I met this question from a faculty booklet which I had trouble on involving metric spaces, stating:

Let $ (X,d) $ be a complete metric space with the sequence of closed sets, $ \{F_n\}_{n \in N} $ satisfying $ X = \bigcup_{n \in N} F_n $. We are to show that $ X = \overline {\bigcup_{n \in N} {\operatorname{int}(F_n)}} $.

I figured the Baire category theorem probably has something to do with it here but applying it simply yields that one of the sets has non empty interior. Is there a way to continue this? Thanks all

  • $\begingroup$ Sorry corrected the closure bar $\endgroup$ – kroner Sep 27 '15 at 0:41
  • $\begingroup$ I know this is a a three year old question and this might be the incorrect place, but do you have a link to the faculty booklet? or can you tell me which university this was? $\endgroup$ – Rab Dec 13 '18 at 20:15

Let $U=X\setminus \overline{\bigcup int(F_n)}$, and apply the Baire category theorem in the space $\overline{U}$.

An explanation by Alex Ravsky. If the set $U$ is non-empty then $\overline{U}$ is a non-empty closed subset of a complete metric space, so it is a complete metric space with respect to the induced metric. Since $\overline{U} = \bigcup_{n \in N} F_n\cap \overline{U}$, by the Baire category theorem there exists $n$ such that a set $F_n\cap \overline{U}$ has non-empty interior in $\overline{U}$. Therefore there exists an open subset $V$ of $X$ such that $\varnothing\ne V\cap \overline{U}\subset F_n\cap\overline{U}$. Since $\varnothing\ne V\cap \overline{U}$ and the set $V$ is open, $\varnothing\ne V\cap U$. Then a non-empty open set $V\cap U$ is contained in $F_n$, so $V\cap U\subset\operatorname{int} F_n$, but $U\cap \operatorname{int} F_n=\varnothing$, a contradiction.

  • $\begingroup$ Thanks but why can I apply it? $\endgroup$ – kroner Sep 27 '15 at 0:47
  • 2
    $\begingroup$ @zbigniew2015: It's a closed subset of a complete metric space. $\endgroup$ – Asaf Karagila Sep 27 '15 at 0:48

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