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4 votes
1 answer
172 views

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

Let ${\frak U}$ be the collection of ultrafilters on $\omega$, and let ${\frak M}$ be the collection of maximal intersecting families on $\omega$. …
4 votes
1 answer
273 views

Supremum of infimum of measure of members of a free ultrafilter

Let $\text{FrU}(\omega)$ be the collection of free ultrafilters on $\omega$. …
4 votes
1 answer
254 views

Minimal cardinality of a filter base of a non-principal uniform ultrafilters

If ${\cal U, V}$ are non-principal uniform ultrafilters on $\kappa$, do we necessarily have $b({\cal U}) = b({\cal V})$? … Thanks to Joseph van Name for making me aware of uniform ultrafilters and the fact that this question is only (potentially) interesting when restricted to these. …
3 votes
2 answers
140 views

Is $(\omega+1)^\omega/{\cal U}$ complete for ${\cal U}$ free ultrafilter?

Let ${\cal U}$ be a free ultrafilter on $\omega$. Is the linearly ordered set $(\omega+1)^\omega/{\cal U}$ complete?
3 votes
1 answer
192 views

${\frak b}$ and ${\frak d}$ in the Rudin-Keisler preordering

If $(Q,\leq)$ is any preordered set (that is, $\leq$ is a reflexive and transitive, but not necessarily anti-symmetric relation), then we say that $S\subseteq Q$ is unbounded if for all $q\in Q$ ther …
1 vote
1 answer
141 views

Is the Rudin-Keisler ordering a continuous relation?

If $X, Y$ are topological, and $R\subseteq X\times Y$ we say that $R$ is continuous (from $X$ to $Y$) if for every $V\subseteq Y$ with $V$ open, we have $$R^{-1}(V) = \{u\in U: \exists v\in V:(u,v)\i …
3 votes
2 answers
245 views

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

Let $\beta\omega$ be collection of all ultrafilters on $\omega$ (principal and non-principal). We endow $\beta\omega$ with an operation $+$ in the following way. …
4 votes
1 answer
221 views

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

$\DeclareMathOperator{\RK}{\mathrm{RK}}$Let $\beta\omega$ be the Stone-Cech compactification of $\omega$ with the discrete topology. We can endow $\beta\omega$ with an addition operation that extends …
4 votes
0 answers
158 views

Finite pre-orders embeddable in the Rudin-Keisler ordering

Let $\NPU(\omega)$ be the set of non-principal ultrafilters on $\omega$. …
4 votes
2 answers
367 views

Multiplicative and additive groups of the field $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}...

Let ${\cal U}$ be a non-principal ultrafilter on $\omega$, and for each $n\in\omega$, let $p_n$ denote the $n$th prime, that is $p_0 = 2, p_1=3, \ldots$ Next we introduce the following standard equiva …
5 votes
1 answer
211 views

Relation between ultrafilters ${\scr U}$ and ${\scr U} \otimes {\scr U}$ [closed]

If ${\scr U}$ and ${\scr V}$ are ultrafilters on non-empty sets $A$ and $B$ respectively, then the tensor product ${\scr U}\otimes{\scr V}$ is the following ultrafilter on $A\times B$: $$\big\{X\subseteq …
4 votes
1 answer
386 views

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

A $P$-point is an ultrafilter $\scr U$on $\omega$ such that for every function $f:\omega\to\omega$ there is $x\in {\scr U}$ such that the restriction $f|_x$ is either constant, or finite-to-one. A $Q …
0 votes
1 answer
196 views

Maximal elements in the Rudin-Keisler ordering

(Ramsey ultrafilters do not necessarily exist.) Question. Does $\text{NPU}(\omega)/\simeq_{RK}$ have maximal elements? …
7 votes
1 answer
192 views

Non-tensor-representable ultrafilters on $\omega$

If ${\cal U}$ and ${\cal V}$ are ultrafilters on non-empty sets $A$ and $B$ respectively, then the tensor product ${\cal U}\otimes{\cal V}$ is the following ultrafilter on $A\times B$: $$\big\{X\subseteq … We say an ultrafilter ${\cal Z}$ on $\omega$ is Tensor-representable if there are non-Keisler-Rudin-equivalent ultrafilters ${\cal U}, {\cal V}$ and a bijection $\psi:\omega^2\to \omega$ such that ${\cal …
3 votes
1 answer
142 views

The Wallman and interval topologies on non-principal ultrafilters with the Rudin-Keisler pre...

If $(P,\leq)$ is a pre-odered set (that is, $\leq$ is a reflexive and transitive relation) and $x\in P$, we set $(\uparrow_{\leq} x) = \{p\in P: p\geq x\}$ and $(\downarrow_{\leq} x) = \{p\in P: p\leq …

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