39
votes

Accepted

### Are all free ultrafilters 'the same' in some sense?

Certain important properties are shared by all free ultrafilters. In many applications of ultrafilters, especially more elementary applications, only these properties are used. In such a situation, it ...

35
votes

Accepted

### Ultrafilters and automorphisms of the complex field

It seems not.
It was shown by Di Prisco and Todorcevic (and reproved later by at least three sets of authors) that if sufficiently large cardinals exist (e.g., a proper class of Woodin cardinals), ...

32
votes

Accepted

### Ultrafilters as a double dual

This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $\mathbb{C}$ which go to zero at infinity. Then the algebra ...

21
votes

### Ultrafilters and diagonal arguments

I think there can be no such argument, for the following reason.
It is consistent with ZF that the statement isn't true, since $x$ might be an amorphous set, a set for which every subset is either ...

21
votes

### Completeness number of ultrafilters

Any countably incomplete ultrafilter has completeness number $\aleph_0$, and if $\kappa$ is measurable then any $\kappa$-complete non-principal ultrafilter on $\kappa$ has completeness number $\kappa$....

20
votes

Accepted

### Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters

Identify $\mathbb{N}$ with $\mathbb{Q}\cap[0,1]$ via a bijection, and consider the subspace $C([0,1])\subset\ell^\infty(\mathbb{N})$ of sequences which extend to a continuous function on $[0,1]$. ...

18
votes

Accepted

### SPOT as a conservative extension of Zermelo–Fraenkel

In plain terms, the conservativity of SPOT over ZF means that if a particular statement S in the language of ZF is provable in SPOT, then ZF can already prove S (with a possibly different proof). Note ...

16
votes

Accepted

### Ultrafilter subtraction and "zero"

Both your guesses are correct. To see this, it's helpful to reformulate the way you're thinking about the subtraction operator on $\beta \mathbb Z$. Beginning with subtraction on $\mathbb Z$, you can ...

15
votes

### Ultrafilters and diagonal arguments

Even without amorphous sets, this cannot happen.
Suppose there are no free ultrafilters on some infinite set $x$, then $\beta x$ has a natural bijection with $x$. Which is strictly smaller than $\...

15
votes

Accepted

### Selective ultrafilter and bijective mapping

No, this fails not only for selective ultrafilters but for all non-principal ultrafilters $\mathcal F$ on $\omega$.
The main ingredient in the proof is the theorem that, if an ultrafilter $\mathcal U$...

14
votes

### Ultrafilters as a double dual

This is an elaboration on Todd Trimble's comment about Tom Leinster's lovely posts about codensity monads. I quite like the codensity monad story; here is my preferred way of telling it.
Suppose you ...

12
votes

Accepted

### Non-tensor-representable ultrafilters on $\omega$

Recall that $\mathcal Z$ is a weak $P$-point if it is not in the closure of any countable subset of $\omega^* \setminus \{\mathcal Z\}$. A weak $P$-point is never the tensor product of two non-...

12
votes

Accepted

### Models of $\mathsf{ZFC}$ with neither $P$- nor $Q$-points

Your definition of P-point should end with "finite-to-one" rather than "injective". As it stands, it defines selective ultrafilters, not P-points.
As far as I know, the consistency of the statement "...

12
votes

Accepted

### Club filter basis in $\omega_1$

That's independent of ZFC.
On the one hand, it's consistent with ZFC that $2^{\aleph_1}=\mathfrak c$, in which case the whole club filter on $\aleph_1$ has cardinality $\mathfrak c$.
On the other ...

12
votes

### SPOT as a conservative extension of Zermelo–Fraenkel

Ali's answer contains all the required technical explanation. My answer is more "sociological background " in nature:
For various reasons, there have been rather strong negative reactions ...

12
votes

Accepted

### Ultrafilter lemma for arbitrary lattice

It is equivalent to AC.
Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial ...

