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I'm reading a proof about the cardinality of the Borel $\sigma$-algebra and it uses some notions from set theory where I'm not all that familiar with. So the idea is that we have a set $U = \bigcup_{\alpha<\omega_1}\mathcal{B}_\alpha$, where the $\mathcal{B}_\alpha$ are defined by recursion. So $\mathcal{B}_0$ is the set of all open intervals in $\mathbb{R}$, $\mathcal{B}_{\alpha+1}$ is the set of all subsets of $\mathbb{R}$ that are either a countable union of members of $\mathcal{B}_\alpha$ or a countable intersection of members of $\mathcal{B}_\alpha$ or the difference of two members of $\mathcal{B}_\alpha$. If $\lambda$ is a limit ordinal, then $\mathcal{B}_\lambda = \bigcup_{\alpha<\lambda}\mathcal{B}_\alpha$.

So somewhere in the proof we want to show that $U$ is a $\sigma$-algebra and thus I need to prove it's closed under countable unions. So suppose $A_n\in U$ for all $n < \omega$. For each $n$ there exists a $\alpha_n<\omega_1$ such that $A_n \in \mathcal{B}_{\alpha_n}$. Now comes the part that I don't get: "Since $cf(\omega_1) > \omega$, there is a $\gamma < \omega_1$ such that $\alpha_n < \gamma$ for all $n<\omega$."

Is there some way around this? Can you get the same conclusion, without using the cofinality?

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In this case, it really just means "the limit of countably many countable ordinals is a countable ordinal". Talking about cofinality is just a fancy way of saying that.

This is a consequence of the axiom of choice, that the countable union of countable sets is countable, and of the fact that the supremum of a set of ordinals is their union.

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