The set of flats of a combinatorial geometry is closed under arbitrary intersection

Question

I am reading Combinatorial Geometry from A Course in Combinatorics by van Lint and Wilson. They define a combinatorial geometry to be an ordered pair $(X,\mathcal{F})$, where $X$ is a set of points and $\mathcal{F}$ is a family of subsets of $X$ called flats. They have to satisfy 4 conditions out of which two are:

1. $\mathcal{F}$ is closed under pairwise (and hence finite) intersection.
2. There are no infinite chains in the poset $\mathcal{F}$.

Then, the authors have gone to remark that these two properties ensure that $\mathcal{F}$ is actually closed under arbitrary intersection. Unfortunately I am unable to understand why.

My attempt: I consider an arbitrary collection of flats, say $\{F_\alpha\}$. I have to prove that $$F:=\bigcap\limits_{\alpha}F_\alpha \in \mathcal{F}$$ It seems I have to assume that $F\notin \mathcal{F}$ and then show that this implies there is some infinite chain in $\mathcal{F}$ and thus get a contradiction.

A hint will be sufficient.

Answer 1

I asked my professor today and got an answer:

Without loss of generality, $F\neq\phi$. It is conventional to include $\phi$ in $\mathcal{F}$.

Consider a countable set $\{F_1, F_2, .., \} \subset \{F_\alpha\}$. Look at the sequence $$F_1 \supset F_1\cap F_2\supset \cdots \supset\bigcap\limits_{i=1}^nF_i\supset\cdots$$ Condition 2 ensures that only finitely many of these are distinct. So, $\exists$ some $k$ such that $$\bigcap\limits_{i=1}^kF_i = \bigcap\limits_{i=1}^{k+1}F_i = \cdots$$ Then, condition 1 tells us that $\mathcal{F}$ is closed under arbitrary intersections.