# The intersection of a locally finite family of open sets is a closed set?

Let $\{ U_i : i \in I \}$ be a locally finite family of open sets, if $I$ is infinite index, and there exist infinite different $U_i$, must $\bigcap_{i \in I} U_i$ be a closed set?

Or, further more,

if $\{ U_i : i \in I \}$ are different open sets, that is to say, for any $i,j \in I, i \neq j$, $U_i \neq U_j$, then must $\bigcap_{i \in I} U_i$ be a closed set?

Or, if $\{ U_i : i \in I \}$ are disjoint open sets, that is to say, for any $i,j \in I, i \neq j$, $U_i \bigcap U_j= \emptyset$, then must $\bigcap_{i \in I} U_i$ be a closed set?

• I think the intersection of any infinite but locally finite family of sets is the empty set. – Rolf Hoyer May 18 '15 at 1:51
• @Rolf Hoyerthe:intersection of any infinite but locally finite family of open sets is the empty set? – David Chan May 18 '15 at 2:32
• Any point has a neighborhood intersecting only finitely many sets in the family. Therefore, there are infinitely many sets in the family not containing it, so our given point is not in the intersection. – Rolf Hoyer May 18 '15 at 2:34
• @Rolf Hoyerthe:I understand. Thank you. – David Chan May 18 '15 at 2:36

2. No, let $U_i$ be composed of a set $U$ which is open and not closed, and open sets which contain $U$. Then the intersection is $U$, which is open and not closed. For instance the sets $\{(0,n)\}_{n \geq 1}$.