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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)

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    $\begingroup$ 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. $\endgroup$ Oct 27, 2021 at 8:41
  • $\begingroup$ BTW a proof by contradiction is unnecessary. $\endgroup$ Oct 27, 2021 at 8:44

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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.

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  • $\begingroup$ Dear Robert Shore, could you please tell the proof by induction you were talking in the answer? $\endgroup$
    – RAHUL
    Oct 27, 2021 at 17:18
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    $\begingroup$ @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. $\endgroup$ Oct 27, 2021 at 18:16

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