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In Folland's book, part (b) of the Baire Category theorem states that

$X$ is not a countable union of nowhere dense sets.

where $X$ is a complete metric space. It doesn't say whether those sets are open or closed, although I suppose we might as well assume closed, since it's a stronger statement and the proof still holds.

Here's my question. In another set of notes, I wrote down

If $X = \bigcup_{i=1}^\infty F_i$ and the $F_i$ are closed, then there is some $i$ such that $F_i \supset B_r(x_0)$, for some $r$ and $x_0$, where $B_r(x_0)$ is a ball of radius $r$ centered at $x_0$.

I see why this follows from the Baire Category theorem, but is there any reason the $F_i$ have to be closed? Can they be open? I suppose if any are open, then the result is trivial. Say $F_i$ is open, then it definitely contains a ball.

I'm weak on topology, so I want to make sure I am understanding this correctly. Thanks.

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  • $\begingroup$ (1). "It doesn't say whether those sets are open or closed. " They may be neither..... (2) $s$ is nowhere dense iff int$(\bar s)=\emptyset.$. If $s$ is open and nowhere dense then $ s=$int$(s)\subset $int$(\bar s)=\emptyset$ so $ s=\emptyset . $ $\endgroup$ Jul 30, 2016 at 20:29

2 Answers 2

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Along the same lines: For the first question, $E_i$ is nowhere dense iff $\overline{E_i}$ is nowhere dense (both saying that $\overline{E_i}$ has empty interior). And if $X$ is not a countable union of $ \overline{E_i}$ then it is not a countable union of $E_i$ either.

For the second question: ${\mathbb R} = (\mathbb R \setminus {\mathbb Q} ) \cup {\mathbb Q} $ but none of the two sets contains an open ball.

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First, a comment: I think you are implicitly assuming that every set is either open or closed. This is not true: consider e.g. the rationals!


On to your questions:

For the first one, there is no restriction on the type of sets involved: they may be open, closed, or neither, so long as they are nowhere dense. (Although in fact it's not hard to prove the general form of the theorem from the version for closed sets only - this is a good exercise!)

For the second, yes, the result would be trivial if we replaced "closed" by "open". Note, however, that the statement is false if we drop the hypothesis of closed: consider e.g. Vitali sets in $\mathbb{R}$.

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  • $\begingroup$ So you're saying they all have to be open or all have to be closed? $\endgroup$
    – Kurt
    Jul 26, 2016 at 19:42
  • $\begingroup$ @Kurt Where do I say anything like that? In question 1, the sets can be of any type - open, closed, or neither! In question 2, the theorem is trivial for open sets, true for closed sets, and false for sets in general. $\endgroup$ Jul 26, 2016 at 19:44
  • $\begingroup$ Oh see I. Still having the problem of thinking a set must be open or closed. Need to shake that habit. I think the choice of terminology just makes me want to think that. $\endgroup$
    – Kurt
    Jul 26, 2016 at 19:46

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