= Countable-family diamond equivalence
{title2=$\Diamond'_S\Longleftrightarrow\Diamond_S$}
Enumerate each countable guessing family and use a <bijection> $\pi:\lambda\times\omega\to\lambda$ with a <club set> of prefix-closure points. The $n$th candidate sequence decodes the $n$th component of the $n$th family entry. If all candidates fail, choose counterexample sets $X_n$ and <club sets> witnessing failure, then code the $X_n$ together using $\pi$. A correct family guess on the common closure <club set> and all failure <club sets> decodes to one of the forbidden correct guesses. The singleton-family implication supplies the reverse direction.
Back to article page
