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$. 2 added 3 characters in body Here is a even simpler example. Let$ {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$. 1 Here is a even simpler example. Let${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\$.