Finite Intersection Property Question

[td88]

I'm having a little trouble distinguishing the line between what the f.i.p implies and what it does not.

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

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 $n \in \mathbb{N}$ 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---------------------------------------
Original

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

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.

 

[Fredrik]

[noparse] Regarding the LaTeX... Start with $$and end with$$. Alternatively, use itex instead of tex. itex is supposed to look better if the math expression appears in the middle of a line of text, but it doesn't always. itex also cuts off the top of some large symbols. You can (and should) preview before you post, but there's a bug that makes old latex images show up in all previews except the first. When that happens, just refresh and resend.[/noparse]

 

[td88]

Thanks for the LaTeX tip, I changed it

 

[SW VandeCarr]

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

F.I.P. is a property of a space that whenever a family of closed subsets is such that any finite subcollection has a non-empty intersection, so does the entire family.

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?

F.I.P. is related to the Axiom of Choice by Tychonoff's theorem which states that the Cartesian product of compact topological spaces is compact.

http://www.indopedia.org/index.php?title=Tychonoff_theorem

 

[blue_raver22]

I would say that the finite intersection property (FIP) is a property of sets that implies that their intersection is non-empty. In other words, if a collection of sets satisfies the FIP, then there exists at least one element that belongs to all of the sets in the collection.

In your specific example, if we have a sequence of closed intervals in R that satisfies the FIP, then this implies that for any finite number of intervals, there exists an element that belongs to all of them. This can be seen by choosing the left endpoint of the first interval for the first element, the right endpoint of the second interval for the second element, and so on. This process can be continued for any finite number of intervals, thus satisfying the condition (\forall n \in \mathbb{N})(\exists x \in \mathbb{R})(x \in \cap_{i=1}^n [a_i, b_i]).

As for the use of the axiom of choice, it is not necessary in this case. The axiom of choice is typically used when we need to choose an element from an infinite number of sets, which is not the case here. In this situation, we are only choosing an element from a finite number of sets, which can be done without the use of the axiom of choice.

I hope this helps clarify the concept of the finite intersection property and its implications.