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Theorem: There is no such $E$. Claim: for each $a\in E$ there exists $b\in E$ with $|b\setminus a|\ge 2$. Proof of Claim: Let $a$ be a counterexample. Then all $b\ne a$ contain exactly one element eac …

Yes. List the pairs $(\alpha,\beta)$ with $\alpha<\beta<\kappa$ as $(\alpha_\lambda,\beta_\lambda), \lambda<\kappa$. Then construct the sets $B_\alpha\in\mathcal B, \alpha<\kappa$ as follows: At stage …

Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$: $$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$ Observe that all but finitely many bloc …

Let $\kappa$ be any infinite cardinal. Let $\cal S=\{\kappa\setminus\{x\}:0<x<\kappa\}$ and $C=\{0\}$. Then all the given conditions are satisfied.

Let $\cal S=\{\{1,2\},\{2,3\},\{3,1\}\}$. Then $\cal S$ has no choice set, whatsoever. So there is no asymmetry -- not every shy set is contained in a choice set, and not every gregarious set contai …