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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

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closed as off topic by Robin Chapman, Yemon Choi, S. Carnahan Jun 23 '10 at 16:54

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Think about $A_n=\{m\in\mathbb{N}:m>n\}$. This is really a bit too basic for MO. –  Robin Chapman Jun 23 '10 at 15:51
    
Seconded (see also the answer below). Voting to close. –  Yemon Choi Jun 23 '10 at 16:16
    
Closed. See wikipedia. –  S. Carnahan Jun 23 '10 at 16:56
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I did look at wikipedia, I just wanted to ensure that what is stated on wikipedia is the same as the formal statement above –  thedude Jun 23 '10 at 17:01
    
@thedude: I think everyone missed your point because the actual question was poorly emphasized. –  François G. Dorais Jun 23 '10 at 18:12

1 Answer 1

up vote 4 down vote accepted

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.

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But there are some reasons to expect that this can be done uniformly in $n$. Namely, when the $a_n$ are compact sets in some topological space and, in particular, when the $a_n$ are finite sets. –  François G. Dorais Jun 23 '10 at 16:33

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