Can arbitrary union of open intervals be written as countable union of open intervals?
Here, I don't even understand what do they mean by arbitrary.
Help me please.
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$\begingroup$ See also: Any open subset of R is a countable union of disjoint open intervals $\endgroup$Martin Sleziak– Martin Sleziak2013-03-04 15:43:39 +00:00Commented Mar 4, 2013 at 15:43
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1 Answer
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In this case arbitrary just means that there is no restriction on the number or kind of open intervals used to form the union. If you allow all possible unions of open intervals, you get precisely the open subsets of $\Bbb R$. The question asks whether you ever need uncountably many open intervals to form some open set in $\Bbb R$, or whether countably many are always sufficient.
HINT: Consider try using just the countable collection $\mathscr{B}=\{(p,q):p,q\in\Bbb Q\text{ and }p<q\}$ of open intervals with rational endpoints. Is every open interval, no matter what its endpoints, a union of members of $\mathscr{B}$?
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$\begingroup$ @ Brian M.Scott: Thanks for the HINT. I got it. $\endgroup$user35039– user350392013-03-03 20:15:30 +00:00Commented Mar 3, 2013 at 20:15
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$\begingroup$ @viru: You’re welcome. $\endgroup$Brian M. Scott– Brian M. Scott2013-03-03 20:18:12 +00:00Commented Mar 3, 2013 at 20:18