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Let $\mathcal I$ be the set of all intervals of $\mathbb R$. There's an exercise that asks to verify that $\mathcal I$ is a semi-algebra. One of the defining properties of a semi-algebra is that it's stable by finite intersections.

Here's an argument to verify that $\mathcal I$ is stable by finite intersections: Let $a, b \in \cap_{i = 1}^n I_i$ such that $a < b$, where $I_i \in \mathcal I$ for $i \in \{1, \dots, n\}$. Then $a, b \in I_i$ for each $i$. That is, $\{x \in \mathbb R: a < x < b\} \subset I_i$ for each $i$. So $\{x \in \mathbb R: a < x < b\} \subset \cap_{i = 1}^nI_i$. Thus, $\cap_{i = 1}^nI_i$ is an interval. (I'm using the definition of an interval given in the top answer to this question: Why is the empty set considered an interval?)

Can I say that $\mathcal I$ is stable by arbitrary intersections? I mean, it seems to me that the above "for each $i$"-argument holds if $i \in \mathcal A$, where $\mathcal A$ is an arbitrary index set.

harisf
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    Yes, $\cal I$ is stable under arbitrary intersections. – David Moews Sep 24 '23 at 16:05
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    Also see this answer for a proof that arbitrary intersection of convex sets is convex (note that intervals are precisely convex sets of $\mathbb{R}$). – Jakobian Sep 25 '23 at 08:43
  • If $I_i$ is an interval with endpoints $a_i, b_i$ (possibly containing one or the other) then $\cap_i I_i$ is an interval with endpoints $\sup a_i, \inf b_i$ (again, possibly containing one or the other), and this is true with an arbitrarily large collection of intervals. – Qiaochu Yuan Sep 25 '23 at 08:45

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