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The original proof by Fichtenholz and Kantorovich is, in my opinion, a nice one. It suffices to find continuum many independent subsets of some countably infinite set $C$; let's take $C$ to be the co …

Like your previous question, Selective ultrafilter and bijective mapping , this fails for all nonprincipal ultrafilters $\mathcal U$ on $\omega$, and for essentially the same reason. If there were suc …

No, there are no such $\mathcal U$ and $\varphi$. Let me start the proof with two simplifying observations. First, $\varphi$ is irrelevant, because bijections of $\omega$ to itself preserve selectivi …

Let $C$ be the set of finite sequences of zeros and ones. I'll exhibit uncountably many infinite subsets of $C$ such that no three form a sunflower. Since $C$ is countably infinite, the example can be …

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 …

Sure, let $K$ be any countably infinite field and let $P$ be the projective plane (or a higher-dimensional projective space) over $K$. Let $S'$ be the set of lines in $P$ (where a line is regarded as …

No; $\mathfrak u'=\mathfrak c$. To prove it, consider any $\mathcal C\subseteq[\omega]^\omega$ with cardinality $<\mathfrak c$. Working modulo finoite subsets of $\omega$ , and closing under (finitary …

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 …

Here's a possibly simpler example than Joel's, in the case of $\omega$. Let $A_0$ be $c$ (where I think of the Cohen real as a subset of $\omega$), and let the rest of the $A_n$'s be any partition of …

I'm not sure the following example is different from the ones already given, but the description is different, and so I hope some readers might find it useful. Fix a countable family $\mathcal F$ of …

If $K$ is a field of cardinality $\aleph_0$, then the points and lines of the projective plane over $F$ constitute a complete regular linear hypergraph. The field $K$ can be recovered (up to isomorphi …

There is no such $n$. Proof: Suppose $n$ and $f$ are as in the problem. Of the $2n+1$ numbers $1,2,4,8, \dots, 2^{2n}$, some three have the same $f$-value. And the smallest of those three is the grea …

EDIT: I'll leave my previous answer up for now (at the end of this one), but here's an easier answer that doesn't need assumptions like CH that go beyond ZFC. It's well-known that there is a family of …

One of the standard examples of an almost disjoint family of cardinality $\mathfrak c$ is the set of paths through the complete binary tree $2^{<\omega}$ (identified with $\omega$ via your favorite bi …