mathbb{N}$ is said to be central if and only if there exists a topological system $(X,T)$ (with $X$ a compact metric space, $T$ a ...

**2**

votes

**0**answers

68 views

### What is known about the krull dimension of an ultrapower ring?

Let $R$ be a ring, $F$ a free ultrafilter on a set $X$ which is not countably complete, and $R_F$ the ultrapower of $R$ with $R \not\cong R_F$. The following two results are from a masters thesis ...

**1**

vote

**0**answers

182 views

### Ultrapowers of ultrapowers

Suppose that you have some structure $S$, and you want to construct an ultrapower of cardinality $\kappa$ to obtain $S^*_\kappa$. Then, say you want to construct a new ultrapower from $S^*_\kappa$, ...

**1**

vote

**0**answers

106 views

### Compactness-like property for universal generalization?

Hi all! I have the following problem. Suppose I have a sequence of models $M_1,M_2,...$, all of which have the same countable domain (call it $D$). $x_1,x_2,...$ is a well-ordering of $D$. $\phi(x)$ ...

**1**

vote

**0**answers

205 views

### Defining filters in closure algebras: reference request

A closure algebra C is a boolean algebra B together with a unary closure operator, and additional axioms, the Kuratowski axioms, that the closure operator must satisfy. (The Wikipedia article prefers ...