Question

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

Answer 1

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.