Are ultralimits the Gromov-Hausdorff limits of a subsequence?

Let (Mi,pi) 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 exists a p∈X and subsequence of (Mi,pi) converging to (X,p) in the pointed Gromov-Hausdorff sense? Answer AttributionSource : Link , Question Author : dg.jan , … Read more

Set of subsequences with the same ultrafilter limit of the original sequence

Let U be a free ultrafilter on the positive integers N and fix U∈U such that U is not cofinite (thanks J.D.Hamkins for the correction.) Consider the natural bijection between (0,1] and the infinite {0,1}-sequences with infinitely many ones (written in base 2). Define also the sequence x=(xn) by xn=0 if n∈U and xn=1 otherwise, … Read more

Selectors for bases of ultrafilters

If U is a selective (Ramsey) ultrafilter on ω and B is its base, then for every sequence A_0\supsetneq A_1\supsetneq A_2\supsetneq\ldots in \mathcal{B} there exists A\in\mathcal{B} such that |A\setminus A_n|\le n+1. I’m interested in the following generalization of this notion. Let \mathcal{B} be a base of a non-principal ultrafilter \mathcal{U} and let \mathcal{F}\subset[\omega]^\omega. Let us … Read more

Finite pre-orders embeddable in the Rudin-Keisler ordering

A pre-ordered set is a pair (P,≤) where P is a set and ≤⊆P×P is a reflexive and transitive relation. Let NPU(ω) be the set of non-principal ultrafilters on ω. The Rudin-Keisler pre-order on NPU(ω) is defined by U≤RKV:⇔(∃f:ω→ω)(∀U∈U)f−1(U)∈V. It is easy to see that ≤RK is reflexive and transitive, but not anti-symmetric. Question. Given … Read more

cardinality of local bases in the non-standard reals

Given a index set S together with a ultrafilter μ on S (such that no set of cardinality <|S| has measure 1). Let the ordered field R(S,μ) denote the ultrapower of R with respect to S, i.e. R(S,μ):=RS/∼, where two maps f,g are called equivalent, if μ({i∈S|f(i)=g(i)})=1. This is again an ordered field. Equip it … Read more

Complexity of ultrafilter limits

Let F be a free ultrafilter on N and, for each A⊆N and n∈N, define dn(A):=|A∩[1,n]|n. Question. Considering P(N)={0,1}N, is it true that the set {A⊆N:F–lim is not Borel? Answer Yes. Let me rather write subsets as sequences: the question is whether the subset L of (a_n) in \{0,1\}^{\mathbf{N}} such that \mathscr{F}\text{-}\lim\sum_{k=1}^n\frac{a_k}{n}=0 is non-Borel. Write … Read more

The example of the idempotent filter or subsets family with finite intersections property

From the answer of Andreas Blass and comments of Ali Enayat on my question Selective ultrafilter and bijective mapping it became clear that for any free ultrafilter U we have U≁. Where \mathcal{F}\otimes\mathcal{G}=\{X\subset A\times B~|~\{a\in A~|~ \{b\in B~|~ (a,b)\in X\} \in\mathcal{G}\}\in\mathcal{F}\} for any two filters \mathcal{F}, \mathcal{G} on sets A, B respectively. And \mathcal{F}\sim\mathcal{G} means … Read more

Does ultrafilter have measure one?

Define a new product measure on cantor space as follows:u({0})=a,u({1})=1-a,where a∈(0,1/2]. Does any ultrafiter U hasn’t measure one? When a=1/2,I know U hasn’t measue one.I guess neither when a∈(0,1/2), but I don’t know how to prove. Answer This question is a bit more subtle than I had originally thought (in the comments), but anyway here’s … Read more

Lattice of differences between ultrafilters

Consider two ultrafilters, $U$ and $V$, on the same cardinal $\kappa$. Let $D(U, V)=\lbrace X\subseteq \kappa: X\in U-V\rbrace$; clearly $D(U, V)$ is a lattice under $\subseteq, \cap, \cup $ since the intersection of two $U$- or $V$-large sets is $U$- or $V$-large, and the union of two $U$- or $V$-small sets is $U$- or $V$-small; … Read more