I just need a quick clarification:


Given a sequence of sets $\{a\_n\}\_{n \in \mathbb{N}}$ in some field $\mathbb{K}$, is saying that it satisfies the finite intersection property equivalent to saying $(\forall n\in \mathbb{N})(\exists x\in \mathbb{K})(x \in \cap\_{i=1}^n a\_n)$


If the previous statement is true, then it seems almost reasonable to say that, because $\{\cap\_{i=1}^n a\_n\}\_{n\in\mathbb{N}}$ is a nested sequence of nonempty sets (because of the f.i.p), $\cap\_{i=1}^{\infty} a\_n \neq \emptyset$, but I know that is not necessarily true


Thanks



Quick Finite Intersection Property Question
What you quote is the finite intersection property.


The point of confusion is in what follows. There is no reason why a nested sequence of nonempty sets can't have empty interesection. Just because we can find an $x\_n\in\cap\_{i=1}^n a\_n$ for every $n$, there is no reason to expect that this can be done uniformly in $n$.


The intervals $[n,\infty)$ also form a nested sequence of nonempty sets with empty intersection.


