The ultrafilters tag has no usage guidance.

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### 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 ...

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### 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 $\beta\mathbb{Z}\setminus\mathbb{Z}$? ...

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### 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 ...

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### Embedding $\beta\mathbb{N}$ into a product of Cantor sets

Let us consider $\beta\mathbb{N}$, the Stone-Czech compactification of the natural numbers (where we do not take $0$ to be a natural number, so the only idempotent elements are nonprincipal ...

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### 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 ...

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### What is the name for a Banach space property closed under ultraproducts?

In Banach space theory, a super-property is a property of a Banach space that is preserved under ultrapowers. (Update (2015-09-28): The property must also be closed under isometric embeddings.) ...

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### 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 ...

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### 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 ...

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343 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 ...

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### 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 ...

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### Are ultralimits the Gromov-Hausdorff limits of a subsequence?

Let $(M_i,p_i)$ be a sequence of $n$-dimensional Riemannian manifolds with lower Ricci curvature bound $-1$. Fix a non-orincipal ultrafilter and let X be the ultralimit of the sequence.
Does there ...

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### 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 ...

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### 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 $S^*_\kappa$, ...

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### 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 $D$. $\phi(x)$ ...

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### 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 ...