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While working through my book I have read that if $V$ is an open set, then $V$ is the union of all possible $\varepsilon$ - neighborhoods of all elements in $V$. This makes sense to me when I talk about the union of rationals, or natural numbers for instance - but what if there are uncountably many $p\in V$? Or to be more specific:

  • What would be a possible way to understand the notion of taking the union an uncountable number of times?

  • How could I formally express this idea?

Nerrit
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l337n00b
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    See, e.g., this question – lulu Jan 03 '23 at 13:30
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    You can define the thing cosntructively the same say you would do for finite or countable union. An element $x$ lies in an arbitrary (finite, countable, uncountable) union of $U_i$ if $x \in U_i$ for some $i$. – nicomezi Jan 03 '23 at 13:48
  • It is unclear what your issue really is: Are you looking for intuition or a formal definition? If you want the former, take a look at Axiom 5 in the ZF set of axioms here. – Moishe Kohan Jan 03 '23 at 16:05
  • @MoisheKohan I'm looking for both. Thanks for the link, will check it out. – l337n00b Jan 03 '23 at 16:07
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    You may be thinking of finite unions as repeated binary unions, thinking of countable unions as some sort of limit of finite ones, and therefore wondering about uncountable unions. If so, it's important to realize that a completely general definition is available, as in the comment by @nicomezi, and is simpler than what you were thinking of. – Andreas Blass Jan 03 '23 at 18:00

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If $S$ is any collection (set) of sets, we write $T=\bigcup S$ (Latex \bigcup S) to mean that any $x\in T$ iff $x$ is a member of at least one of the sets in the collection $S.$ That is, $$\bigcup S=\{x: \exists \sigma\in S:\,x\in\sigma\}.$$ The cardinality of $S$ is irrelevant. We do not "take the union" as some kind of sequence. It just exists.

DanielWainfleet
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