# Finite intersection property and proper filter.

Let set $$X\ne\phi$$ and let $$\mathscr{F}$$ be a proper filter defined on $$X$$. Then it implies that $$\mathscr{F}$$ has the finite intersection property

Proof:-

Assume that for some $$\mathscr{F}$$ defined over $$X$$ this doesn't hold true we have that,

$$\exists A_{1},A_{2}...,A_{n}\in\mathscr{F}:\bigcap_{1\le k\le n}A_{k}=\phi$$ Now since proper filters are closed under intersections, we have that $$\phi\in\mathscr{F}$$ which contradicts the fact that $$\mathscr{F}$$ is a proper filter

Now my questions are

$$(1)$$- Are there any mistakes in my proof?

$$(2)$$- Are proper filters closed under intersections only for finitely many elements, what happens if we take $$X$$ to be a infinite set, then can I have a proper filter which have infinitely many elements in it?(I request to justify it via an example)

• Normally one of the axioms for a filter is $\forall A,B \in \mathcal F: A \cap B \in \mathcal F$. You have to extend this to finite intersections by a simple induction proof. The finite part is usually not an axiom but a derived property. So "filters are closed under finite intersections" is more accurate (properness is irrelevant here) and needs to be shown or quoted from a theorem in your text. The properness only plays a role when you note the intersection cannot be empty. Oct 27, 2021 at 8:41
• BTW a proof by contradiction is unnecessary. Oct 27, 2021 at 8:44

## 1 Answer

To be precise, you should probably prove by induction that a filter has the finite intersection property but you have the right idea.

It’s easy to show that the cofinite subsets of $$\Bbb N$$ constitute a filter. Let $$A_k= \{ n \in \Bbb N \mid n \gt k \}$$. Then $$\cap A_k = \varnothing$$. This shows that filters need not be closed under countable intersections.

• Dear Robert Shore, could you please tell the proof by induction you were talking in the answer? Oct 27, 2021 at 17:18
• @RAHUL By the definition of a filter, the intersection of two elements is itself in the filter. That's the base step. Now assume you know the intersection of $n$ elements of a filter is always in the filter and consider the intersection of $n+1$ elements of the filter. That's the intersection of $n$ elements of the filter (which you know by your inductive hypothesis is itself an element of the filter) and $1$ element of the filter. Thus, the intersection of $n+1$ elements of the filter is also the intersection of two elements of the filter, so it must also be an element of the filter. QED. Oct 27, 2021 at 18:16