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Yes. Just take $\mathcal{A}$ to be $[\omega]^2$ together with the powerset of the even integers.

I think they exist for all $k \geq 2$. Here is a proof. Fix an enumeration of $[\omega]^2$. For the first step in the process add an arbitrary $k$-set $q_1=q(p_1)$ containing the first pair $p_1$ of …

Here is how to reduce the problem to a finite colouring problem. Let $G'=G_0 \cup \dots \cup G_K$. For each $t \in \mathbb{N}$, the $tG'$ be the subgraph of $G^\infty$ consisting of $t$ consecutive co …

Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A_{c,p}$ is a clutter $\mathcal C$ with ground set $\ …

For all $n \geq 2$, here is a simple winning strategy for Spoiler on the $n^2 \times n^2$ board, that requires at most $n^2-1$ moves to win. I assume that Solver plays first, but the strategy can eas …

To answer Jon Noel's question in the comments, there is no such example for finite hypergraphs. Claim. Let $H=(V,E)$ be a finite hypergraph such that $|e| > 1$ for all $e \in E$ and for all distinc …

No. For each element $x \in \kappa$, let $g(x)$ be the set of elements in $\frak{E}$ that contain $x$. By assumption, for all $x \in \kappa$, we have $|g(x)| \leq \kappa$. We may clearly assume $\e …

Your formal version does not look correct. For each boy $b$, there should be a total order $\leq_b$ on the set of girls $G$ (this is the preference order for $b$) and for each girl $g$, there should …

No, this is not possible. Here is an elaboration of Eric Wofsey's comment. Suppose it is possible and let $M$ be a maximal (under inclusion) matching of $G$ (this exists by Zorn's lemma). Then $| …

I think this is actually a pretty interesting question, so I'm not sure why it has been voted down. Let me briefly expand on Qiaochu's comment. Suppose I ask the related question: Does the notion o …