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A topology is closed under arbitrary union of open sets, and therefore a fortiori also closed under countable union of open sets.

Why do we require arbitrary union rather than just countable union?

What might be an example that would illustrate why we require arbitrary union? Or are they just equivalent when talking about unions of sets?

The most similar question I've found to this one is regarding the validity of arbitrary union without assuming topology as prior.

Edit: This question was not addressed in answers given to a similar seeming post regarding finite intersection, rather than arbitrary or countable intersection.

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  • $\begingroup$ I looked at this. The answers given to that question do not address or resolve the above question. $\endgroup$ – Howerut Sep 12 '15 at 9:47
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The concepts are not equivalent: Let $X$ be an uncountable set and say that a subset $U\subseteq X$ is open if $U=X$ or $U$ is countable (which includes finite or empty). That would be a "topology" in your modified sense, but not in the "traditional" sense.

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  • $\begingroup$ I think this is a sufficient motivation to have arbitrary rather than countable unions in the definition of a topology. Thank you! $\endgroup$ – Howerut Sep 12 '15 at 9:58
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    $\begingroup$ Another motivation is that the discrete topology on an uncountable set $X$ is generated by the one-point sets. We need arbitrary unions to be able to form $X$ as a union of one-point sets. $\endgroup$ – Patrick Stevens Sep 12 '15 at 10:09
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    $\begingroup$ There are other types of structures where we do restrict to countable unions.For example the family of Borel sets generated by a topology, or the family of Lebesgue-measurable subsets of the reals.A countable union of Borel sets is a Borel set ,but an uncountable union might not be. $\endgroup$ – DanielWainfleet Sep 12 '15 at 10:13

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