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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 …