I'm having a little trouble distinguishing the line between what the f.i.p implies and what it does not.(adsbygoogle = window.adsbygoogle || []).push({});

**EDIT2** Hopefully this will make things more clear

What I'm really interested in is a formal definition of the f.i.p regardless of the set in question or the field.

Given the sequence [itex] \{a_n\}_{n \in \mathbb{N}} [/itex] in some field [itex] \mathbb{K} [/itex], is having the f.i.p equivalent to saying [itex] (\forall n \in \mathbb{N})(\exists x \in \mathbb{K})(x \in \cap_{i=1}^n a_n) [/itex]

The question about the sequence was that I was afraid (even though it seems pretty clear to me that this should follow from the previous statement) that it's construction assumed that the intersection of the entire set was non-empty as the sequence itself is infinite and says for all [itex] n \in \mathbb{N} [/itex] pick an element from the nth intersection.

My question about the axiom of choice, was that, if I'm just given a sequence of nonempty sets (assuming no other properties like order complete) and I just say form a new sequence by picking some random element from each set in the old sequence, does this require the Axiom of Choice?

Thanks

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Original

Specifically, if we are working in R and [itex] \{[a_i, b_i]\}_{i \in \mathbb{N}} [/itex] satisfies the finite intersection property, then does that imply that [itex] (\forall n \in \mathbb{N})(\exists x \in \mathbb{R})(x \in \cap_{i=1}^n [a_i, b_i]) [/itex], and, if it does, then can we use the axiom of choice (I believe this is needed...) to define a sequence [itex] \{c_n\}_{n \in \mathbb{N}} [/itex] where [itex] c_n \in \cap_{i=1}^n [a_i, b_i] [/itex]?

Thanks

**EDIT** I stated R above to make things simple, but it may help to know that in reality I'm trying to prove that sequential (Cauchy) completeness in a totally ordered Archimedean field implies that every sequence of bounded closed interval with the finite intersection property has a nonempty intersection.

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# Finite Intersection Property Question

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