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Here is a even simpler example. Let ${ x_{\alpha} }_{\alpha\in \{x_{\alpha} \}_{\alpha\in \omega_1}$ be an increasing chain in the Turing degrees. For every $\alpha$, let $B_{\alpha}={y\mid B_{\alpha}=\{y\mid y\geq_T x_{\alpha}}$x_{\alpha}\}$. Each $B_{\alpha}$ is a boldface $\Sigma^0_3$ set. Then for any countable ordinal $\beta$, $\bigcap_{\alpha<\beta}B_{\alpha}$ is not empty but $\bigcap_{\alpha<\omega_1}B_{\alpha}=\emptyset$. |
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Here is a even simpler example. Let $ {x_{\alpha}}_{\alpha\in { x_{\alpha} }_{\alpha\in \omega_1}$ be an increasing chain in the Turing degrees. For every $\alpha$, let $B_{\alpha}={y\mid y\geq_T x_{\alpha}}$. Each $B_{\alpha}$ is a boldface $\Sigma^0_3$ set. Then for any countable ordinal $\beta$, $\bigcap_{\alpha<\beta}B_{\alpha}$ is not empty but $\bigcap_{\alpha<\omega_1}B_{\alpha}=\emptyset$. |
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Here is a even simpler example. Let ${x_{\alpha}}_{\alpha\in \omega_1}$ be an increasing chain in the Turing degrees. For every $\alpha$, let $B_{\alpha}={y\mid y\geq_T x_{\alpha}}$. Each $B_{\alpha}$ is a boldface $\Sigma^0_3$ set. Then for any countable ordinal $\beta$, $\bigcap_{\alpha<\beta}B_{\alpha}$ is not empty but $\bigcap_{\alpha<\omega_1}B_{\alpha}=\emptyset$. |
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