In the following demostration of the 9th chapter of "Introduction to Set Theory" by Karel Hrbacek and Thomas Jech is written that the collection $\mathcal{B}$ of the all bounded subsets of a infinite regular cardinal $\aleph_\alpha=\omega_\alpha$ is egual to $\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$.
Could someone help me to prove it?
Suppose that $X\subseteq\omega_\alpha$ is bounded. This means that there is some $\delta<\omega_\alpha$ with $X\subseteq\delta\subseteq\omega_\alpha$, so $X\in\mathcal{P}(\delta)$ and $X\in\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$, which shows $\mathcal B\subseteq\bigcup_{\delta<\omega_\alpha}\mathcal P(\delta)$.
Suppose now that $X\in\bigcup_{\delta<\omega_\alpha}\mathcal P(\delta)$. Then there is a $\xi<\omega_\alpha$ with $X\subseteq\xi$. Since $\xi\in\omega_\alpha$ and $\omega_\alpha$ is transitive we have $\xi\subseteq\omega_\alpha$, so we also have $X\subseteq \xi\subseteq\omega_\alpha$, or in other words $X$ is bounded subset of $\omega_\alpha$. This shows that $\bigcup_{\delta<\omega_\alpha}\mathcal P(\delta)\subseteq\mathcal B$.
Since we proved a double inclusion the two sets are equal.