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My understanding of the closure of a set $S$ under an operation $\oplus$ is that applying $\oplus$ to elements of $S$ yields only other elements of $S$. However, in my undergraduate topology class we recently came across the concept that, although a given topology $\mathcal{T}$ is closed under union and intersection, while the union of infinitely many elements of $\mathcal{T}$ is necessarily a member of $\mathcal{T}$, the intersection is not. Is there a concise way to express this? (i.e., " $\mathcal{T}$ is infinitely closed under the operation $\cup$", or something like that)

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The way this is usually expressed is to say that $\mathcal{T}$ is "closed under arbitrary unions". See e.g. this question (or google the phrase).

Similarly, if we only allowed (say) countable unions, we would say $\mathcal{T}$ is "closed under countable unions". For instance, in a given topology the class of $F_\sigma$ sets is closed under countable unions.

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    $\begingroup$ ...and the other half is usually "closed under finite intersections". $\endgroup$ – John Hughes Jan 18 '17 at 11:34

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