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Countable intersection of F-sigma sets

  1. Jan 22, 2013 #1

    jbunniii

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    My question concerns [itex]F_\sigma[/itex] subsets of [itex]\mathbb{R}[/itex]. An [itex]F_\sigma[/itex] set is one which can be expressed as a countable union of closed sets.

    I have several books that state that a countable intersection of [itex]F_\sigma[/itex] sets need not be an [itex]F_\sigma[/itex] set (indeed, such sets have their own designation, [itex]F_{\sigma\delta}[/itex]), but none of them gives a counterexample. Does anyone know one offhand? This isn't homework/coursework; I'm just curious and haven't been able to come up with one. I would prefer a concrete example if possible, not just an existence proof.

    Relevant facts: Clearly a countable UNION of [itex]F_\sigma[/itex] sets is [itex]F_\sigma[/itex]. All open and closed sets in [itex]\mathbb{R}[/itex] are [itex]F_\sigma[/itex]. [itex]\mathbb{Q}[/itex] is an [itex]F_\sigma[/itex] set as it is a countable union of singletons. The only concrete example of a set I know is not [itex]F_\sigma[/itex] is [itex]\mathbb{R}\setminus\mathbb{Q}[/itex], the set of irrationals.
     
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  3. Jan 22, 2013 #2

    micromass

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    [tex]\mathbb{R}\setminus \mathbb{Q} = \bigcap_{q\in \mathbb{Q}} \mathbb{R}\setminus \{q\}[/tex]
     
  4. Jan 22, 2013 #3

    jbunniii

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    Yes, I just had that realization while thinking about it again. I had been trying some more convoluted way of finding a sequence of open sets that decreased to [itex]\mathbb{R}\setminus\mathbb{Q}[/itex], but missed the most obvious one!
     
  5. Jan 22, 2013 #4

    jbunniii

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    The above construction also shows that [itex]\mathbb{R}\setminus\mathbb{Q}[/itex] is a [itex]G_\delta[/itex] set (i.e. a countable intersection of open sets), which also follows immediately from the fact that its complement [itex]\mathbb{Q}[/itex] is a countable union of singletons, hence a [itex]F_\sigma[/itex] set.

    So [itex]\mathbb{R}\setminus\mathbb{Q}[/itex] is in some sense an atypical member of [itex]F_{\sigma\delta}[/itex] because it is also in [itex]G_\delta[/itex]. Similarly, [itex]\mathbb{Q}[/itex] is an atypical member of [itex]F_{\sigma\delta}[/itex] because it is also in [itex]F_\sigma[/itex].

    This led me to look for an example of a "pure" [itex]F_{\sigma\delta}[/itex] set, one which is in neither [itex]F_\sigma[/itex] nor [itex]G_\delta[/itex].

    I think that [itex]S = (\mathbb{Q} \cap (-\infty,0)) \cup ((\mathbb{R}\setminus\mathbb{Q}) \cap (0,\infty))[/itex] is such a set. (i.e. the union of all negative rationals and all positive irrationals.) Here is my reasoning:

    If [itex]S[/itex] were a [itex]F_\sigma[/itex] set, then [itex]S \cap (0,\infty)[/itex] would also be [itex]F_\sigma[/itex], but that is not the case, because [itex]S \cap (0,\infty) = ((\mathbb{R}\setminus\mathbb{Q}) \cap (0,\infty))[/itex]. Similarly, [itex]S[/itex] is not a [itex]G_\delta[/itex] set. However, I can form a descending sequence of [itex]F_\sigma[/itex] sets whose intersection is [itex]S[/itex] by starting with [itex](Q \cap(-\infty,0)) \cup (0,\infty)[/itex] and removing one positive rational at a time.

    Then I started thinking about what a "pure" [itex]F_{\sigma\delta\sigma}[/itex] set would look like, and it made me tired.
     
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