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I was going through a rudimentary course in mathematical analysis covering Metric spaces and the book opens up with the idea of open sets. While mentioning the property of open sets it cites that if I is any indexing set then union of open sets Gi where i belongs to I is also an open set . It arouse questions in my mind as to why it couldn't simply state that if G1,G2...Gn are open sets then their union is also an open set . I suspect that I am missing a finer point here and is curious to know about the deeper things, if any , associated with this .

user12817
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  • you may want the union of an infinite number of open sets, a number that may not even be countable. the union notation: $G=\cup_{n \in I} G_n$ resolves this at a notational level. however there are reasons to be aware of the issue you have raised. for example a similar notation for the Cartesian product of an arbitrary indexed family of sets: $G=\prod_{n \in I} G_n$ touches on the Axiom of Choice. – David Holden Aug 04 '14 at 16:51

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The finer point you are missing is the following: if you state $G_1,\ldots,G_n$, you only have a finite number of sets. But the property of open sets is that any (finite or infinite) union of open sets is still open.

Also, note that if you state $G_1, G_2, \ldots$, you are assuming that the number of sets is countable. Using a general indexing set $I$ allows you to consider uncountably infinite collections.

M Turgeon
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