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Let $2^{<\omega}$ be the binary tree and assign to each branch $x$ the family $F_x$ of finite sets that intersect it. If $x_1$, $x_2$, $\ldots$ $x_k$ is a finite set of (distinct) branches then there …

The following is a ZFC example, due to Michael Hrušák, of a MAD family without sunflowers of cardinality $3$. Start with the standard AD family $\mathcal{B}=\{B_f:f\in{}^\omega2\}$ of branches through …

Consider the reverse lexicographic order $\prec$ on $\lambda\times\lambda^+$ ($(\alpha,\beta)\prec(\gamma,\delta)$ iff $\beta<\delta$ or $\beta=\delta$ and $\alpha<\gamma$); its order type is equal to …

I think the problem is slightly more basic. Fremlin has the fully correct $$ D \Vdash \check D\in\dot{\mathcal{G}} $$ Also, Fremlin did not fix one generic $G$ at the outset; he works with names and t …

The positive solution uses an equivalent of the Axiom of Choice: for every infinite set $A$ there is a bijection $f:A\to A\times A$. In the basic Fraenkel Model (section 4.3 in Jech's Axiom of Choice …

Assume $K_n$ is a minor of $G$. Each vertex of $K_n$ corresponds to a connected subset of vertices of $G$ as it can only have been obtained by contracting edges. Each edge of $K_n$ can only have been …

In topological language: for any closed subset, $F$, of $\beta\mathbb{N}\setminus\mathbb{N}$ the family $I_F=\{A:A^*\cap F=\emptyset\}$ is an ideal. Your property translates into: the closed set $F$ i …

I believe not: let $f$ be any colouring and take a maximal equivalence relation $\sim$ on $\omega$ with the property that $m\sim n$ implies $f(\{m,n\})=0$. Note that $\sim$ can be extreme: the identit …

The method from this answer leads to $B_\alpha\le\binom{\alpha+1}{2}$ for finite $\alpha$. Split $\omega$ into $\alpha+1$ infinite sets, say $X_i=\{n:n\equiv i \pmod{\alpha+1}\}$. For each $p\in[\alph …

To confirm Andreas' suspicion: Balcar and Vojtáš proved in Almost Disjoint Refinement of families of subsets of $\mathbb{N}$ that every ultrafilter on $\mathbb{N}$ has an almost disjoint refinement. T …

Your argument is basically Kurepa's proof from his thesis Ensembles ordonnées et ramifiés, see page 96 (a footnote has Aronszajn's construction). As noted in the comments you need to show that what yo …

For $\kappa=\aleph_0$ yes: there are (many) models with ultrafilters of character less than $\mathfrak{c}$. Let $E\subseteq[\omega]^\omega$ be a base for an ultrafilter, say $|E|=\aleph_1<\mathfrak{c} …

A construction that uses the equality $\kappa^\lambda=\kappa$ directly runs as follows. By that equality the set $H$ has cardinality $\kappa$, so we can find a surjection $f:\kappa\to H$ such that for …