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Let $V=\mathbb Z,E=\{\{m\in\mathbb Z:m\geq n\}:n\in\mathbb Z\}$. It's clear every transversal basis has at most one element, but no basis $\{k\}$ is good, since $\{k+1\}$ is contained intersects more sets from $E$.
Let $V=\mathbb Z,E=\{\{m\in\mathbb Z:m\geq n\}:n\in\mathbb Z\}$. It's clear every transversal basis has at most one element, but no basis $\{k\}$ is good, since $\{k+1\}$ is contained intersects more sets from $E$.
Let $V=\mathbb Z,E=\{\{m\in\mathbb Z:m\geq n\}:n\in\mathbb Z\}$. It's clear every transversal basis has at most one element, but no basis $\{k\}$ is good, since $\{k+1\}$ intersects more sets from $E$.
Let $V=\mathbb Z,E=\{\{m\in\mathbb Z:m\geq n\}:n\in\mathbb Z\}$. It's clear every transversal basis has at most one element, but no basis $\{k\}$ is good, since $\{k+1\}$ is contained intersects more sets from $E$.
