Let X be a topological space and F a collection of closed subsets. Define G as the complements of sets in F. Then a union of sets in G is the complement of an intersection of the corresponding sets in F. So 2 F has FIP if and only if G has no finite subcover of X. Also F has the total intersection property.

Thank you!!

Guest Apr 17, 2022

#1**0 **

Do you want the proof of this statement?

Or do you think this statement is wrong and you want to disprove it?

MaxWong Apr 17, 2022

#2**+1 **

I didn't expect a topology question on this site haha

Anyways, this statement looks correct to me, and I will attempt to prove it.

Let \(\mathcal F=\{F_1, F_2, \cdots\}\) and \(\mathcal G = \{F_1^c, F_2^c, \cdots\}\).

(=>) Suppose that F has F.I.P. That means for any finite subset \(\{F_1', F_2', \cdots, F_N'\} \subseteq \mathcal F\), \(\displaystyle \bigcap_{i = 1}^N F_i'\neq \varnothing\).

Further suppose on the contrary that G has a finite subcover of X, i.e., there exists a finite subset \(\{\left(F_1'\right)^c,\left(F_2'\right)^c,\cdots,\left(F_N'\right)^c\} \subseteq \mathcal G\) where \(\displaystyle\bigcup_{i = 1}^N \left(F_i'\right)^c=X\). Taking complement on both sides, that means there exists a finite subset \(\{F_1', F_2', \cdots, F_N'\} \subseteq \mathcal F\) such that \(\displaystyle \bigcap_{i = 1}^N F_i' = \varnothing\). That contradicts the F.I.P. of F. Therefore, by contradiciton, G has no finite subcover of X.

(<=) Suppose that G has no finite subcover of X. That means for any finite subset \(\{\left(F_1'\right)^c,\left(F_2'\right)^c,\cdots,\left(F_N'\right)^c\} \subseteq \mathcal G\), we have \(\displaystyle\bigcup_{i = 1}^N \left(F_i'\right)^c\neq X\). Again, taking complement on both sides yields for any for any finite subset \(\{F_1', F_2', \cdots, F_N'\} \subseteq \mathcal F\), we have \(\displaystyle \bigcap_{i = 1}^N F_i' \neq \varnothing\). Then F has F.I.P.

That finishes the proof that F has F.I.P. if and only if G has no finite subcover of X.

MaxWong
Apr 17, 2022