Cardinality of $\sigma$-algebra generated by an infinite family of sets Let $\mathcal{F}$ be an infinite family of subset of $X$ of cardinality $\kappa$ (thus $\kappa$ is an infinite cardinal). From the recursive description of generated $\sigma$-algebra, I know that the $\sigma$-algebra $\langle \mathcal{F}\rangle$ which is generated by $\mathcal{F}$ has cardinality at most $\kappa^{\aleph_0}$. On the other hand, since $\langle \mathcal{F}\rangle $ contains $\mathcal{F}$,  $\langle \mathcal{F}\rangle$ has cardinality at least $\kappa$. Thus we have $\kappa\leq |\langle \mathcal{F}\rangle |\leq \kappa^{\aleph_0}$. Is is true that $|\langle \mathcal{F}\rangle |=\kappa^{\aleph_0}$? I'm not very familiar with cardinal arithmetic, thanks for any help.

Yes. If $B$ is a sigma complete infinite boolean algebra, then $|B| = |B|^{\aleph_0}$. You can find a proof in the Handbook of Boolean algebra Vol. 1, theorem 12.2.
