How to proof that these statements are equivalent:

  • Every intersection of countably many dense open sets is dense.
  • The interior of every union of countably many closed nowhere dense sets is empty.

Moreover, why does it follow for Baire Spaces and why is it strictly weaker that:

  • $X=\bigcup_{k=1}^\infty A_k\quad\Rightarrow\quad\exists k_0\in\mathbb{N},x_0\in X: \operatorname{cl}(A_{k_0})\in\mathcal{N}_{x_0}$

...in words, the space is not meagre iff whenever it is a given by a countable union of subsets then at least one of them is large.


$A_n, n\in\Bbb N$, are countably many closed nowhere dense sets if and only if $D_n:=X-A_n$ are open dense sets (More generally $A_n$ is nowhere dense iff $D_n$ has a dense interior). Now $$\overline{\bigcap_n D_n}=\overline{X-\bigcup_n A_n} =X-\text{int}\left(\bigcup_nA_n\right)$$ So $\bigcap D_n$ is dense iff $\bigcup A_n$ has empty interior.

Note that in the second statement is turned into an equivalent one by omitting the word "closed". In one direction this is trivial. For the other direction, note that the closure of a nowhere dense set is a closed nowhere dense set. And if the union of their closures has empty interior, then so does the union of the original sets.

  • $\begingroup$ Seems like everything boils down to the fact: $\operatorname{cl}(A^c)=\operatorname{int}(A)^c\text{ resp. }\operatorname{cl}(A)^c=\operatorname{int}(A^c)$ $\endgroup$ – C-Star-W-Star Jan 27 '14 at 16:09
  • $\begingroup$ Where of course the second equation follows from the first by $A=(A^c)^c$ :-) $\endgroup$ – Stefan Hamcke Jan 27 '14 at 16:21
  • $\begingroup$ yeah of course ^^ just looks nicer if both stated ;-) $\endgroup$ – C-Star-W-Star Jan 27 '14 at 16:24
  • $\begingroup$ Is there a proper description for a set to be dense at some point s.t. its complement won't be dense at that point? Sth like: A everywhere dense iff its complement nowhere dense $\endgroup$ – C-Star-W-Star Jan 27 '14 at 16:25
  • 1
    $\begingroup$ @Freeze_S: If a Baire space $B$ is the union of countably many sets, then at least one of them has a closure with non-empty interior. But its own interior could still be empty. Take for example $$\Bbb R=\Bbb R\setminus\Bbb Q\cup\bigcup\{\{q\}\mid q\in\Bbb Q\}$$ The set of irrationals as well as each singleton $\{q\}$ has empty interior. $\endgroup$ – Stefan Hamcke Jan 27 '14 at 20:59


Take the complement. The complement of a dense open set is a closed nowhere dense set.


Assume that $\bigcap_n U_n$ is dense where $(U_n)_{n\in I}$ are open sets. Then

the interior of $$\left(\bigcap_n U_n\right)^c = \bigcup_n U_n^c$$ is empty (since its complement is a dense set). But $U_n^c$ are closed nowhere dense sets (closed because they are complements of open sets and nowhere dense, since they are closed and have empty interior). The other direction goes the same.


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