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.

  • $\begingroup$ Try to take $\mathcal {F} $ to be a sigma algebra itself. Then it is obvious what the cardinality of the generated sigma algebra is. $\endgroup$ – PhoemueX Dec 1 '15 at 15:22
  • $\begingroup$ @PhoemueX In this case, I think we have $\kappa^{\aleph_0}=\kappa$. Note that there exists no $\sigma$-algebra with cardinality $\aleph_0$. Thus if $\mathcal{F}$ itself is a $\sigma$-algebra, it must has cardinality greater than $\aleph_0$. In this case, we may have $\kappa=\kappa^{\aleph_0}$. Indeed, if $\kappa=c$, the equality holds. $\endgroup$ – Xiang Yu Dec 1 '15 at 15:49
  • $\begingroup$ Deciding whether $\kappa =\kappa^\aleph_0$ seems to be a difficult problem (math.stackexchange.com/questions/184746/…), possibly even undecidable in ZFC. If we assume the generalized continuum hypothesis, this holds for all uncountable sets. But in general, it seems to be hard. Note also that every infinite sigma algebra has at least the cardinality of the continuum. $\endgroup$ – PhoemueX Dec 1 '15 at 17:47

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.

  • $\begingroup$ Thanks for your answer. From wikipedia en.wikipedia.org/wiki/Complete_Boolean_algebra, I know what is a complete Boolean algebra. But what does sigma complete infinite Bool algebra mean? $\endgroup$ – Xiang Yu Dec 2 '15 at 7:35
  • $\begingroup$ It means that every countable subset of $B$ has a supremum. So every sigma algebra is an example. $\endgroup$ – hot_queen Dec 2 '15 at 10:35
  • $\begingroup$ It's a weaker notion than complete Boolean algebra, isn't it? $\endgroup$ – Xiang Yu Dec 2 '15 at 10:40
  • $\begingroup$ Yes but the proof just uses countable completeness. $\endgroup$ – hot_queen Dec 6 '15 at 21:03

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