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Is a non-open, countable intersection of open sets closed?

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What is a non-open intersection? –  Git Gud Mar 12 '13 at 20:45
To clarify, you are asking 'If the countable intersection of open sets is not open, does this imply that the intersection is closed?' –  Dan Rust Mar 12 '13 at 20:45
Yes, that's correct. –  Cleaner Mar 12 '13 at 20:45
@GitGud: Suppose that $\mathscr{U}$ is a family of open sets such that $\bigcap\mathscr{U}$ is not open. The OP is asking whether it is then the case that $\bigcap\mathscr{U}$ must be closed. –  Brian M. Scott Mar 12 '13 at 20:49
@Cleaner I see this is your fourth question here and you haven't accepted any answers so far. Please consider going through your questions and accepting answers to the question in wich you got at least a satisfying answer. –  Git Gud Mar 12 '13 at 20:55

3 Answers 3

up vote 8 down vote accepted

Not necessarily: in $\Bbb R$ take the intersection of the sets $\left(-\frac1n,1\right)$ for $n\in\Bbb Z^+$: you get $[0,1)$.

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Maybe. A countable intersection of open sets is called a G$_\delta$-set. Examples of these (in the real line) include

  • $[ 0 , 1 ] = \bigcap_{n} ( - \frac{1}{n} , 1+ \frac{1}{n} )$;
  • $( 0 , 1 ] = \bigcap_{n} ( 0 , 1+ \frac{1}{n} )$;
  • the set $\mathbb{R} \setminus \mathbb{Q}$ of all irrational numbers: $\bigcap_{q \in \mathbb{Q}} ( \mathbb{R} \setminus \{ q \} )$.
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Consider the intersection of the open intervals $\left(0, 1+\frac{1}{n}\right)$, where $n$ ranges over the positive integers.

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