11
votes

Accepted

### Are free ultrafilters as posets product-irreducible?

No. Every nonprincipal ultrafilter $U$, considered as a partial under $\subseteq$, is a nontrivial product order. To see this, suppose that $U$ is a nonprincipal ultrafilter on $\kappa$.
Partition $\...

11
votes

Accepted

### What is the Turing degree associated with an ultrafilter $U$?

Great question! This is something that Uri Andrews, Mingzhong Cai, David Diamondstone, and I looked at in a recent (still unpublished) paper.
First of all, let's note that there's an important ...

11
votes

Accepted

### Ultrafilter on the ordinal $\omega^\omega$

The relevant general construction is the sum of a family $\{\mathcal V_i:i\in I\}$ of an indexed family of ultrafilters, with respect to an ultrafilter $\mathcal U$ on the index set $I$. If $\mathcal ...

11
votes

Accepted

### Maximal intersecting families on $\omega$ that are not ultrafilters

Let $U,V,W$ be three distinct ultrafilters on $\omega$. Let $M$ be the family of those subsets of $\omega$ that belong to at least two of $U,V,W$. Then $M$ is a maximal intersecting family, it is not ...

11
votes

Accepted

### Is the Tukey order well-founded

I'll show that the Tukey order is consistently ill-founded. I suspect it is open whether it is consistently well-founded. As you are aware, the well-foundedness of the Tukey order is not known to be ...

10
votes

Accepted

### Cardinality of a set of pairwise non-order-isomorphic ultrafilters on $\omega$

Let me prove that a two ultrafilters $\mathcal{U}_{1},\mathcal{U}_{2}$ on $\omega$ are isomorphic as posets if and only if they are Rudin-Kielser equivalent. Suppose that $\phi:\mathcal{U}_{1}\...

10
votes

### Are all free ultrafilters 'the same' in some sense?

"Fast" ultrafilters and "slow" ultrafilters are used in this paper and have different properties.
The definition of "slow" ultrafilter: let $\mathcal P$ be the set of ...

9
votes

### Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters

Eric Wofsey's answer is very nice and simple. Nevertheless this might be interesting. Very old results of Kakutani [Concrete Representation of Abstract (M)-Spaces, Ann. of Math. 42 (1941)] show that ...

9
votes

Accepted

### Multiplicative and additive groups of the field $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z})/\simeq_{\cal U}$

1. $(K,+)$ and $(\mathbb R,+)$ are isomorphic.
The additive group of any field $K$ is a vector space over its prime field ($\mathbb F_p$ or $\mathbb Q$), hence it is determined up to isomorphism by ...

9
votes

Accepted

### When do two ultrafilters yield isomorphic ultrapowers?

If one allows an arbitrary signature, the answer to this question is fairly well-known. Consider the structure $(\lambda,A)_{A\subseteq \lambda}$. Let $(M,R_A)_{A\subseteq \lambda}$ be the common ...

9
votes

Accepted

### Addition and Rudin-Keisler ordering in $\beta \omega$

For Question 1: First, there are idempotent ultrafilters, so some ultrafilters satisfy 1 in a very strong form. But 1 does not hold in general. The reason is that the semigroup $\beta\omega-\omega$ ...

9
votes

Accepted

### "Completion property" in $(\beta\omega,+)$

No. Kunen proved that there are many weak P-points in $\beta\omega-\omega$, which means they are nonprincipal ultrafilters that are not in the closure of any countable set of other nonprincipal ...

9
votes

Accepted

### Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?

If $\kappa=\aleph_0$ then yes: every ultrafilter is $\aleph_0$-complete.
If $\kappa>\aleph_0$ then no, if $\lambda X$ is nonempty. Split $X$ into countably many sets $\{X_n:n\in\mathbb{N}\}$, of ...

9
votes

Accepted

### Supremum of infimum of measure of members of a free ultrafilter

The answer is: zero.
The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with ...

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