# Maximality with respect to the Finite Intersection Property

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• Sep 25th 2013, 07:21 AM
Aryth
Maximality with respect to the Finite Intersection Property
I'm working on a problem, but I just can't figure it out.

Let $\displaystyle X$ be any set and suppose that $\displaystyle \mathcal{M}$ is a collection of subsets of $\displaystyle X$ that is maximal with respect to the finite intersection property. Prove the following statements are true.

1. The intersection of any finite nonempty subcollection of $\displaystyle \mathcal{M}$ is a member of $\displaystyle \mathcal{M}$.

2. Any subset of $\displaystyle X$ that is not disjoint with every member of $\displaystyle \mathcal{M}$ is contained in $\displaystyle \mathcal{M}$.

I'm working on the first one, but I have found nothing that indicates that $\displaystyle \mathcal{M}$ is closed under finite intersection. Any help is greatly appreciated.
• Sep 25th 2013, 09:18 AM
emakarov
Re: Maximality with respect to the Finite Intersection Property
For 1, an obvious idea is this: if the intersection of a finite subcollection is not in M, add it. If the result has the FIP, this would contradict the maximality of M. So, it is left to prove that the result of addition does, in fact, have the FIP.
• Sep 25th 2013, 09:35 AM
Aryth
Re: Maximality with respect to the Finite Intersection Property
Let's say we take a finite subcollection, say $\displaystyle \mathcal{N}$, of $\displaystyle \mathcal{M}$ and let $\displaystyle U = \bigcap N$ where $\displaystyle N$ are the sets from $\displaystyle \mathcal{N}$. Can we say that, since $\displaystyle U\subseteq N$ and $\displaystyle N$ has nonempty intersection with every member of $\displaystyle \mathcal{M}$, $\displaystyle U$ must also have nonempty intersection with members from $\displaystyle \mathcal{M}$?
• Sep 25th 2013, 10:13 AM
emakarov
Re: Maximality with respect to the Finite Intersection Property
Quote:

Can we say that, since $\displaystyle U\subseteq N$ and $\displaystyle N$ has nonempty intersection with every member of $\displaystyle \mathcal{M}$, $\displaystyle U$ must also have nonempty intersection with members from $\displaystyle \mathcal{M}$?

Yes, but not just because $\displaystyle U\subseteq N$ and $\displaystyle N$ has nonempty intersection with every member of $\displaystyle \mathcal{M}$.

Let $\displaystyle U=\bigcap\mathcal{N}$ and suppose adding $\displaystyle U$ to $\displaystyle \mathcal{M}$ breaks the FIP, i.e., there exist a finite $\displaystyle \mathcal{K}\subseteq\mathcal{M}\cup\{U\}$ such that $\displaystyle \bigcap\mathcal{K}=\emptyset$. Then $\displaystyle \mathcal{K}$ must contain $\displaystyle U$ because $\displaystyle \mathcal{M}$ has the FIP. Then what can be said about $\displaystyle \bigcap\left(\left(\mathcal{K}\setminus\{U\}\right )\cup\mathcal{N}\right)$?
• Sep 25th 2013, 10:32 AM
Aryth
Re: Maximality with respect to the Finite Intersection Property
Honestly, I'm not entirely certain...
• Sep 25th 2013, 10:55 AM
emakarov
Re: Maximality with respect to the Finite Intersection Property
The idea is that for the purposes of intersection, $\displaystyle \mathcal{N}$ can replace $\displaystyle U$. If you have any collection that contains $\displaystyle U$, you can remove $\displaystyle U$, add $\displaystyle \mathcal{N}$, and the intersection won't change. This is because $\displaystyle U=\bigcap\mathcal{N}$. Moreover, $\displaystyle \mathcal{N}$ is finite, so using it to replace $\displaystyle U$ in any finite collection also produces a finite collection. All this is to say that, in the context of post #4, $\displaystyle \left(\mathcal{K}\setminus\{U\}\right)\cup\mathcal {N}$ is finite and its intersection is empty, just like for $\displaystyle \mathcal{K}$. But $\displaystyle \left(\mathcal{K}\setminus\{U\}\right)\cup\mathcal {N}\subseteq\mathcal{M}$, which contradicts the FIP of $\displaystyle \mathcal{M}$.
• Sep 25th 2013, 10:58 AM
Aryth
Re: Maximality with respect to the Finite Intersection Property
That's what confused me. I thought that it looked weird. I understand that now. Thanks a lot for your help.